From 8d3c41567297774c6c3d3ae7c3279b7028b738d8 Mon Sep 17 00:00:00 2001 From: Hearot Date: Mon, 30 Jun 2025 19:46:21 +0200 Subject: [PATCH] feat(eti): aggiunge la prima parte degli esercizi del corso --- .gitignore | 1 + .../commands.tex | 75 + .../format.tex | 46 + .../Elementi di Teoria degli Insiemi/main.pdf | Bin 0 -> 477090 bytes .../Elementi di Teoria degli Insiemi/main.tex | 1493 +++++++++++++++++ .../preamble.tex | 61 + .../quiver.sty | 40 + 7 files changed, 1716 insertions(+) create mode 100644 Terzo anno/Elementi di Teoria degli Insiemi/commands.tex create mode 100644 Terzo anno/Elementi di Teoria degli Insiemi/format.tex create mode 100644 Terzo anno/Elementi di Teoria degli Insiemi/main.pdf create mode 100644 Terzo anno/Elementi di Teoria degli Insiemi/main.tex create mode 100644 Terzo anno/Elementi di Teoria degli Insiemi/preamble.tex create mode 100644 Terzo anno/Elementi di Teoria degli Insiemi/quiver.sty diff --git a/.gitignore b/.gitignore index d3392c9..d3bfaee 100644 --- a/.gitignore +++ b/.gitignore @@ -12,3 +12,4 @@ *.blg *.run.xml *-eps-converted-to.pdf +*.prob \ No newline at end of file diff --git a/Terzo anno/Elementi di Teoria degli Insiemi/commands.tex b/Terzo anno/Elementi di Teoria degli Insiemi/commands.tex new file mode 100644 index 0000000..6ff14e4 --- /dev/null +++ b/Terzo anno/Elementi di Teoria degli Insiemi/commands.tex @@ -0,0 +1,75 @@ +\newcommand{\eu}{\mathrm{e}} +\newcommand{\im}{\mathrm{i}} + +\newcommand{\degree}{\,^{\circ}} + +\newcommand{\transpose}[1]{{#1}^{\mathsf{T}}} + +\newcommand{\Int}{\int\limits_{-\infty}^{\infty}} + +\newcommand{\rint}[2]{\int{#1}\dd{#2}} + +\newcommand{\Rint}[4]{\int\limits_{#1}^{#2}{#3}\dd{#4}} + +\DeclareMathOperator{\ot}{ot} +\DeclareMathOperator{\cof}{cof} +\DeclareMathOperator{\TC}{TC} +\newcommand{\ORD}{\text{ORD}} + +\makeatletter +\newcommand*\bigcdot{\mathpalette\bigcdot@{.5}} +\newcommand*\bigcdot@[2]{\mathbin{\vcenter{\hbox{\scalebox{#2}{$\m@th#1\bullet$}}}}} +\makeatother + +\newcommand{\Ham}{\hat{\mathcal{H}}} + +\renewcommand{\Tr}{\mathrm{Tr}} + +\newcommand{\christoffelsecond}[4]{\dfrac{1}{2}g^{#3 #4}(\partial_{#1} g_{#2 #4} + \partial_{#2} g_{#1 #4} - \partial_{#4} g_{#1 #2})} + +\newcommand{\riemanncurvature}[5]{\partial_{#3} \Gamma_{#4 #2}^{#1} - \partial_{#4} \Gamma_{#3 #2}^{#1} + \Gamma_{#3 #5}^{#1} \Gamma_{#4 #2}^{#5} - \Gamma_{#4 #5}^{#1} \Gamma_{#3 #2}^{#5}} + +\newcommand{\covariantriemanncurvature}[5]{g_{#1 #5} R^{#5}{}_{#2 #3 #4}} + +\newcommand{\riccitensor}[5]{g_{#1 #5} R^{#5}{}_{#2 #3 #4}} + +\renewcommand{\emptyset}{\varnothing} +\renewcommand{\iff}{\longleftrightarrow} +\renewcommand{\implies}{\longrightarrow} +\renewcommand{\impliedby}{\longleftarrow} + +\DeclareMathOperator{\dom}{dom} +\DeclareMathOperator{\imm}{imm} +\newcommand{\rel}{\mathcal{R}} +\newcommand{\FF}{\mathcal{F}} +\newcommand{\bigtimes}{\varprod} +\DeclareMathOperator{\Fun}{Fun} +\DeclareMathOperator{\id}{id} +\newcommand{\PP}{\mathcal{P}} + +\newcommand{\NN}{\mathbb{N}} +\newcommand{\ZZ}{\mathbb{Z}} +\newcommand{\QQ}{\mathbb{Q}} +\newcommand{\RR}{\mathbb{R}} +\newcommand{\UU}{\mathbb{V}} +\newcommand{\VV}{\mathbb{V}} + +\newcommand{\se}{\text{se }} +\newcommand{\altrimenti}{\text{altrimenti}} +\newcommand{\inv}{^{-1}} + +\newcommand{\AC}{$\mathsf{AC}$} +\newcommand{\PA}{$\mathsf{PA}$} +\newcommand{\cc}{\mathfrak{c}} +\DeclareMathOperator{\Fin}{Fin} +\newcommand{\PPFin}{\PP^{\mathsf{fin}}} +\DeclareMathOperator{\FSeq}{FSeq} +\DeclareMathOperator{\bige}{Big} +\DeclareMathOperator{\Sym}{\mathfrak{S}} + +\newcommand{\OO}{\mathfrak{O}} +\newcommand{\basis}{\mathcal{B}} + +\newcommand{\restr}[2]{ + #1\arrowvert_{#2} +} diff --git a/Terzo anno/Elementi di Teoria degli Insiemi/format.tex b/Terzo anno/Elementi di Teoria degli Insiemi/format.tex new file mode 100644 index 0000000..bdfa7ad --- /dev/null +++ b/Terzo anno/Elementi di Teoria degli Insiemi/format.tex @@ -0,0 +1,46 @@ +\usepackage{titlesec} +\usepackage[many]{tcolorbox} +\usepackage{environ} + +\titlespacing*{\chapter}{0cm}{-2.0cm}{0.50cm} +\titlespacing*{\section}{0cm}{0.50cm}{0.25cm} + +\setlength{\parindent}{0pt} +\setlength{\parskip}{1ex} + +\newtcbtheorem[list inside={prob}]{problem}{Problema}% + {enhanced, + colback = black!5, %white, + colbacktitle = black!5, + coltitle = black, + boxrule = 0pt, + frame hidden, + borderline west = {0.5mm}{0.0mm}{black}, + fonttitle = \bfseries\sffamily, + breakable, + before skip = 3ex, + after skip = 3ex, + boxsep = 2mm +}{problem} + +\newenvironment{solution} + {\let\qed\relax\begin{proof}[Soluzione]} + {\end{proof}} + +\newenvironment{altsolution} + {\let\qed\relax\begin{proof}[Soluzione alternativa]} + {\end{proof}} + +\newif\ifhideproofs +% \hideproofstrue + +\ifhideproofs + \usepackage{environ} + \NewEnviron{hide}{} + \let\solution\hide + \let\endsolution\endhide + \let\altsolution\hide + \let\endaltsolution\endhide +\fi + +\tcbuselibrary{skins, breakable} diff --git a/Terzo anno/Elementi di Teoria degli 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Si mostri che $(a, b)_4$ e +$(a, b)_5$ invece non la soddisfano. +\end{problem} + +\begin{solution} +Per il principio di sostituzione, se $a = a'$ e $b = b'$, $(a, b) = (a', b')$ è +sicuramente vero. Mostriamo che per $(a, b)_K$, $(a, b)_1$, $(a, b)_2$ e $(a, b)_3$ vale anche il viceversa, ossia che +$(a, b) = (a', b') \implies a = a' \land b = b'$. Per $(a, b)_4$ e +$(a, b)_5$ mostriamo invece un controesempio alla proprietà caratterizzante di +una coppia ordinata. + +\begin{enumerate}[(a.)] + \item Se $(a, b)_K = (a', b')_K$, allora $\{a\} = \bigcap (a, b)_K = \bigcap (a', b')_K = \{a'\}$, e ciò è possibile solo se $a = a'$. Analogamente $\{a, b\} = + \bigcup (a, b)_K = \bigcup (a', b')_K = \{a, b'\}$, dove si è utilizzato che + $a' = a$. Se $b = a$, $b' \in \{a, b'\} = \{a, b\} = \{a\}$ implica che $b' = a = b$. Se invece $b \neq a$, $b \in \{a, b\} = \{a, b'\}$: se per assurdo $b \neq b'$, + allora si avrebbe necessariamente $a = b$, \Lightning. Quindi $b = b'$, da cui + la tesi. + \item Sia $(a, b)_1 = (a', b')_1$. Dividiamo la dimostrazione in più casi. + + \begin{itemize} + \item Se $b = \emptyset$, allora, affinché $\{a', \emptyset\}$ sia elemento di $(a, b)_1$, deve valere $a' = a$ o $a' = \{\emptyset\}$. Dividiamo ancora in due casi. + + \begin{itemize} + \item Sia dunque $a' = \{\emptyset\}$. Se per assurdo $a \neq \{\emptyset\}$, si avrebbe necessariamente $b' = \emptyset$, e dunque $\{\{a, \emptyset\}, \{\emptyset, \{\emptyset\}\}=(a, b)_1=(a', b')_1=\{\{\emptyset, \{\emptyset\}\}$, che è impossibile dato che $a \neq \{\emptyset\}$. Quindi $a = \{\emptyset\} = a'$. Affinché $\{b', \{\emptyset\}\}$ sia elemento di $(a, b)_1$, deve valere anche $b = b'$ o $b = \emptyset$; dacché $b' = \emptyset$, in entrambi i casi si ottiene + $b = b'$. + + \item Se invece $a' = a$ con $a' \neq \{\emptyset\}$, allora l'unico elemento in $(a, b)_1$ che contiene $\{\emptyset\}$ è $\{\emptyset, \{\emptyset\}\}$, e in $(a', b')_1$ è $\{b', \{\emptyset\}\}$, da cui necessariamente $b' = \emptyset = b$. Dunque $a = a'$ e $b = b'$. + \end{itemize} + + In entrambi i casi si ha dunque $a = a'$ e $b = b'$. + + \item Se $a = \{\emptyset\}$, affinché $\{b', \{\emptyset\}\}$ sia elemento di $(a, b)_1$, deve valere $b' = \emptyset$ o $b' = b$. Dividiamo ancora in due casi. + + \begin{itemize} + \item Sia dunque $b' = \emptyset$. Se per assurdo + $b \neq \emptyset$, si avrebbe necessariamente $a' = \{\emptyset\}$, e dunque $\{\{b, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}=(a, b)_1=(a', b')_1=\{\{\emptyset, \{\emptyset\}\}$, che è impossibile dato che $b \neq \emptyset$. Quindi $b = \emptyset = b'$. Affinché $\{a', \emptyset\}$ sia elemento di $(a, b)_1$, deve valere anche $a = a'$ o $a' = \{\emptyset\}$; dacché $a = \{\emptyset\}$, in entrambi i casi si ottiene + $a = a'$. + + \item Se invece $b' = b$ con $b' \neq \emptyset$, allora l'unico elemento in $(a, b)_1$ che contiene $\emptyset$ è $\{\{\emptyset\}, \emptyset\}$, e in $(a', b')_1$ è $\{a', \emptyset\}$, da cui necessariamente $a' = \emptyset = a$. Dunque $a = a'$ e $b = b'$. + \end{itemize} + + In entrambi i casi si ha dunque $a = a'$ e $b = b'$. + + \item Se $a \neq \{\emptyset\}$ e $b \neq \emptyset$, allora l'unico elemento in $(a, b)_1$ che contiene $\{\emptyset\}$ è $\{b, \{\emptyset\}\}$, e poiché $\{b', \{\emptyset\}\} \in (a', b')_1 = (a, b)_1$, da cui necessariamente $b' = b$, dacché $b \neq \{\emptyset\}$. Analogamente l'unico elemento in $(a, b)_1$ che contiene $\emptyset$ è $\{a, \emptyset\}$, e poiché $\{a', \emptyset\} \in (a', b')_1 = (a, b)_1$, da cui necessariamente $a' = a$, dacché $a \neq \emptyset$. + \end{itemize} + + In tutti i casi si ottiene $a = a'$ e $b = b'$, da cui la tesi. + + \item Sia $(a, b)_2 = (a', b')_2$. In $(a, b)_2$ l'unico elemento contenente un + insieme vuoto è $\{\{a\}, \emptyset\}$ (né $\{a\}$ né $\{b\}$ sono vuoti, uno contiene $a$, l'altro $b$), e analogamente in $(a', b')_2$ è + $\{\{a'\}, \emptyset\}$. Poiché $\{a\}$ e $\{a'\}$ sono entrambi non vuoti + e $(a, b)_2 = (a' b')_2$, necessariamente $a = a'$. Analogamente, l'unico + elemento di $(a, b)_2$ che non contiene un insieme vuoto è $\{\{b\}\}$, + mentre in $(a', b')_2$ è $\{\{b'\}\}$, da cui $b = b'$, e dunque la tesi. + + \item Sia $(a, b)_3 = (a', b')_3$. Distinguiamo più casi. + + \begin{itemize} + \item Se $a = \emptyset$ e $b = \emptyset$, $(a, b)_3 = \{\{\emptyset\}\}$. + Necessariamente $b = \emptyset$, altrimenti $(a', b')_3$ conterrebbe un + insieme di due elementi, mentre $(a, b)_3$ contiene un solo singoletto. + Allo stesso modo $a = \emptyset$, altrimenti $(a', b')_3$ conterrebbe due + elementi, mentre $(a, b)_3$ ha un solo singoletto. + + \item Se $a = \emptyset$, ma $b \neq \emptyset$, allora $b' \neq \emptyset$: se infatti per assurdo $b' = \emptyset$, allora $(a', b')_3$ conterrebbe solo singoletti, mentre $(a, b)_3$ contiene un insieme di due elementi, dacché $b \neq \emptyset$. Dal momento che in $(a, b)_3$ l'unico insieme di due elementi è $\{b, \emptyset\}$, mentre in $(a', b')_3$ è $\{b', \emptyset\}$, da cui $b = b'$. Infine, anche i singoletti $\{a\}$ e $\{a'\}$ dei due insiemi sono uguali, e quindi $a = a'$. + + \item Se $a \neq \emptyset$, ma $b = \emptyset$, allora necessariamente $b' = \emptyset = b$: altrimenti $(a', b')_3$ conterrebbe un insieme di due elementi, mentre $(a, b)_3$ contiene solo singoletti. Poiché $a \neq \emptyset$, $(a, b)_3$ ha due elementi, e dunque $a' \neq \emptyset$: + altrimenti $(a', b')_3$ avrebbe un singolo elemento, mentre $(a, b)_3$ ne + ha due. Allora, poiché $a \neq \emptyset$, $a' \neq \emptyset$ e $b = b' = \emptyset$, + deduciamo che necessariamente deve valere $a = a'$. + + \item Se $a \neq \emptyset$ e $b \neq \emptyset$, allora $b' \neq \emptyset$: altrimenti $(a', b')_3$ conterrebbe solo singoletti, mentre $(a, b)_3$ contiene + un insieme di due elementi. Dal momento che $b \neq \emptyset$, $b' \neq \emptyset$ e in + $(a, b)_3$ l'unico insieme di due elementi è $\{b, \emptyset\}$, mentre + in $(a', b')_3$ è $\{b', \emptyset\}$, allora necessariamente $b = b'$. + Analogamente l'unico singoletto contenuto in $(a, b)_3$ deve essere uguale + a quello contenuto in $(a', b')_3$, ossia $\{a\} = \{a'\}$, da cui $a = a'$. + \end{itemize} + + In tutti i casi si ottiene $a = a'$ e $b = b'$, da cui la tesi. + + \item $(\{\emptyset\}, \{\emptyset\})_4 = \{\{\emptyset\}, \{\{\emptyset\}\}\} = (\{\{\emptyset\}\}, \emptyset)_4$; tuttavia $\{\emptyset\} \neq \emptyset$, perché $\emptyset \in \{\emptyset\}$, ma $\emptyset \notin \emptyset$. Dunque + $(a, b)_4$ non soddisfa la proprietà caratterizzante delle coppie ordinate. + + \item $(\{\emptyset\}, \{\emptyset\})_5 = \{\{\{\emptyset\}, \emptyset\}, \{\emptyset\}\} = (\emptyset, \{\{\emptyset\}, \emptyset\})_5$, ma come abbiamo + appena visto $\{\emptyset\} \neq \emptyset$. Dunque + $(a, b)_4$ non soddisfa la proprietà caratterizzante delle coppie ordinate. +\end{enumerate} +\end{solution} + +\begin{problem}{Formula esplicita per la coppia di Kuratowski}{problem-2} + Si scriva una formula esplicita per $(a, b) = \{\{a\}, \{a, b\}\}$. +\end{problem} + +\begin{solution} Trasformiamo tramite opportune sostituzioni la formula +originale in una formula esplicita: + + \begin{itemize}[$\leadsto$] + \item[] $(a, b) = \{\{a\}, \{a, b\}\}$ + \item $\forall x (x \in (a, b) \iff x \in \{\{a\}, \{a, b\}\})$ (estensionalità) + \item $\forall x (x \in (a, b) \iff (x = \{a\} \lor x = \{a, b\})$ + \item $\forall x (x \in (a, b) \iff (\forall y (y \in x \iff y \in \{a\}) \lor \forall y (y \in x \iff y \in \{a, b\}))$ (estensionalità) + \item $\forall x (x \in (a, b) \iff (\forall y (y \in x \iff y = a) \lor \forall y (y \in x \iff (y = a \lor y = b)))$. + \end{itemize} +\end{solution} + +\begin{problem}{Dominio e immagine di una relazione sono insiemi}{problem-3} + Si dimostri che $\dom(\rel) = \{ x \mid \exists y ((x, y) \in \rel) \}$ e $\imm(\rel) = \{ y \mid \exists x ((x, y) \in \rel) \}$ sono effettivamente insiemi. +\end{problem} + +\begin{solution} + Si osserva che $\bigcup (a, b) = \{a, b\}$, e quindi $\bigcup \left( \bigcup \rel \right)$ + -- che è un insieme per l'assioma dell'unione -- contiene indistintamente elementi del dominio e dell'immagine di $\rel$. Possiamo allora utilizzare l'assioma + di separazione per mostrare che $\dom(\rel)$ e $\imm(\rel)$ sono insiemi: + + \begin{enumerate}[(i.)] + \item $\dom(\rel) = \{ x \in \bigcup \left( \bigcup \rel \right) \mid \exists y ((x, y) \in \rel) \}$, + \item $\imm(\rel) = \{ y \in \bigcup \left( \bigcup \rel \right) \mid \exists x ((x, y) \in \rel) \}$. + \end{enumerate} +\end{solution} + +\begin{problem}{Forme equivalenti dell'assioma di scelta (1)}{problem-4} + Si dimostrino che le seguenti asserzioni sono tra loro equivalenti: + + \begin{enumerate}[(i.)] + \item $\forall I \, \forall \sigma ((\exists x (x \in I) \land \sigma\,\text{$I$-successione} \land \forall j (j \in I \implies \exists y(y \in \sigma(j)))) \implies \exists z (z \in \bigtimes_{i \in I} \sigma(i)))$ (assioma di scelta, \AC), + + \item $\forall \FF ((\forall c (c \in \FF \implies \exists d (d \in c)) \land \forall a \forall b ((a \in \FF \land b \in \FF \land \lnot(\forall y(y \in a \iff y \in b))) \implies \lnot \exists z(z \in a \land z \in b))) \implies \exists X( \forall F(F \in \FF \implies \exists a(a \in X \land a \in F \land \forall b((b \in X \land b \in F) \implies \forall w(w \in a \iff w \in b))))))$ ($\forall \FF$ famiglia di insiemi non vuoti a due a due disgiunti, $\exists X$ tale che $\forall F \in \FF$ vale che $\abs{X \cap F} = 1$), + + \item $\forall f ((f \in \Fun(A, B) \land f\text{ surgettiva}) \implies \exists g (g \in \Fun(B, A) \land f \circ g = \id_B))$ (per funzioni surgettive esiste un'inversa destra), + + \item $\forall X (\exists z(z \in X) \implies \exists f (f \in \Fun(\PP(X) \setminus \{\emptyset\}, X) \land \forall y(y \in \PP(X) \setminus \{\emptyset\} \implies f(y) \in y)))$ (esiste sempre una funzione di scelta sulle parti). + \end{enumerate} +\end{problem} + +\begin{solution} + Mostriamo l'equivalenza di (i.) con tutte le altre asserzioni\footnote{ + Avremmo potuto dimostrare anche solo 4 implicazioni invece che 6, seguendo + un ciclo di implicazioni della forma (i.) $\implies$ (ii.) $\implies$ (iii.) $\implies$ (iv.) $\implies$ (i.), ma per comodità e spirito di esercizio si + è preferito riportare tutte le equivalenze con la formulazione originale + di \AC. + }. + + \begin{itemize} + \item[\fbox{(i.) $\implies$ (ii.)}] Sia $\FF$ una famiglia di insiemi non vuoti a due a due disgiunti. Se $\FF$ è vuota, la tesi è banale. Se $\FF$ non + è vuota, si può considerare la $\FF$-successione $\id_\FF$. Poiché gli elementi + di $\FF$ sono non vuoti, $\bigtimes_{f \in \FF} \id_\FF(f) = \bigtimes_{f \in \FF} f \neq \emptyset$ per l'assioma di scelta. Sia dunque $g \in \bigtimes_{f \in \FF} f$. Consideriamo $X = \{ y \in \bigcup \FF \mid \exists x (x \in \FF \land y = g(x)) \} = g(\FF)$, che è un'insieme per l'assioma di separazione. \smallskip + + Sia $F \in \FF$. Allora $g(F) \in X$. Mostriamo che se $y$ è un insieme tale per cui $y \in X$ e $y \in F$, allora $y = g(F)$. Dacché $y \in X$, esiste + $z \in \FF$ tale per cui $y = g(z)$. Poiché $g \in \bigtimes_{f \in \FF} f$, + $g(z) \in z$; allo stesso tempo $g(z) = y \in F$. Se $z$ fosse diverso da $F$, + $g(z)$ sarebbe un elemento di $z \cap F$, che per ipotesi è l'insieme vuoto, + \Lightning. Quindi $z = F$, e dunque $y = g(F)$. Dunque $\abs{X \cap F} = 1$. + + \item[\fbox{(ii.) $\implies$ (i.)}] Sia $\sigma$ una $I$-successione tale per cui + $\sigma(i)$ è non vuoto per ogni $i \in I$. Consideriamo la famiglia di + insiemi $\FF := \{ y \in \PP(\PP(\PP(\bigcup \{I, \bigcup \imm \sigma\}))) \mid \exists i (i \in I \land y = \{i\} \times \sigma(i)) \}$. Allora $\FF$ è una famiglia di insiemi non vuoti + a due a due disgiunti (infatti due elementi distinti di $\FF$ hanno elementi con prima coordinata diversa, e quindi sono diversi per la proprietà + caratterizzante delle coppia ordinata di Kuratowski). + Dunque esiste un insieme $X$ tale per cui $\abs{X \cap (\{i\} \times \sigma(i))} = 1$ per ogni $i \in I$. Si può allora costruire una funzione + $f : I \to \bigcup_{i \in I} \sigma(i)$ tale per cui $f(i) \in \sigma(i)$ + prendendo come $f(i)$ come l'unico elemento in $\abs{X \cap (\{i\} \times \sigma(i))}$ -- tale funzione è un insieme perché questa definizione è + traducibile in formula e perché tale $f \in \Fun(I, \bigcup_{i \in I} \sigma(i))$, dove quest'ultimo è un insieme (e dunque si può applicare + l'assioma di separazione). Allora $\bigtimes_{i \in I} \sigma(i) \neq \emptyset$, ovverosia l'assioma di scelta è vero. + + \item[\fbox{(i.) $\implies$ (iii.)}] Sia $f : A \to B$ una funzione surgettiva. + Sia $\sigma : B \to \PP(A)$ tale per cui $\sigma(b) = f\inv(b)$. Poiché + $f$ è surgettiva, $\sigma(b) \neq \emptyset$ per ogni $b \in B$. Allora, + per l'assioma di scelta, $\bigtimes_{b \in B} \sigma(b) \neq \emptyset$. + Sia $g \in \bigtimes_{b \in B} \sigma(b)$. Dal momento che + $g(b) \in f\inv(b)$, $f(g(b)) = b$ per ogni $b \in B$, ovverosia + $g$ è un'inversa destra di $f$. + + \item[\fbox{(iii.) $\implies$ (i.)}] Sia $\sigma$ una $I$-successione tale per + cui $\sigma(i) \neq \emptyset$ per ogni $i \in I$. Consideriamo + l'insieme $A = \{ y \in I \times \bigcup \imm \sigma \mid \exists i (i \in I \land \exists b (b \in \sigma(i) \land y = (i, b) )) \}$. Possiamo allora + costruire una funzione $f : A \to I$ tale per cui $(i, b) \mapsto i$, ovverosia + $f = \pi_1 \circ \iota$, dove + $\iota$ è l'immersione di $A$ in + $I \times \bigcup \imm \sigma$ e $\pi_1$ è la proiezione indotta dalla + prima coordinata di $I \times \bigcup \imm \sigma$. Chiaramente $f$ è + surgettiva dal momento che $\sigma(i) \neq \emptyset$ per ogni $i \in I$. + Allora $f$ ammette un'inversa destra $g : I \to A$. Se allora $\pi_2$ è la proiezione indotta dalla seconda coordinata di $I \times \bigcup \imm \sigma$, $h = \pi_2 \circ \iota \circ g$ è una funzione da $I$ a $\bigcup \imm \sigma$ + tale per cui, per costruzione, $h(i) \in \sigma(i)$, ovverosia + $h \in \bigtimes_{i \in I} \sigma(i)$, che dunque non è vuoto. + + \item[\fbox{(i.) $\implies$ (iv.)}] Consideriamo la $\left(\PP(X) \setminus \{\emptyset\}\right)$-successione $\id_{\PP(X) \setminus \{\emptyset\}}$. Poiché a $\PP(X)$ è + stato tolto l'insieme vuoto $\emptyset$, ogni suo elemento è non vuoto. + Pertanto $\bigtimes_{Y \in \PP(X) \setminus \{\emptyset\}} Y \neq \emptyset$ + per l'assioma di scelta, ossia esiste $f : \PP(X) \setminus \{\emptyset\} \to \bigcup \imm \id_{\PP(X) \setminus \{\emptyset\}} = \bigcup_{Y \in \PP(X) \setminus \{\emptyset\}} Y = X$ tale per cui $f(y) \in \id_{\PP(X) \setminus \{\emptyset\}}(y) = y$ per $y \in \PP(X) \setminus \{\emptyset\}$, ossia + $f$ è la funzione di scelta cercata. + + \item[\fbox{(iv.) $\implies$ (i.)}] Sia $\sigma$ una $I$-successione tale per cui + $\sigma(i) \neq \emptyset$ per ogni $i \in I$. Consideriamo allora + $X = \bigcup \imm \sigma$, per il quale vale $\sigma(i) \subseteq X$ per + ogni $i \in I$, ovverosia $\sigma(i) \in \PP(X) \setminus \{\emptyset\}$ (infatti i $\sigma(i)$ sono \underline{non} vuoti). Allora esiste una funzione di scelta $f : \PP(X) \setminus \{\emptyset\} \to X$. Sia $h : I \to \bigcup_{i \in I} \sigma(i)$ definita in modo tale che + $h(i) = f(\sigma(i))$. Dacché $f$ è una funzione di scelta, $h(i) = f(\sigma(i)) \in \sigma(i)$, ossia $h$ è ben definita sul suo codominio ed + è un elemento di $\bigtimes_{i \in I} \sigma(i)$, che dunque non è vuoto. + \end{itemize} +\end{solution} + +\begin{problem}{Composizione di relazioni e di funzioni}{problem-5} + Si dia una opportuna definizione di composizione per le relazioni $\RR$ e $\RR'$ con $\imm(\rel) = \dom(\rel')$ e si mostri + che per tale definizione la composizione di relazioni che sono funzioni è ancora una funzione e che tale composizione coincide con l'usuale composizione di funzioni. +\end{problem} + +\begin{solution} + Siano $\rel$ e $\rel'$ due relazioni. Definiamo $\rel' \circ \rel$ in modo tale che + i suoi elementi siano le coppie $(x, y)$ con $x \in \dom(\rel)$ e $y \in \imm(\rel')$ + tale per cui $\exists z (z \in \imm(\rel) \land (x, z) \in \rel \land (z, y) \in \rel')$, dove ricordiamo che $\imm(\rel) = \dom(\rel')$. In altre parole: + \[ + \rel' \circ \rel = \{ c \in \dom(\rel) \times \imm(\rel') \mid \varphi(c) \}, + \] + dove: + \[ + \varphi(c) = \exists x \exists y (x \in \dom(\rel) \land y \in \imm(\rel') \land c = (x, y) \land \exists z (z \in \imm(\rel) \land (x, z) \in \rel \land (z, y) \in \rel')). + \] + Osserviamo innanzitutto che $\rel' \circ \rel$ è un insieme, per l'assioma + della separazione. Successivamente, notiamo che $\rel' \circ \rel$ è una relazione, + essendo un insieme di coppie ordinate, che $\imm(\rel' \circ \rel) \subseteq \imm(\rel')$ e che $\dom(\rel' \circ \rel) = \dom(\rel)$ -- dove si è usato ancora + che $\imm(\rel) = \dom(\rel')$. \smallskip + + Siano ora $g := \rel$ e $f := \rel'$ funzioni. Mostriamo che $f \circ g$ è + una funzione. Sia $x \in \dom(f \circ g) = \dom(g)$ e supponiamo che, date $y$ e $y'$ in $\imm(\rel' \circ \rel) \subseteq \imm(\rel')$, $(x, y)$ e $(x, y')$ appartengano + entrambe a $f \circ g$. Allora esistono $z$ e $z'$ in $\imm(\rel)$ tali per cui + $(x, z)$, $(x, z')$ appartengano entrambe a $g$ e $(z, y)$, $(z', y')$ appartengano + a $f$. Dal momento che $g$ è una funzione, dalla prime due appartenenze si + ricava $z = z'$; dacché anche $f$ è una funzione, dalle seconde due, sostituendo $z = z'$, si ricava $y = y'$, ovverosia $f \circ g$ è una funzione. \smallskip + + Mostriamo che tale composizione coincide con l'usuale composizione di funzioni, + ovverosia verifichiamo che $(f \circ g)(x) = f(g(x))$ per ogni $x \in \dom(g)$. + Sappiamo che $(x, g(x)) \in g$ e che $(g(x), f(g(x))) \in f$, allora -- + per definizione di $f \circ g$ -- $(x, f(g(x))) \in f \circ g$. Poiché + $f \circ g$ è una funzione, si deve allora avere necessariamente + $(f \circ g)(x) = f(g(x))$. +\end{solution} + +\begin{problem}{La classe delle funzioni da $A$ a $B$ è un insieme}{problem-6} + Si dimostri che $\Fun(A, B) = \{ f \mid f \text{ funzione da } A \text{ a } B \}$ è un insieme. +\end{problem} + +\begin{solution} + Poiché $f \in \Fun(A, B)$ ha dominio $A$ e immagine $B$, allora + $f \subseteq A \times B$, ovverosia $f \in \PP(A \times B)$. Dunque, + $\Fun(A, B)$ è un insieme per l'assioma di separazione, dacché: + \[ + \Fun(A, B) = \{ f \in \PP(A \times B) \mid f \text{ funzione da } A \text{ a } B \}, + \] + dove usiamo che ``$f$ funzione da $A$ a $B$'' è effettivamente una formula ammissibile (ossia si può sviluppare usando i simboli primitivi della nostra teoria). +\end{solution} + +\begin{problem}{$\{a, b, c\}$ è un insieme}{problem-7} + Si dimostri che $\{a, b, c\}$ è un insieme. +\end{problem} + +\begin{solution} + Usiamo gli assiomi di ZF per dimostrare che $\{a, b, c\}$ è un insieme, + sapendo che $a$, $b$ e $c$ sono insiemi. Per l'assioma della coppia esistono + gli insiemi $\{a, b\}$ e $\{c, c\} = \{c\}$. Ancora per l'assioma della coppia + esiste $\{\{a, b\}, \{c\}\}$. Allora per l'assioma dell'unione esiste + $\bigcup \{\{a, b\}, \{c\}\} = \{a, b, c\}$, come desideravamo. +\end{solution} + +\begin{problem}{Forme equivalenti dell'assioma di scelta (2)}{problem-8} + Si mostri che l'assioma di scelta è equivalente alla seguente asserzione: + \[ + \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j} = \bigcup_{f \in \Fun(I, J)} \,\bigcap_{i \in I} A_{i, f(i)}, \qquad \forall (A_{i,j} \mid i \in I, \, j \in J). \quad (*) + \] +\end{problem} + +\begin{solution} + Mostriamo le due implicazioni separatamente. + + \begin{itemize} + \item [\fbox{\AC $\implies$ $(*)$}] Sia $x \in \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j}$. Allora per ogni $i_0 \in I$, esiste $j_0 \in J$ tale per cui + $x \in A_{i_0, j_0}$. In altre parole l'insieme $B_i = \{j \in J \mid x \in A_{i,j}\}$ \underline{non} è vuoto per ogni $i \in I$. Per l'assioma di scelta + $\bigtimes_{i \in I} B_i$ \underline{non} è vuoto. Sia dunque $h \in \bigtimes_{i \in I} B_i$. Per costruzione $x \in A_{i, h(i)}$ per ogni + $i \in I$, e dunque $x \in \bigcap_{i \in I} A_{i, f(i)} \subseteq \bigcup_{f \in \Fun(I, J)} \,\bigcap_{i \in I} A_{i, f(i)}$. Pertanto vale + $x \in \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j} \implies x \in \bigcup_{f \in \Fun(I, J)} \,\bigcap_{i \in I} A_{i, f(i)}$. \smallskip + + Sia $x \in \bigcup_{f \in \Fun(I, J)} \,\bigcap_{i \in I} A_{i, f(i)}$. Allora esiste $f \in \Fun(I, J)$ tale per cui $x \in A_{i, f(i)} \subseteq \bigcup_{j \in J} A_{i,j}$ per ogni $i \in I$. Pertanto $x \in \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j}$. Dunque vale $x \in \bigcup_{f \in \Fun(I, J)} \,\bigcap_{i \in I} A_{i, f(i)} \implies x \in \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j}$. \smallskip + + Infine, per l'assioma di estensionalità, si conclude che: + \[ + \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j} = \bigcup_{f \in \Fun(I, J)} \,\bigcap_{i \in I} A_{i, f(i)}. + \] + + \item [\fbox{$(*)$ $\implies$ \AC}] Mostriamo che $(*)$ implica l'esistenza di + una funzione di scelta sulle parti di un qualsiasi insieme \underline{non} vuoto, che sappiamo essere equivalente ad \AC. Sia dunque $X$ un insieme \underline{non} vuoto. Consideriamo $I = \PP(X) \setminus \{\emptyset\}$ + e $J = X$. Sia $A_{i, j} = \{X\}$ se $j \in i$, e $\emptyset$ altrimenti. Allora vale che: + \[ + \bigcap_{i \in I} \bigcup_{j \in J} A_{i,j} = \bigcap_{\substack{Y \subseteq X \\ Y \neq \emptyset}} \bigcup_{x \in X} A_{Y\!\!, \,x} = + \bigcap_{\substack{Y \subseteq X \\ Y \neq \emptyset}} \{X\} = \{X\}, + \] + dove la penultima uguaglianza è dovuta al fatto che gli $Y$ sono stati + scelti non vuoti (e quindi esiste almeno un elemento $x$ dentro ogni $Y$, + per cui $A_{Y\!\!, \,x} = \{X\}$). \smallskip + + Per $(*)$ vale allora che: + \[ \{X\} = \bigcap_{\substack{Y \subseteq X \\ Y \neq \emptyset}} \bigcup_{x \in X} A_{Y\!\!, \,x} = \bigcup_{f \in \Fun(\PP(X) \setminus \{\emptyset\}, X)} \bigcap_{\substack{Y \subseteq X \\ Y \neq \emptyset}} A_{Y\!\!, \,f(Y)}, \] + ovverosia esiste $f \in \Fun(\PP(X) \setminus \{\emptyset\}, X)$ tale per + cui $X \in A_{Y\!\!, \,f(Y)}$ per ogni $Y \in \PP(X) \setminus \{\emptyset\}$. + Questo è possibile solo se $A_{Y\!\!, \,f(Y)} = \{X\}$ per ogni $Y$, ovverosia + per costruzione se $f(Y) \in Y$. Dunque $f$ è una funzione di scelta sulle parti di $X$, + da cui la tesi. + \end{itemize} +\end{solution} + +\begin{problem}{La classe degli insiemi equipotenti è propria}{problem-9} + Si mostri che, dato $y \neq \emptyset$, $E_y = \{ x \mid \abs{x} = \abs{y} \}$ \underline{non} è un insieme. +\end{problem} + +\begin{solution} + Sia $x$ un insieme. Poiché $y \neq \emptyset$, esiste $z \in y$. Per l'assioma + della separazione, $y \setminus \{z\} = \{ x \in y \mid x \neq z \}$ è un + insieme. Per l'assioma della coppia esistono $\{x, x\} = \{x\}$ e + $\{y \setminus \{z\}, \{x\}\}$, dunque, per l'assioma dell'unione, + esiste $\bigcup \{y \setminus \{z\}, \{x\}\} = (y \setminus \{z\}) \cup \{x\}$. + $(y \setminus \{z\}) \cup \{x\}$ è in bigezione con $y$ tramite la mappa + che manda tutti gli elementi diversi da $x$ in sé stessi e $x$ in $z$, + quindi $\abs{(y \setminus \{z\}) \cup \{x\}} = \abs{y}$, ovverosia + $(y \setminus \{z\}) \cup \{x\} \in E_y$. Dunque + $x \in (y \setminus \{z\}) \cup \{x\} \in E_y$, ovverosia ogni insieme + è elemento di un elemento in $E_y$. Pertanto, se $E_y$ fosse un insieme, + per l'assioma dell'unione si avrebbe $\bigcup E_y = \UU$, ovverosia la classe + di tutti gli insiemi sarebbe un insieme, ma questo è falso, \Lightning. Dunque + $E_y$ non è un insieme. +\end{solution} + +\begin{problem}{Una bigezione esplicita tra $[0, 1)$ e $(0, 1)$}{problem-10} + Si trovi una bigezione esplicita tra $[0,1)$ e $(0,1)$. +\end{problem} + +\begin{solution} + Sfruttando l'idea che abbiamo utilizzato per costruire una bigezione tra + $\RR$ e $\RR \setminus \{0\}$ possiamo costruire direttamente una bigezione + tra $[0, 1)$ e $(0, 1)$, come segue: + \[ + f(x) = \begin{cases} + \frac{1}{2} & \se x = 0, \\ + x & \se x \neq 0 \text{ e } 1/x \notin \NN, \\ + \frac{1}{\frac{1}{x}+1} & \se x \neq 0 \text{ e } 1/x \in \NN. + \end{cases} + \] + In questo modo, i numeri non naturali vengono mandati in loro stessi, + $0$ viene mandato in $\frac{1}{2}$ (e dacché $1 \notin [0, 1)$ ciò non + crea problemi), $\frac{1}{2}$ viene mappato a $\frac{1}{3}$, + $\frac{1}{3}$ a $\frac{1}{4}$, etc... -- seguendo l'analoga filosofia + adottata nel caso di $\RR$ e $\RR \setminus \{0\}$, in cui fissavamo + i non naturali e mandavamo i naturali nei loro successivi. +\end{solution} + +\begin{problem}{Prodotto, unione a intersezione nulla, spazi di funzioni e parti di insiemi equipotenti sono equipotenti}{problem-11} + Siano $A$ e $A'$ tali che $\abs{A} = \abs{A'}$. Siano + $B$ e $B'$ tali che $\abs{B} = \abs{B'}$. Si mostri allora che: + + \begin{enumerate}[(i.)] + \item $\abs{A \times B} = \abs{A' \times B'}$, + \item $\abs{A \cup B} = \abs{A' \cup B'}$, se $A \cap B = A' \cap B' = \emptyset$, + \item $\abs{\Fun(A, B)} = \abs{\Fun(A', B')}$, + \item $\abs{\PP(A)} = \abs{\PP(A')}$. + \end{enumerate} + Mostrare che prodotto, parti, unione, spazio delle funzioni di insiemi equipotenti sono equipotenti. +\end{problem} + +\begin{solution} + Sia $f : A \to A'$ una bigezione, così come $g : B \to B'$. + Dimostriamo la tesi punto per punto. + + \begin{enumerate}[(i.)] + \item La mappa $A \times B \ni (a, b) \mapsto (f(a), g(b)) \in A' \times B'$ è + una funzione la cui inversa è $A' \times B' \ni (a', b') \mapsto (f\inv(a'), g\inv(b')) \in A \times B$, dunque è una bigezione. Allora $\abs{A \times B} = \abs{A' \times B'}$. + + \item Dal momento che $A \cap B = \emptyset$, la funzione $h : A \cup B \to A' \cup B'$ tale per cui: + \[ + h(x) = \begin{cases} + f(x) & \se x \in A, \\ + g(x) & \se x \in B. + \end{cases} + \] + è ben definita. \smallskip + + Chiaramente $h$ è suriettiva, dal momento che lo sono sia $f$ + che $g$. Se $x$ e $y$ in $A \cup B$ sono tali che $h(x) = h(y)$, allora $h(x)$ e $h(y)$ devono appartenere entrambi ad + $A'$ o entrambi a $B'$ dal momento che $A' \cap B' = \emptyset$. Per lo + stesso motivo $x$ e $y$ devono necessariamente appartenere entrambe o $A$ o a $B$, e dunque deve valere o $f(x) = f(y)$ (se $x$, $y \in A$) o + $g(x) = g(y)$ (se $x$, $y \in B$), da cui si deduce a prescindere che vale + $x = y$, da cui l'iniettività di $h$. Poiché $h$ è iniettiva e suriettiva, + $h$ è una bigezione. Dunque $\abs{A \cup B} = \abs{A' \cup B'}$. + + \item Costruiamo $F : \Fun(A, B) \to \Fun(A', B')$ tale per cui + $F(h) = g \circ h \circ f\inv$. Mostriamo che $F$ è effettivamente una + bigezione. + + \[\begin{tikzcd} + A && B \\ + \\ + {A'} && {B'} + \arrow["h", from=1-1, to=1-3] + \arrow["f"', tail, two heads, from=1-1, to=3-1] + \arrow["g", tail, two heads, from=1-3, to=3-3] + \arrow["{F(h)}"', from=3-1, to=3-3] + \end{tikzcd}\] + + Sia $G : \Fun(A', B') \to \Fun(A, B)$ tale per cui $G(k) = g\inv \circ k \circ f$. Mostriamo che $F$ e $G$ sono una l'inversa destra dell'altra: + + \begin{itemize} + \item $F(G(k)) = F(g\inv \circ k \circ f) = g \circ g\inv \circ k \circ f \circ f\inv = \id_{B'} \circ k \circ \id_{A'} = k$, ovverosia + $G$ è inversa destra di $F$. + + \item $G(F(h)) = G(g \circ h \circ f\inv) = g\inv \circ g \circ h \circ f\inv \circ f = \id_B \circ h \circ \id_A = h$, ovverosia $F$ è + inversa destra di $G$. + \end{itemize} + + Allora $G$ è l'inversa di $F$, e dunque $F$ è una bigezione. Si conclude + pertanto che $\abs{\Fun(A, B)} = \abs{\Fun(A', B')}$. + + \item Dal momento che $\abs{\PP(A)} = \abs{\Fun(A, \{0, 1\})}$, + che $\abs{\PP(A')} = \abs{\Fun(A', \{0, 1\})}$ e che -- dal punto (iii.) -- + $\abs{\Fun(A, \{0, 1\})} = \abs{\Fun(A', \{0, 1\})}$, allora, per + transitività, vale che $\abs{\PP(A)} = \abs{\PP(A')}$. \smallskip + + In alternativa, possiamo costruire $F : \PP(A) \to \PP(A')$ tale per cui $F(C) = \{ y \in A' \mid \exists x (x \in C \land y = f(x))\}$ per $C \subseteq A$, ovverosia + $F(C)$ è l'immagine di $C$ tramite $f$. $F$ ammette come inversa + $G : \PP(A') \to \PP(A)$ tale per cui $G(D) = \{ z \in A \mid f(z) \in D \}$ + per $D \subseteq A'$, + ovverosia la controimmagine di $D$ tramite $f$. \smallskip + + Mostriamo che $F$ e $G$ sono l'una l'inversa destra dell'altra: + + \begin{itemize} + \item $G(F(C)) = G(\{ y \in A' \mid \exists x (x \in C \land y = f(x))\}) + = \{ z \in A \mid f(z) \in \{ y \in A' \mid \exists x (x \in C \land y = f(x))\} \} = \{ z \in A \mid f(z) \in A' \land \exists x (x \in C \land f(z) = f(x)) \}$. \smallskip + + Dal momento che $f(z) \in A'$ è sempre vera e che $f(z) = f(x) \iff z=x$ per l'iniettività di $f$, + l'insieme $G(F(C))$ si riscrive come $\{ z \in A \mid \exists x (x \in C \land z = x) \} = \{ z \in A \mid z \in C \} = C$, dunque $F$ è inversa + destra di $G$. + + \item $F(G(D)) = F(\{ z \in A \mid f(z) \in D \}) = \{ y \in A' \mid \exists x (x \in \{ z \in A \mid f(z) \in D \} \land y = f(x))\} = \{ y \in A' \mid \exists x (x \in A \land f(x) \in D \land y = f(x)) \}$. \smallskip + + Sostituendo $y = f(x)$ in $f(x) \in D$ ($\leadsto y \in D$), portando fuori $y \in D$ fuori dalle condizioni di esistenza (è indipendente dalla variabile $x$) e sfruttando che $\exists x (x \in A \land y = f(x))$ per $y \in A'$ è sempre vera per la suriettività di $f$, + si riscrive $F(G(D))$ come $\{ y \in A' \mid y \in D \land \exists x (x \in A \land y = f(x)) \} = \{ y \in A' \mid y \in D \} = D$, dunque + $G$ è inversa destra di $F$. + \end{itemize} + + Allora $G$ è l'inversa di $F$, e dunque $F$ è una bigezione. Si conclude + pertanto ancora che $\abs{\PP(A)} = \abs{\PP(A')}$. + \end{enumerate} +\end{solution} + +\begin{problem}{Una bigezione da $\NN$ ad $A \subseteq \NN$ infinito}{problem-12} + Sia $A \subseteq \NN$ un insieme infinito. Si dimostri che la funzione $f : \NN \to A$ + definita di seguito per ricorsione numerabile è una bigezione: + \[ + f(n) = \begin{cases} + \min(A) & \se n = 1, \\ + \min(A \setminus \{ f(1), \ldots, f(n-1) \}) & \altrimenti. + \end{cases} + \] +\end{problem} + +\begin{solution} + Mostriamo che, dato $n \in \NN$, $f(n) < f(n+1)$, + da cui si deduce che $f$ è strettamente crescente e dunque iniettiva. Sia $B = A \setminus \{ f(1), \ldots, f(n-1) \}$. Allora $f(n) = \min(B)$ + e $f(n+1) = \min(B \setminus \{ f(n) \}$. In + particolare $f(n+1)$ è dunque un numero naturale + distinto da $f(n)$ e appartenente a $B$; essendo + allora $f(n)$ il minimo di $B$, $f(n) < f(n+1)$. \smallskip + + Infine mostriamo che $f$ è surgettiva. Supponiamo + per assurdo che $f$ non lo sia: allora + $\imm f \neq A$, dunque $A \setminus \imm f$ -- + essendo sottinsieme di $\NN$ -- ammette un minimo $a$. + Sicuramente $a \neq \min(A)$, perché $f(1) = \min(A)$; + in particolare $a > \min(A)$. + Allora $[\min(A), a)_A$ non è vuoto ed ammette + dunque un massimo $a' \in A$. Poiché $a' < a$, + essendo $a$ il minimo di $A \setminus \imm f$, + necessariamente $a' \in \imm f$, ovverosia + esiste $n \in \NN$ tale per cui $f(n) = \min (A \setminus \{f(1), \ldots, f(n-1)\}) = a'$. Allora + $f(n+1) = \min (A \setminus \{f(1), \ldots, f(n-1), a'\})$. Dal momento che $a \notin \imm f$, sicuramente + $a \in A \setminus \{f(1), \ldots, f(n-1), a'\}$. + Essendo $a$ il successore di $a'$ in $A$, allora + $a$ è anche il minimo di $a \in A \setminus \{f(1), \ldots, f(n-1), a'\}$, e dunque $f(n+1) = a$, ovverosia + $a \in \imm f$, \Lightning. Quindi $f$ è surgettiva. \smallskip + + Dal momento che $f$ è sia iniettiva che surgettiva, + allora $f$ è bigettiva. +\end{solution} + +\begin{problem}{Una funzione iniettiva da $\Fun(\NN, \{0, 1\})$ in $[0, 1)$}{problem-13} + Si dimostri che la funzione da $\Fun(\NN, \{0, 1\})$ in $[0, 1)$ che associa a $f$ la serie $\sum_{n \in \NN} \frac{f(n)}{10^n}$ è iniettiva. +\end{problem} + +\begin{solution} + Siano $f$ e $g$ elementi di $\Fun(\NN, \{0, 1\})$ tali per cui $\sum_{n \in \NN} \frac{f(n)}{10^n}$ $=$ $\sum_{n \in \NN} \frac{g(n)}{10^n}$. Supponiamo + $f \neq g$. Allora $\{ n \in \NN \mid f(n) \neq g(n) \}$, che dunque non è vuoto, ammette per il principio del buon ordinamento un minimo + $n_0$ per cui $f(n_0) \neq g(n_0)$. Allora, eliminando i termini uguali precedenti + a $n_0$ e moltiplicando per $10^{n_0}$ si ottiene: + \[ + f(n_0) + \sum_{n \in \NN} \frac{f(n_0 + n)}{10^n} = g(n_0) + \sum_{n \in \NN} \frac{g(n_0 + n)}{10^n}, + \] + da cui: + \[ + f(n_0) - g(n_0) = \sum_{n \in \NN} \frac{g(n_0 + n) - f(n_0 + n)}{10^n}. + \] + Stimando $f(n_0) - g(n_0)$ sapendo che $f(n_0 + n)$, $g(n_0 + n)$ appartengono + a $\{0, 1\}$, si ottiene: + \[ + -\frac{1}{9} = \sum_{n \in \NN} -\frac{1}{10^n} \leq f(n_0) - g(n_0) \leq \sum_{n \in \NN} \frac{1}{10^n} = \frac{1}{9}. + \] + Poiché $f(n_0) - g(n_0)$ è un naturale tra $-1$, $0$ e $1$, ciò implica che + sia $0$, ossia $f(n_0) = g(n_0)$, \Lightning. Quindi $f = g$, ossia + la funzione indicata nella tesi è iniettiva. +\end{solution} + +\begin{problem}{Se $X$ è infinito, allora $\abs{\Fin(X)} = \abs{\FSeq(X)} = \abs{X}$ (\AC)}{problem-14} + Si assuma di sapere che \AC~è equivalente ad affermare che $\abs{X \times X} = \abs{X}$ per ogni $X$ infinito. Assumendo allora \AC, si dimostri che, dato $X$ infinito, + $\abs{\Fin(X)} = \abs{\FSeq(X)} = \abs{X}$. +\end{problem} + +\begin{solution} + Poiché vale \AC, esiste una funzione di scelta $f : \PP(X) \setminus \{\emptyset\} \to X$, ovvero sia $f$ è una funzione tale per cui $f(Y) \in Y$ per $Y \in \PP(X) \setminus \{\emptyset\}$. \smallskip + + Mostriamo in ordine i seguenti risultati: + \begin{tasks}[label=(\roman*.), label-width=19.4064pt](3) + \task $\abs{X} \leq \abs{\Fin(X)}$, + \task $\abs{\Fin(X)} \leq \abs{\FSeq(X)}$, + \task $\abs{\FSeq(X)} \leq \abs{X}$. + \end{tasks} + + \begin{enumerate}[(i.)] + \item Sia $F_1 : X \to \Fin(X)$ tale per cui + $x \mapsto \{x\}$. Chiaramente $F_1$ è iniettiva, + dunque $\abs{X} \leq \abs{\Fin(X)}$. + + \item Poiché $X$ è infinito, in particolare $X$ \underline{non} è + vuoto. Sia dunque $c$ un elemento di $X$. + Sia $Y \in \Fin(X)$ \underline{non} vuoto. Se $n = \abs{Y}$, allora + si può definire per ricorsione numerabile la funzione $f_Y : n \to Y$ tale per cui: + \[ + f_Y(i) = \begin{cases} + f(Y) & \se i = 0, \\ + f(Y \setminus \{f_Y(0), \ldots, f_Y(i-1)\}) & \altrimenti. + \end{cases} + \] + Possiamo dunque definire $F_2 : \Fin(X) \to \FSeq(X)$ in modo + tale che $F_2(\emptyset) = (c)$ e $F_2(Y) = (c, f_Y(0), \ldots, f_Y(\abs{Y} - 1))$. Mostriamo che $F_2$ è iniettiva. \smallskip + + Innanzitutto se $F_2(Y) = F_2(Y')$, deve chiaramente + valere o $Y = Y' = \emptyset$ o $\abs{Y} = \abs{Y'} = n$. + Allora, per la proprietà caratterizzante delle coppie ordinate, + deve valere $f_Y(0) = f_{Y'}(0)$, ..., $f_Y(n-1) = f_{Y'}(n-1)$. + Poiché $f_Y$ e $f_{Y'}$ sono funzioni surgettive, vale che: + \[ Y = \imm f_Y = \{ f_Y(0), \ldots, f_Y(n-1) \} = \{ f_{Y'}(0), \ldots, f_{Y'}(n-1) \} = \imm f_{Y'} = Y'. \]\ + Dunque $F_2$ è iniettiva, da cui $\abs{\Fin(X)} \leq \abs{\FSeq(X)}$. + + \item Osserviamo innanzitutto che $\FSeq(X) = \bigcup_{n \in \NN} X^n$. + Poiché $X$ \underline{non} è vuoto esiste un elemento $c \in X$. + Allora $\abs{X} \leq \abs{\{c\} \times X} \leq \abs{\NN \times X} \leq \abs{X \times X} = \abs{X}$, dove si è usato che $X$ si immerge + in modo naturale in $\{c\} \times X$, che $\{c\} \leq \NN$ dacché + $\NN$ è \underline{non} vuoto, e che $X$, essendo infinito, ammette + una funzione surgettiva su $\NN$. Pertanto, per il teorema di + Cantor-Bernstein, vale che $\abs{\NN \times X} = \abs{X}$. \smallskip + + Dal momento che $X$ è infinito, sappiamo che $\abs{X^n} = \abs{X}$ + applicando ricorsivamente $\abs{X \times X} = \abs{X}$, che sappiamo + essere vero grazie all'assioma di scelta. Quindi $\bige(X^n, X)$ è + \underline{non} vuoto. Allora, ancora per l'assioma di scelta, + $\bigtimes_{n \in \NN} \bige(X^n, X)$ è \underline{non} vuoto. + Sia $(f_i \mid i \in \NN) \in \bigtimes_{n \in \NN} \bige(X^n, X)$. + Costruiamo $F_3 : \FSeq(X) \to \NN \times X$ in + modo tale che $F_3((x_1, \ldots, x_n)) = (n, f_n((x_1, \ldots, x_n)))$. \smallskip + + Mostriamo che $F_3$ è iniettiva. Se $F_3((x_1, \ldots, x_n)) = F_3((y_1, \ldots, y_m))$, allora $(n, f_n((x_1, \ldots, x_n))) = (m, f_n((y_1, \ldots, y_m)))$. Per la proprietà caratterizzante delle coppie ordinate, + dalle prime coordinate si ricava $m = n$, e poi dalle seconde si + deduce che $f_n((x_1, \ldots, x_n)) = f_n((y_1, \ldots, y_n))$, + e dunque che $(x_1, \ldots, x_n) = (y_1, \ldots, y_n)$. Dunque + $F_3$ è iniettiva, da cui $\abs{\FSeq(X)} \leq \abs{\NN \times X}$. \smallskip + + Per quanto detto prima $\abs{\NN \times X} = \abs{X}$, dunque + $\abs{\FSeq(X)} \leq \abs{X}$. + \end{enumerate} + + Infine, applicando il teorema di Cantor-Bernstein sulla catena: + \[ + \abs{X} \leq \abs{\Fin(X)} \leq \abs{\FSeq(X)} \leq \abs{X}, + \] + si ottiene esattamente la tesi. +\end{solution} + +\begin{problem}{$\abs{X} \leq \abs{Y}$ $\implies$ $\abs{\Fin(X)} \leq \abs{\Fin(Y)}$ e $\abs{\FSeq(X)} \leq \abs{\FSeq(Y)}$}{problem-15} + Siano $X$ e $Y$ insiemi non vuoti tali per cui $\abs{X} \leq \abs{Y}$. Si + mostri che $\abs{\Fin(X)} \leq \abs{\Fin(Y)}$ e che + $\abs{\FSeq(X)} \leq \abs{\FSeq(Y)}$. +\end{problem} + +\begin{solution} + Sia $f : X \to Y$ una funzione iniettiva. Allora $f$ induce + due funzioni $F : \Fin(X) \to \Fin(Y)$ e $S : \FSeq(X) \to \FSeq(Y)$, dove $F(Z) = f(Z)$, per $Z \in \Fin(X)$, e + $S((x_1, \ldots, x_n)) = (f(x_1), \ldots, f(x_n))$ per + $x_1$, ..., $x_n$ appartenenti a $X$ e $n$ variabile tra + i numeri naturali. \smallskip + + Poiché $f$ è iniettiva, $F$ lo è -- infatti $F(Z) = F(Z') \implies F\inv(F(Z)) = F\inv(F(Z')) \implies Z = Z'$; dunque + $\abs{\Fin(X)} \leq \abs{\Fin(Y)}$. \smallskip + + Inoltre, se $\hat x = (x_1, \ldots, x_n)$ e $\hat y = (y_1, \ldots, y_m)$ sono + sequenze finite in $\FSeq(X)$ con $m$ e $n$ numeri naturali, + allora $S(\hat x) = S(\hat y)$ implica sicuramente $m = n$. + D'altra parte, per la proprietà caratterizzante delle coppie ordinate, $S(\hat x) = S(\hat y)$ implica $f(x_i) = f(y_i)$ + per ogni $i$ naturale tra $1$ e $n$. Dacché $f$ è iniettiva, + questo vuol dire che $x_i = y_i$ per ogni tale $i$, e dunque + che $\hat x = \hat y$, ovverosia $S$ è iniettiva. Pertanto + $\abs{\FSeq(X)} \leq \abs{\FSeq(Y)}$. +\end{solution} + +\begin{problem}{Se $A \cap A' = \emptyset$, $\abs{\Fun(A, B) \times \Fun(A', B)} = \abs{\Fun(A \cup A', B)}$}{problem-16} + Siano $A$ e $A'$ due insiemi per cui $A \cap A' = \emptyset$. Sia anche $B$ un insieme. Si mostri + allora che $\abs{\Fun(A, B) \times \Fun(A', B)} = \abs{\Fun(A \cup A', B)}$. +\end{problem} + +\begin{solution} + Siano $f : A \to B$ e $g : A' \to B$ due funzioni. Dal momento che + $A \cap A' = \emptyset$, $f \cup g$ è ancora una funzione -- infatti + non vi sono collisioni nel dominio. Allora $F : \Fun(A, B) \times \Fun(A', B) \to \Fun(A \cup A', B)$ dove $F((f, g)) = f \cup g$ è ben definita. \smallskip + + Sia $G : \Fun(A \cup A', B) \to \Fun(A, B) \times \Fun(A', B)$ definita + in modo tale che $G(h) = (\restr{h}{A}, \restr{h}{A'})$, dove + $\restr{f}{X'} := f \cap (X' \times \imm f)$ è la restrizione di $f : X \to Y$ + sul dominio $X' \subseteq X$. Mostriamo + che $F$ e $G$ sono l'una l'inversa destra dell'altra. + + \begin{itemize} + \item $G(F((f, g)) = G(f \cup g) = ((\restr{f \cup g}{A}), (\restr{f \cup g}{A'}))$. Poiché $A \cap A' = \emptyset$, non vi sono collisioni + nel dominio, e dunque $G(F((f, g))) = (f, g)$ dacché $f$ ha dominio + $A$ e $g$ ha dominio $A'$. Dunque $F$ è inversa destra di $G$. + + \item $F(G(h)) = F((\restr{h}{A}, \restr{h}{A'})) = \restr{h}{A} \cup \restr{h}{A'} = h$. Dunque $G$ è inversa destra di $F$. + \end{itemize} + + Si conclude dunque che $G$ è l'inversa di $F$, ovverosia che $F$ è + una bigezione. Dunque $\abs{\Fun(A, B) \times \Fun(A', B)} = \abs{\Fun(A \cup A', B)}$. +\end{solution} + +\begin{problem}{Calcolo di cardinalità riguardanti $\NN$ e $\RR$ (\AC)}{problem-17} + Sia $n$ un numero naturale. Si calcoli la cardinalità dei seguenti insiemi, applicando l'assioma di scelta solo dove indicato: + + \begin{tasks}[label=(\alph*.),label-width=20.25777pt](4) + \task $\RR^n$, + \task $\Fin(\RR)$, + \task $\FSeq(\RR)$, + \task $\FSeq\!\uparrow\!(\RR)$, + \task $\Fun(\NN, \NN)$, + \task $\Fun\!\uparrow\!(\NN, \NN))$, + \task $\Sym(\NN)$, + \task $\Fun(\NN, \RR)$, + \task $\Fun\!\uparrow\!(\NN, \RR)$, + \task $[\NN]^{\aleph_0}$, + \task $[\RR]^{\aleph_0}$ (\AC), + \task $\Fun(\RR, \RR)$, + \task $C^0(\RR)$, + \task $\OO(\RR^2)$. + \end{tasks} +\end{problem} + +\begin{solution} + Risolviamo ogni punto del problema. + + \begin{enumerate}[(a.)] + \item Sappiamo che $\abs{\RR \times \RR} = \cc$. + Posto allora che $\abs{\RR^{n-1}} = \cc$, vale + che $\abs{\RR^n} = \abs{\RR^{n-1} \times \RR} = \abs{\RR \times \RR} = \cc$. Si conclude allora + per induzione che $\abs{\RR^n} = \cc$. + + \item[(b., c.)] Poiché $\RR$ è infinito, + $\abs{\Fin(\RR)} = \abs{\FSeq(\RR)} = \abs{\RR} = \cc$ + (vd. \textit{Problema 14}). + \addtocounter{enumi}{2} + + \item Poiché $\FSeq\!\uparrow\!(\RR) \subseteq \FSeq(\RR)$, sicuramente $\abs{\FSeq\!\uparrow\!(\RR)} \leq \cc$ per il punto (c.). D'altra + parte $\RR$ si immerge naturalmente in + $\FSeq\!\uparrow\!(\RR)$ (un numero è una $1$-sequenza, crescente per mancanza di altri numeri), + dunque $\cc \leq \abs{\FSeq\!\uparrow\!(\RR)}$. + Allora, per il teorema di Cantor-Bernstein, + $\abs{\FSeq\!\uparrow\!(\RR)} = \cc$. + + \item Poiché $\abs{2} \leq \abs{\NN}$, allora + $\cc = \abs{2^\NN} \leq \abs{\NN^\NN} \leq \abs{\left(2^\NN\right)^\NN} = \abs{2^{\NN \times \NN}} \leq + \abs{2^\NN} = \cc$, dove abbiamo usato anche che $\abs{\NN \times \NN} = \abs{\NN}$. Dunque, + per il teorema di Cantor-Bernstein, $\abs{\Fun(\NN, \NN)} = \cc$. + + \item Dacché $\Fun\!\uparrow\!(\NN, \NN) \subseteq \Fun(\NN, \NN)$, allora $\abs{\Fun\!\uparrow\!(\NN, \NN)} \leq \cc$. \smallskip + + Poiché $\NN$ è infinito, $\abs{\Fin(\NN)} = \abs{\NN}$. + Allora, dacché $\abs{\PP(\NN)} = \cc$, $\abs{\PP(\NN) \setminus \Fin(\NN)} = \cc$ (vd. \textit{Problema 19}). + Sia $A \subseteq \PP(\NN) \setminus \Fin(\NN)$. Possiamo + definire per ricorsione numerabile + la funzione $f_A : \NN \to \NN$ in modo tale che: + \[ + f_A(n) = \begin{cases} + \min(A) & \se n = 1, \\ + \min(A \setminus \{f(1), \ldots, f(n)\}) & \altrimenti. + \end{cases} + \] + Chiaramente $f_A$ è crescente e ha $A$ come immagine (vd. \textit{Problema 12}) -- da + cui $f_A = f_B \implies A = B$ (vd. \textit{Problema 14}, (ii.) per una semplice dimostrazione). Dunque + la mappa $F : \PP(\NN) \setminus \Fin(\NN) \to \Fun\!\uparrow\!(\NN, \NN)$ tale per cui $F(Y) = f_Y$ è iniettiva, da cui + si deduce che $\cc = \abs{\PP(\NN) \setminus \Fin(\NN)} \leq + \abs{\Fun\!\uparrow\!(\NN, \NN)}$. \smallskip + + Infine, per il teorema di Cantor-Bernstein, si deduce che + $\abs{\Fun\!\uparrow\!(\NN, \NN)} = \cc$. + + \item Dal momento che $\Sym(\NN) \subseteq \Fun(\NN, \NN)$, allora sicuramente $\abs{\Sym(\NN)} \leq \cc$. Da (f.) + sappiamo che $\abs{\PP(\NN) \setminus \Fin(\NN)} = \cc$. + Sia $A$ un sottinsieme infinito di $\NN$. Definiamo + per ricorsione numerabile la funzione $f_A : \NN \to A$ + tale per cui: + \[ + f_A(n) = \begin{cases} + \min(A) & \se n = 1, \\ + \min(A \setminus \{f(1), \ldots, f(n-1)\}) & \altrimenti. + \end{cases} + \] + Sappiamo che $f_A$ è bigettiva (vd. \textit{Problema 12}). + Definiamo allora $\sigma_A : \NN \to \NN$ per ricorsione numerabile in modo tale che: + \[ + \sigma_A(n) = \begin{cases} + n & \se n \in \NN \setminus A, \\ + f_A(n-1) & \se f_A\inv(n) \text{ è pari}, \\ + f_A(n+1) & \se f_A\inv(n) \text{ è dispari}. + \end{cases} + \] + In altre parole $\sigma_A$ ``modifica'' l'identità + $\id_{\NN}$ trasponendo a due a due i numeri consecutivi + della lista $( f_A(1), f_A(2), \ldots )$. \smallskip + + Definiamo $F : \PP(\NN) \setminus \Fin(\NN) \to \Sym(\NN)$ + in modo tale che $F(A) = \sigma_A$. Mostriamo che + $F$ è iniettiva. Se $\sigma_A = \sigma_B$, allora + $\sigma_A$ e $\sigma_B$ hanno lo stesso insieme di + punti fissi, ovverosia, poiché $f_A$ è bigettiva, + $\NN \setminus A = \NN \setminus B$, da cui + $A = B$. Pertanto $\cc = \abs{\PP(\NN) \setminus \Fin(\NN)} \leq \abs{\Sym(\NN)}$. \smallskip + + Infine, per il teorema di Cantor-Bernstein si conclude che + $\abs{\Sym(\NN)} = \cc$. + + \item Si osserva che $\cc = \abs{2^\NN} \leq \abs{\NN^\NN} \leq \abs{\RR^\NN} = \abs{(2^\NN)^\NN} = \abs{2^{\NN \times \NN}} = \abs{2^\NN} = \cc$. Allora, per il teorema di + Cantor-Bernstein, $\abs{\Fun(\NN, \RR)} = \cc$. + + \item Poiché $\Fun\!\uparrow\!(\NN, \RR) \subseteq \Fun(\NN, \RR)$, sicuramente $\abs{\Fun\!\uparrow\!(\NN, \RR)} \leq \cc$. \smallskip + + Sia $\iota$ l'immersione di $\NN$ in $\RR$. Poiché $\iota$ + mantiene l'ordinamento, la funzione $F : \Fun\!\uparrow\!(\NN, \NN) \to \Fun\!\uparrow\!(\NN, \RR)$ tale per cui + $F(f) = f \circ \iota$ è ben definita e iniettiva. + Dunque $\cc = \abs{\Fun\!\uparrow\!(\NN, \NN)} \leq \abs{\Fun\!\uparrow\!(\NN, \RR)}$. \smallskip + + Si conclude dunque che, per il teorema di Cantor-Bernstein, + $\abs{\Fun\!\uparrow\!(\NN, \RR)} = \cc$. + + \item Poiché i sottinsiemi infiniti di $\NN$ sono + sono numerabili, $[\NN]^{\aleph_0} = \PP(\NN) \setminus \Fin(\NN)$. Dacché $\abs{\PP(\NN)} = \cc$ + e $\abs{\Fin(\NN)} = \aleph_0$ (vd. \textit{Problema 14}), allora si sta togliendo + un insieme al più numerabile ad un insieme + che ha la cardinalità del continuo. Pertanto, + per il \textit{Problema 19}, $\abs{[\NN]^{\aleph_0}} = \cc$. + + \item Sia $F : \RR \to [\RR]^{\aleph_0}$ definita + in modo tale che $F(x) = \{x-1/n \mid n \in \NN\}$. + Chiaramente $F$ è iniettiva, da cui + $\cc \leq \abs{[\RR]^{\aleph_0}}$. \smallskip + + Sia $A \in [\RR]^{\aleph_0}$. Poiché $A$ è numerabile, + esiste una funzione iniettiva $f_A : \NN \to \RR$ + con $\imm f_A = A$. In particolare $B_A = \{ f \in \Fun(\NN, \RR) \mid \imm f = A \}$ \underline{non} è vuoto. Pertanto, + applicando l'assioma di scelta, $\bigtimes_{A \in [\RR]^{\aleph_0}} B_A$ è \underline{non} vuoto. + Sia $F$ dunque una funzione in $\bigtimes_{A \in [\RR]^{\aleph_0}} B_A$. $F$ è iniettiva, + infatti $F(A) = F(A') \implies \imm F(A) = \imm F(A')$, + da cui $A = A'$. Pertanto $\abs{[\RR]^{\aleph_0}} \leq \abs{\RR^\NN}$, e per il punto (h.) $\abs{\RR^\NN} = \cc$. + Pertanto $\abs{[\RR]^{\aleph_0}} \leq \cc$. \smallskip + + Applicando il teorema di Cantor-Bernstein si conclude che + $\abs{[\RR]^{\aleph_0}} = \cc$. + + \item Poiché $\cc = \abs{\{0\} \times \RR} \leq \abs{\NN \times \RR} \leq \abs{\RR \times \RR} = \cc$, per il teorema di Cantor-Bernstein vale che + $\abs{\NN \times \RR} = \cc$. Dacché $\abs{2^\RR} \leq \abs{\RR^\RR} = \abs{(2^\NN)^\RR} = \abs{2^{\NN \times \RR}} = \abs{2^\RR}$, allora, ancora per il teorema di Cantor-Bernstein, + $\abs{\RR^\RR} = \abs{2^\RR} > \cc$. + + \item Dal momento che $\QQ$ è denso in $\RR$, una funzione + continua è completamente determinata dalla sua restrizione su $\QQ$. + Pertanto la mappa $F : C^0(\RR) \to \Fun(\QQ, \RR)$ tale per cui + $F(f) = \restr{f}{\QQ}$ è iniettiva. Dunque $\abs{C^0(\RR)} \leq \abs{\RR^\QQ} = \abs{\RR^\NN}$. Per (h.) $\abs{\Fun(\NN, \RR)} = \cc$, + dunque $\abs{C^0(\RR)} \leq \cc$. Allo stesso tempo la mappa + $G : \RR \to C^0(\RR)$ tale per cui $G(k) = [x \mapsto x+k]$ è + iniettiva, e dunque $\cc \leq \abs{C^0(\RR)}$. Per il teorema + di Cantor-Bernstein si deduce allora che $\abs{C^0(\RR)} = \cc$. + + \item Dal\footnote{ + Il procedimento presentato si può facilmente generalizzare + per dimostrare che $\abs{\OO(\RR^n)} = \cc$ per ogni $n \in \NN$. + } momento che la topologia euclidea su $\RR^2$ è più fine + di quella cofinita, per ogni $(x, y) \in \RR^2$ vale + $\RR^2 \setminus \{(x, y)\} \in \OO(\RR^2)$. Dunque + la mappa $F : \RR \times \RR \to \OO(\RR^2)$ tale per cui + $(x, y) \mapsto \RR^2 \setminus \{(x, y)\}$ è ben definita. Si osserva che + $F$ è chiaramente iniettiva, dunque $\cc = \abs{\RR \times \RR} \leq \abs{\OO(\RR^2)}$. \smallskip + + Dacché $\QQ^2$ è denso nello spazio metrico $\RR^2$, $\basis = \{ z \in \PP(\RR^2) \mid +\exists q_1 \exists q_2 \exists r (q_1 \in \QQ \land q_2 \in \QQ \land r \in \QQ^+ \land z = B_r(q_1, q_2)) \}$ è una base + di $\RR^2$. Dunque per ogni $A \in \OO(\RR^2)$ l'insieme + $B_A = \{ z \in \QQ \times \QQ \times \QQ^+ \mid \exists q_1 \exists q_2 \exists r (q_1 \in \QQ \land q_2 \in \QQ \land r \in \QQ^+ \land z = (q_1, q_2, r) \land B_r(q_1, q_2) \in \PP(A) \}$ è \underline{non} vuoto. \smallskip + + Sia $f : \OO(\RR^2) \to \PP(\QQ \times \QQ \times \QQ^+)$ + tale per cui $f(A) = B_A$. Allora $f$ è chiaramente + iniettiva, e dunque $\abs{\OO(\RR^2)} \leq \abs{\PP(\QQ \times \QQ \times \QQ^+)} = \abs{\PP(\NN)} = \cc$. \smallskip + + Infine, per il teorema di Cantor-Bernstein, si conclude che + $\abs{\OO(\RR^2)} = \cc$. + \end{enumerate} +\end{solution} + +\begin{problem}{Una relazione numerabile ha dominio e immagine al più numerabile (senza fare uso di \AC)}{problem-18} + Sia $\rel$ una relazione con $\abs{\rel} = \aleph_0$. Si dimostri allora che + $\abs{\dom(\rel)} \leq \aleph_0$ e che $\abs{\imm(\rel)} \leq \aleph_0$, senza + impiegare l'assioma di scelta. +\end{problem} + +\begin{solution} + Sia $f : \rel \to \NN$ una bigezione tra + $\rel$ e $\NN$. Sia $x \in \dom(\rel)$. Definiamo allora + $A_x = \{ z \in \rel \mid \exists y (y \in \imm \rel \land z = (x, y))\}$, + che è un insieme per l'assioma di separabilità. + Poiché $x \in \dom(\rel)$, $A_x$ \underline{non} è vuoto. Pertanto, + per il principio del buon ordinamento, $f(A_x)$ ammette un minimo. \smallskip + + Sia $F : \dom(\rel) \to \NN$ tale per cui $F(x) = \min (f(A_x))$. Mostriamo + che $F$ è iniettiva. $F(x) = F(y) \implies \min (f(A_x)) = \min (f(A_y))$. + Siano $m$, $n \in \imm \rel$ con $(x, m) \in \rel$ e $(y, n) \in \rel$ tali per cui + $\min (f(A_x)) = f((x, m))$ e $\min (f(A_y)) = f((y, n))$; allora + $f((x, m)) = f((y, n))$. Dacché $f$ è una bigezione, $(x, m) = (y, n)$. + Pertanto, per la proprietà caratterizzante delle coppie ordinate, + $x = y$. Dunque $F$ è iniettiva, da cui $\abs{\dom(\rel)} \leq \abs{\NN} = \aleph_0$. \smallskip + + Analogamente, se $y \in \imm(\rel)$, si può definire tramite l'assioma + di separabilità l'insieme $B_y = \{ z \in \rel \mid \exists x (x \in \dom \rel \land z = (x, y))\}$. In questo modo possiamo costruire + $G : \imm(\rel) \to \NN$ tale per cui $G(y) = \min (f(B_y))$, che + si mostra allo stesso modo essere iniettiva, da cui $\abs{\imm(\rel)} \leq \abs{\NN} = \aleph_0$. +\end{solution} + +\begin{problem}{Se $A$ è al più numerabile e $B$ ha almeno la cardinalità del continuo, $\abs{B \setminus A} = \abs{B}$}{problem-19} + Si assuma già di sapere che, dato $C \subseteq \RR \times \RR$ con $\abs{C} \leq \aleph_0$, + vale che $\abs{(\RR \times \RR) \setminus C} = \abs{\RR \times \RR} = \cc$. Si deduca che, dati $B$ con $\abs{B} \geq \cc$ + e $A$ con $\abs{A} \leq \aleph_0$, allora $\abs{B \setminus A} = \abs{B}$. +\end{problem} + +\begin{solution} + Possiamo assumere senza perdita di generalità che $A$ sia sottinsieme + di $B$. Infatti $B \setminus A = B \setminus (A \cap B)$ e + $\abs{A \cap B} \leq \abs{A} \leq \aleph_0$, dunque le ipotesi sono ancora + vere scegliendo $A \cap B$ al posto di $A$, dando lo stesso risultato. \smallskip + + Poiché $\abs{B} = \cc$, esiste una funzione bigettiva $f : B \to \RR \times \RR$. + In particolare $\abs{f(A)} = \abs{A} \leq \aleph_0$ e $f(A) \subseteq f(B) = \RR \times \RR$. Allora, per ipotesi vale $\abs{f(B) \setminus f(A)} = \abs{f(B)}$, + da cui $\abs{B} =\abs{f(B)} = \abs{f(B) \setminus f(A)} = \abs{B \setminus A}$. \smallskip + + Poiché $\abs{B} \geq \cc$, esiste una funzione iniettiva $f : \RR \times \RR \to B$. + In particolare $\abs{f\inv(A)} \leq \abs{A} = \aleph_0$ e $f\inv(A) \subseteq f\inv(B) = \RR \times \RR$. Allora, per ipotesi vale $\abs{f\inv(B) \setminus f\inv(A)} = \abs{f\inv(B)}$, da cui $\abs{B} \geq \abs{f\inv(B)} = \abs{f\inv(B) \setminus f\inv(A)} $ +\end{solution} + +\begin{problem}{Cardinalità di $A \setminus B$ e $A \cup B$ se $\abs{B} < \abs{A}$ e $A$ è infinito (\AC)}{problem-20} + Sia $A$ un insieme infinito e sia $B$ tale per cui $\abs{B} < \abs{A}$. Si dimostri che + $\abs{A \setminus B} = \abs{A \cup B} = \abs{A}$, assumendo l'assioma di scelta. +\end{problem} + +\begin{solution} + +\end{solution} + +\begin{problem}{Proprietà dei numeri naturali (1) -- Definizione equivalente dell'ordinamento stretto}{problem-21} + Siano $n$, $m \in \omega$. Si dimostri che $n \in m \iff n \subsetneq m$. +\end{problem} + +\begin{solution} + Mostriamo per induzione che $\varphi(m)$ è vera per ogni $m \in \omega$, dove: + \[ + \varphi(m) = \forall n (n \in m \implies n \subsetneq m). + \] + \begin{enumerate} + \item[$\boxed{\varphi(0)}$] Dal momento che $0 = \emptyset$ non + ammette elementi, + $\Psi(0)$ è vacuamente vera. + + \item[$\boxed{\varphi(m+1)}$] Se $n \in m+1$, allora + $n \in m$ o $n = m$. Se $n \in m$, allora, + per l'ipotesi induttiva, vale che $n \subsetneq m$. + Dal momento che $m+1 = m \cup \{m\}$, allora + $n \subsetneq m+1$. Se invece $n = m$, per + definizione di $m+1$ vale $n \subsetneq m+1$. Pertanto + $\varphi(m+1)$ vale se vale $\varphi(m)$. + \end{enumerate} + Pertanto, applicando il principio di induzione, se $m$ è un elemento di $\omega$, $n \in m \implies n \subsetneq m$. Mostriamo infine per induzione che + $\Psi(m)$ è vera per ogni $m \in \omega$, dove: + \[ + \Psi(m) = \forall n (n \in \omega \implies (n \in m \iff n \subsetneq m)). + \] + \begin{enumerate} + \item[$\boxed{\Psi(0)}$] Dal momento che $0 = \emptyset$ non + ammette elementi né sottinsiemi propri, + $\Psi(0)$ è vacuamente vera. + + \item[$\boxed{\Psi(m+1)}$] Una direzione è già verificata + dal momento che vale $\varphi(m+1)$. Sia pertanto + $n \subsetneq m+1 = m \cup \{m\}$ con $n \in \omega$. Se valesse + $m \in n$, allora -- poiché vale $\varphi(n)$ (qui è cruciale che $n$ sia un numero naturale!) -- + $m \subsetneq n$, da cui $m+1 = m \cup \{m\} \subseteq n$; + questo è tuttavia assurdo, dacché implicherebbe + $n = m+1$, \Lightning. Pertanto $m \notin n$, da cui + $n \subseteq m$. Se $n = m$, allora $n = m \in m+1$ per + costruzione; altrimenti $n \subsetneq m$, da cui, + per l'ipotesi induttiva, si ricava che + $n \in m$, e dunque che $n \in m+1$. Pertanto + $\Psi(m+1)$ vale se vale $\Psi(m)$, + \end{enumerate} + Infine si conclude che $\Psi(m)$ è vera per ogni $m \in \omega$ applicando + il principio d'induzione, ovverosia è vera la tesi. +\end{solution} + +\begin{problem}{Proprietà dei numeri naturali (2) -- Massimo e minimo tra $n$ e $m$}{problem-22} + Siano $n$, $m \in \omega$. Si dimostri che $\min \{n, m\} = n \cap m$ e che $\max \{n, m\} = n \cup m$. +\end{problem} + +\begin{solution} + Se $n \leq m$, allora $n \subseteq m$ per il \textit{Problema 21}. Dunque $\min \{n, m\} = n = n \cap m$ e $\max \{n, m\} = m = n \cup m$. Analogamente se $m \leq n$, allora $m \subseteq n$, da cui $\min \{n, m\} = m = n \cap m$ e $\max \{n, m\} = n = n \cup m$. +\end{solution} + +\begin{problem}{Proprietà dei numeri naturali (3) -- $\omega$ è un insieme transitivo}{problem-23} + Si mostri che $\omega$ è transitivo, ovvero si dimostri che gli elementi degli elementi di $\omega$ sono elementi di $\omega$. +\end{problem} + +\begin{solution} + Mostriamo per induzione che $\Psi(m)$ + è vera per ogni $m \in \omega$, dove: + \[ + \Psi(m) = \forall n(n \in m \implies n \in \omega). + \] + \begin{enumerate} + \item[$\boxed{\Psi(0)}$] Dal momento che $0 = \emptyset$ non + ammette elementi, + $\Psi(0)$ è vacuamente vera. + + \item[$\boxed{\Psi(m+1)}$] Sia $n \in m + 1$. Allora + $n = m$ o $n \in m$. Nel primo caso, $n$ appartiene + a $\omega$ (infatti $m \in \omega$); nel secondo caso, + per ipotesi induttiva, $n \in \omega$. Dunque $\Psi(m+1)$ + vale se $\Psi(m)$. + \end{enumerate} + Pertanto, applicando il principio di induzione si ricava + che $\Psi(m)$ è vera per ogni $m \in \omega$, ovverosia + $\omega$ è un insieme transitivo. +\end{solution} + +\begin{problem}{Proprietà dei numeri naturali (4) -- $\hat{x} \in \omega \implies x \in \omega$}{problem-24} + Si mostri che se $\hat{x} := x \cup \{x\}$ appartiene a + $\omega$, allora $x \in \omega$. +\end{problem} + +\begin{solution} + Sappiamo grazie al \textit{Problema 23} che $\omega$ è un insieme + transitivo. Dal momento che $x \in \hat x$ e + $\hat x \in \omega$, allora $x \in \omega$, essendo + elemento di un elemento di $\omega$. +\end{solution} + +\begin{problem}{Infinito $\implies$ Dedekind-infinito (\AC)}{problem-25} + Sia $A$ un insieme infinito. Assumendo l'assioma di scelta, + si dimostri che $A$ è Dedekind-infinito, ovverosia si mostri + che esiste $B \subsetneq A$ tale per cui $\abs{B} = \abs{A}$. +\end{problem} + +\begin{solution} + Poiché vale \AC{} e $A$ è infinito, esiste una funzione iniettiva + $f : \omega \to A$. Dacché $\abs{\omega} = \abs{\omega \setminus \{0\}}$ tramite + la bigezione indotta dal successore $S : n \mapsto n+1$, allora + $\abs{f(\omega)} = \abs{f(\omega \setminus \{0\})}$. Detto allora + $B = f(\omega \setminus \{0\})$, chiaramente $A \neq B$ dal momento + che $f$ è iniettiva. Dunque $B$ è l'insieme desiderato. +\end{solution} + +\begin{problem}{Caratterizzazioni delle bigezioni su insiemi finiti}{problem-26} + Sia $A$ un insieme finito e sia $f : A \to A$ una funzione. Si mostri che + sono equivalenti le seguenti affermazioni: + \begin{enumerate}[(i.)] + \item $f$ è bigettiva, + \item $f$ è iniettiva, + \item $f$ è surgettiva. + \end{enumerate} +\end{problem} + +\begin{solution} + Mostriamo il ciclo di implicazioni (i.) $\implies$ (ii.) $\implies$ (iii.) $\implies$ (i.), in modo tale da dimostrare tutte le equivalenze. + + \begin{itemize} + \item[\fbox{(i.) $\implies$ (ii.)}] Se $f$ è bigettiva, allora $f$ è + per definizione anche iniettiva. + + \item[\fbox{(ii.) $\implies$ (iii.)}] Se $f : A \to A$ è iniettiva, + Se $f$ \underline{non} fosse surgettiva, allora $f(A)$ sarebbe un sottinsieme proprio + di $A$, e dunque si avrebbe $\abs{f(A)} < \abs{A}$, violando il principio + dei cassetti. Dunque $f$ è surgettiva. + + \item[\fbox{(iii.) $\implies$ (i.)}] Poiché + $A$ è finito\footnote{ + Se $A$ è \underline{finito}, $f$ è ammette sempre un'inversa destra, anche senza + utilizzare \AC. + }, $f$ ammette un'inversa destra $g : A \to A$, che è iniettiva. Allora, + poiché (ii.) $\implies$ (iii.), $g$ è surgettiva, e dunque $g$ è bigettiva. + Pertanto $f$ è anch'essa bigettiva essendo l'inversa di $g$. + \end{itemize} +\end{solution} + +\begin{problem}{$A$, $B$ finiti $\implies$ $A \cap B$, $A \cup B$, $A \times B$, $\PP(A)$ finiti}{problem-27} + Siano $A$ e $B$ due insiemi finiti. Si mostri che i seguenti insiemi sono + finiti: + \begin{tasks}[label=(\alph*.),label-width=17.21605pt](4) + \task $A \cap B$, + \task $A \cup B$, + \task $A \times B$, + \task $\PP(A)$. + \end{tasks} +\end{problem} + +\begin{solution} + Mostriamo le varie richieste ordinatamente. + \begin{enumerate}[(a.)] + \item Poiché $A \cap B \subseteq A$ e $A$ è finito, allora + $A \cap B$ è sottinsieme di un insieme finito e dunque è + finito. + + \item Mostriamo per induzione che vale $\Psi(m)$ per ogni $m \in \omega$, dove: + \[ + \Psi(m) = \forall n \forall A \forall B ((\abs{A} = m \land \abs{B} = n) \implies \exists k (k \in \omega \land \abs{A \cup B} = k)). + \] + \begin{enumerate} + \item[$\boxed{\Psi(0)}$] Se $\abs{A} = 0$, allora $A$ necessariamente + è l'insieme vuoto. Dunque $\abs{A \cup B} = \abs{B} = m$. Pertanto + $\Psi(0)$ è vera. + + \item[$\boxed{\Psi(m+1)}$] Sappiamo che $\abs{A} = m+1$. Sia $f : A \to m+1$ + una bigezione. Se poniamo $A' = f\inv(m+1 \setminus \{m\}$, allora + $\abs{A'} = \abs{m}$. Pertanto, per l'ipotesi induttiva esiste $k \in \omega$ + tale per cui $\abs{A' \cup B} = k$; in particolare esiste una bigezione + $g : A' \cup B \to k$. Se $f\inv(\{m\}) \in B$, $A' \cup B = A \cup B$ e + dunque $\abs{A \cup B} = k$. Altrimenti, possiamo estendere $g$ ad + $h : A \cup B \to k+1$ ponendo $\restr{h}{A' \cup B} = g$ e + $h(f\inv(\{m\})) = k$. Dal momento che $h$ è l'incollamento di due bigezioni, + la cui seconda va da $\{f\inv(\{m\})\}$ in $\{k\}$, a dominio e immagini + disgiunte, $h$ è una bigezione, e dunque $\abs{A \cup B} = k+1$. Pertanto + $\Psi(m+1)$ vale se vale $\Psi(m)$. + \end{enumerate} + Si conclude dunque per il principio di induzione che $A \cup B$ è sempre + finito se $A$ e $B$ lo sono. + + \item Se $\abs{B} = n$, allora esiste una bigezione $f : n \to B$. + Costruiamo allora $\sigma : n \to A \times B$ in modo tale che + $\sigma(i) = A \times \{f(i)\}$. In particolare, dacché + $\abs{A} = \abs{A \times \{f(i)\}}$ per ogni $i \in n$, $\sigma(i)$ è + sempre finito. Osserviamo che $A \times B = \bigcup \imm \sigma = \bigcup_{i \in n} A \times \{f(i)\}$. Allora $A \times B$ è unione finita di insiemi finiti, + e dunque è finita per il punto (b.)\footnote{ + In realtà il punto (b.) dimostra solo che l'unione di due insiemi finiti + è finita, ma si può estendere la tesi a un numero finito di insiemi finiti applicando il principio di induzione. + }. + + \item Dal momento che $\abs{A} = m \implies \abs{\PP(A)} = \abs{\PP(m)}$, + è sufficiente mostrare che $\PP(m)$ è finito. Mostriamo + dunque per induzione che vale $\varphi(m)$ per ogni $m \in \omega$, dove: + \[ + \varphi(m) = \forall A ((\abs{A} = m) \implies \exists j(j \in \omega \land \abs{\PP(A)} = j)). + \] + \begin{enumerate} + \item[$\boxed{\varphi(0)}$] $\abs{A} = 0$ è possibile se e solo se + $A = 0$. Allora $\PP(0) = 1$, che è finito. + + \item[$\boxed{\varphi(m+1)}$] È sufficiente mostrare la tesi per + $m+1$ dacché, per il \textit{Problema 11}, insiemi equipotenti hanno + parti equipotenti. Si osserva che $\PP(m+1) = \PP(m \cup \{m\}) = \bigcup_{i \in m+1} \PP(m \setminus \{i\} \cup \{m\})$. Ogni $m \setminus \{i\} \cup \{m\}$ + ha la cardinalità di $m$, e dunque, per l'ipotesi induttiva, + $\PP(m \setminus \{i\} \cup \{m\})$ è finito. Allora, per (b.)\footnotemark[\value{footnote}], $\PP(m+1)$ è finito. Dunque + $\varphi(m+1)$ vale se vale $\varphi(m)$. + \end{enumerate} + Si conclude dunque per il principio di induzione che $\PP(A)$ è sempre finito se + $A$ è finito. + \end{enumerate} +\end{solution} + +\begin{problem}{La somma e il prodotto su $\omega$ sono ben definiti}{problem-28} + Siano $m$ e $n \in \omega$. Ricordiamo allora che, dati $A$ e $B$ insiemi qualsiasi + tali per cui $\abs{A} = m$, $\abs{B} = n$ e $A \cap B = \emptyset$, si definiscono + somma e prodotto come segue: + \[ + m + n = \abs{A \cup B}, \quad m \cdot n = \abs{m \times n}. + \] + Si mostri che entrambe le definizioni sono ben poste. +\end{problem} + +\begin{solution} + Per il \textit{Problema 27}, sia $A \cup B$ che $m \times n$ sono finiti, + essendo $A$, $B$, $m$ e $n$ finiti. Pertanto il prodotto è già ben definito. + Per il \textit{Problema 11}, scambiando $A$ e $B$ con insiemi equipotenti ancora + disgiunti, si ottiene ancora un insieme equipotente a $A \cup B$, e dunque + anche la somma è ben definita. +\end{solution} + +\begin{problem}{Associatività della somma su $\omega$}{problem-29} + $+$ è associativa, cioè: + \[ + \forall x, y, z, (x+y)+z=x+(y+z). + \] +\end{problem} + +\begin{solution} + +\end{solution} + +\begin{problem}{Distributività della somma rispetto al prodotto su $\omega$}{problem-30} + $+$ è distributiva rispetto a $\cdot$, cioè: + \[ + \forall x, y, z, x(y+z)=x\cdot y+x\cdot z. + \] +\end{problem} + +\begin{solution} + +\end{solution} + +\begin{problem}{La somma e il prodotto di elementi non nulli è non nulla}{problem-31} + Se $x$, $y \neq 0$, allora $x+y \neq 0$ e + $x \cdot y \neq 0$. +\end{problem} + +\begin{solution} + +\end{solution} + +\begin{problem}{Relazione d'ordine totale sui modelli di \PA}{problem-32} + Se $(N, 0, S, +, \cdot)$ è un modello di \PA{}, ossia un + modello degli assiomi + di Peano al primo + ordine, allora + la relazione: + \[ + x < y \iff \exists z (z \neq 0 \land x+y=y) + \] + è tale per cui: + + \begin{enumerate}[(i.)] + \item $<$ è un ordine totale su $N$, + \item $x < y$, + allora $\forall z (z \in N \implies x+z \alpha$. + \item $\alpha + \beta \geq \beta$. + \item $\alpha + \beta + 1 > \beta$. + \item $\alpha \cdot \beta > \alpha$, se anche $\alpha \neq 0$. + \end{enumerate} +\end{problem} + +\begin{problem}{Disuguaglianze strette sugli ordinali con ordinale fisso a sinistro}{problem-52} + Siano $\alpha$ e $\gamma < \gamma'$ ordinali. Allora si mostri che: + + \begin{enumerate}[(i.)] + \item $\alpha + \gamma < \alpha + \gamma'$. + \item $\alpha \cdot \gamma < \alpha \cdot \gamma'$, se anche $\alpha \neq 0$. + \end{enumerate} +\end{problem} + +\begin{problem}{Distributività a destra degli ordinali}{problem-53} + Siano $\alpha$, $\beta$ e $\gamma$ ordinali. Si mostri allora che: + \[ \alpha \cdot (\beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma. \] +\end{problem} + +\begin{problem}{Forme normali di Cantor di $(\omega + 3) \cdot n$, $(\omega + 3)^2$, $(\omega + 3)^n$ e $(\omega + 3)^{\omega + 3}$}{problem-54} + Si calcolino le forme normali di Cantor di: + + \begin{tasks}[label=(\roman*.), label-width=19.4064pt](4) + \task $(\omega + 3) \cdot n$, + \task $(\omega + 3)^2$, + \task $(\omega + 3)^n$, + \task $(\omega + 3)^{\omega + 3}$. + \end{tasks} +\end{problem} + +\begin{problem}{Equivalenza tra la ricorsione transfinita per casi e la ricorsione transfinita con una sola dichiarazione}{problem-55} + Si mostri che sono equivalenti la ricorsione transfinita per casi e quella che impiega invece una sola funzione classe. +\end{problem} + +\begin{problem}{Assorbimento a sinistra per ordinali finiti di $\omega$}{problem-56} + Si mostri che $n + \omega = \omega$ per ogni $n \in \omega$. +\end{problem} + +\begin{problem}{Caratterizzazione degli ordinali che rispettano \underline{sulla somma} la proprietà di assorbimento a sinistra per ordinali più piccoli}{problem-57} + Sia $\alpha \neq 0$ un ordinale. Si mostri che sono equivalenti: + + \begin{enumerate}[(i.)] + \item $\beta + \gamma < \alpha$ se $\beta$, $\gamma < \alpha$ ($\alpha$ è additivamente chiuso). + \item $\beta + \alpha = \alpha$ se $\beta < \alpha$ ($\alpha$ rispetta \underline{sulla somma} la proprietà di assorbimento a sinistra per ordinali più piccoli). + \item $\exists \, \delta$ ordinale per cui $\alpha = \omega^\delta$. + \end{enumerate} +\end{problem} + +\begin{problem}{Caratterizzazione degli ordinali che rispettano \underline{sul prodotto} la proprietà di assorbimento a sinistra per ordinali più piccoli}{problem-58} + Sia $\alpha \neq 0$ un ordinale. Si mostri che sono equivalenti: + + \begin{enumerate}[(i.)] + \item $\beta \cdot \gamma < \alpha$ se $\beta$, $\gamma < \alpha$ ($\alpha$ è moltiplicativamente chiuso). + \item $\beta \cdot \alpha = \alpha$ se $\beta < \alpha$ ($\alpha$ rispetta \underline{sul prodotto} la proprietà di assorbimento a sinistra per ordinali più piccoli). + \item $\exists \, \delta$ ordinale per cui $\alpha = \omega^{(\omega ^ \delta)}$. + \end{enumerate} +\end{problem} + +\begin{problem}{Moltiplicare a destra per $\omega$ trasforma l'ordinale nella più piccola potenza di $\omega$ che lo maggiora}{problem-59} + Sia $\alpha$ un ordinale per cui $\omega^\delta \leq \alpha < \omega^{\delta+1}$. Allora $\alpha \cdot \omega = \omega^{\delta + 1}$. +\end{problem} + +\begin{problem}{$\alpha \cdot \omega = \beta \cdot \omega$ se e solo se $\alpha$ e $\beta$ sono contenuti tra due stesse potenze successive di $\omega$}{problem-60} + Siano $\alpha$, $\beta$ ordinali. Se $\alpha \cdot \omega = \beta \cdot \omega$, allora esiste $\delta$ ordinale per cui + $\omega^\delta \leq \alpha$, $\beta < \omega^{\delta + 1}$. +\end{problem} + +\begin{problem}{Condizioni sull'assorbimento della somma}{problem-61} + Siano $\alpha$, $\beta \neq 0$. Allora vale una e una sola delle seguenti affermazioni: + + \begin{enumerate}[(i.)] + \item $\alpha + \beta = \beta$. + \item $\beta + \alpha = \alpha$. + \item $\exists \, \delta$ con $\omega^\delta \leq \alpha$, $\beta < \omega^{\delta + 1}$. + \end{enumerate} +\end{problem} + +\begin{problem}{Proprietà fondamentali dell'algebra cardinale}{problem-62} + Siano $\kappa$, $\mu$, $\ni$ cardinali. Allora: + + \begin{enumerate}[(i.)] + \item $(\kappa + \mu) + \nu = \kappa + (\mu + \nu)$. + \item $(\kappa \cdot \mu) \cdot \nu = \kappa \cdot (\mu \cdot \nu)$. + \item $\kappa + \mu = \mu + \kappa$. + \item $\kappa \cdot \mu = \mu \cdot \kappa$. + \item $\kappa \cdot (\mu + \nu) = \kappa \cdot \mu + \kappa \cdot \nu$. + \item $\kappa^\mu \cdot \kappa^\nu = \kappa^{\mu + \nu}$. + \item $(\kappa^\mu)^\nu = \kappa^{\mu \cdot \nu}$. + \end{enumerate} +\end{problem} + +\begin{problem}{$\aleph_\alpha \leq \beth_\alpha$ per ogni ordinale $\alpha$}{problem-63} + Sia $\alpha$ un ordinale. Si mostri che $\aleph_\alpha \leq \beth_\alpha$. +\end{problem} + +\begin{problem}{Proprietà fondamentali della somma infinita di cardinali}{problem-64} + Sia $I$ un insieme. Si mostri allora che: + + \begin{enumerate}[(i.)] + \item $\sum_{i \in I} 1 = \abs{I}$. + \item $\sum_{i \in I} \kappa = \kappa \cdot \abs{I}$. + \item $\sum_{i \in I} \kappa_i = \sum_{i \in I} \kappa_{\sigma(i)}$ per ogni permutazione $\sigma \in \Sym(I)$. + \item Se $\kappa_i \leq \mu_i$ per ogni $i \in I$, allora $\sum_{i \in I} \kappa_i \leq \sum_{i \in I} \mu_i$. + \end{enumerate} +\end{problem} + +\begin{problem}{Proprietà fondamentali del prodotto infinito di cardinali}{problem-65} + Sia $I$ un insieme. Si mostri allora che: + + \begin{enumerate}[(i.)] + \item $\prod_{i \in I} \kappa_i = \prod_{i \in I} \kappa_{\sigma(i)}$ per ogni permutazione $\sigma \in \Sym(I)$. + \item $\prod_{i \in I} \kappa^{\mu_i} = \kappa^{\sum_{i \in I} \mu_i}$, e quindi $\prod_{i \in I} \kappa = \kappa^{\abs{I}}$. + \item $\left( \prod_{i \in I} \kappa_i \right)^\mu = \prod_{i \in I} \kappa_i^\mu$. + \item Se $I = \bigsqcup_{j \in J} I_j$, allora $\prod_{i \in I} \kappa_i = \prod_{j \in J} \prod_{i \in I_j} \kappa_i$ (associatività generalizzata). + \end{enumerate} +\end{problem} + +\begin{problem}{Le cofinalità sono regolari: $\cof(\cof(A)) = \cof(A)$}{problem-66} + Sia $(A, <)$ un insieme totalmente ordinato. Si mostri allora che: + + \[ \cof(\cof(A)) = \cof(A). \] +\end{problem} + +\begin{problem}{$\cof(\alpha + \beta) = \cof(\beta)$}{problem-67} + Siano $\alpha$ e $\beta$ ordinali. Allora $\cof(\alpha + \beta) = \cof(\beta)$. +\end{problem} + +\begin{problem}{Ogni cardinalità regolare è la cofinalità di un ordinale arbitrariamente grande}{problem-68} + Sia $\mu$ un cardinale regolare. Si mostri allora che per ogni cardinale $\kappa$ esiste $\nu \geq \kappa$ (come ordinali) + tale per cui $\cof(\nu) = \mu$. +\end{problem} + +\begin{problem}{Per $\lambda$ ordinale limite, $\cof(\aleph_\lambda) = \cof(\beth_\lambda) = \cof(\lambda)$}{problem-69} + Sia $\lambda$ un ordinale limite, si mostri allora che $\cof(\aleph_\lambda) = \cof(\beth_\lambda) = \cof(\lambda)$. +\end{problem} + +\begin{problem}{$V_\alpha$ è chiuso per sottinsiemi, unioni e chiusure transitive}{problem-70} + Sia $V_\alpha$ il livello $\alpha$-esimo della gerarchia di von Neumann. Si mostri che: + + \begin{enumerate}[(i.)] + \item Se $x \subseteq y \in V_\alpha$, allora $x \in V_\alpha$. + \item Se $A \in V_\alpha$, allora $\bigcup A \in V_\alpha$. + \item Se $A \in V_\alpha$, allora $\TC(A) \in V_\alpha$, dove $\TC(A)$ è la chiusura transitiva di $A$. + \end{enumerate} +\end{problem} + +\begin{problem}{$V_* = \bigcup_{\alpha \in \ORD} V_\alpha$ è una classe propria}{problem-70} + Sia $V_* := \bigcup_{\alpha \in \ORD} V_\alpha$. Si mostri che $V_*$ è una classe propria. +\end{problem} + +\begin{problem}{Caratterizzazione della validità degli assiomi di coppia, delle parti, dell'infinito + e della scelta per i livelli della gerarchia di von Neumann} + Sia $\alpha$ un ordinale. Si mostri che: + + \begin{enumerate}[(i.)] + \item Vale l'assioma della coppia in $V_\alpha$ se e solo se $\alpha$ è limite. + \item Vale l'assioma delle parti in $V_\alpha$ se e solo se $\alpha$ è limite. + \item Vale l'assioma dell'infinito in $V_\alpha$ se e solo se $\alpha > \omega$. + \item Vale l'assioma della scelta in $V_\alpha$ se e solo se $\alpha$ è limite. + \end{enumerate} +\end{problem} + +\end{document} diff --git a/Terzo anno/Elementi di Teoria degli Insiemi/preamble.tex b/Terzo anno/Elementi di Teoria degli Insiemi/preamble.tex new file mode 100644 index 0000000..1a3a713 --- /dev/null +++ b/Terzo anno/Elementi di Teoria degli Insiemi/preamble.tex @@ -0,0 +1,61 @@ +% \usepackage{fontspec} + +% ================================================== + +\usepackage[italian]{babel} +\usepackage[margin = 1in]{geometry} + +% ================================================== + +\usepackage{mathrsfs} +\usepackage{amsfonts} +\usepackage{amsmath} +\usepackage{amsthm} +\usepackage{amssymb} +\usepackage{physics} +\usepackage{dsfont} +\usepackage{esint} +\usepackage{marvosym} +\usepackage{kpfonts} +\usepackage{quiver} + +% ================================================== + +\usepackage{enumerate} +\usepackage[shortlabels, inline]{enumitem} +\usepackage{framed} +\usepackage{csquotes} + +% ================================================== + +\usepackage{float} +\usepackage{tabularx} +\usepackage{xcolor} +\usepackage{multicol} +\usepackage{subcaption} +\usepackage{caption} +\captionsetup{format = hang, margin = 10pt, font = small, labelfont = bf} + +\usepackage[round, authoryear]{natbib} + +\usepackage{hyperref} +\definecolor{links}{rgb}{0.36,0.54,0.66} +\hypersetup{ + colorlinks = true, + linkcolor = black, + urlcolor = blue, + citecolor = blue, + filecolor = blue, + pdfauthor = {Author}, + pdftitle = {Title}, + pdfsubject = {subject}, + pdfkeywords = {one, two}, + pdfproducer = {LaTeX}, + pdfcreator = {pdfLaTeX}, + } + +\usepackage{lmodern} +\renewcommand{\familydefault}{\sfdefault} + +\usepackage{sfmath} +\usepackage{hyperref} diff --git a/Terzo anno/Elementi di Teoria degli Insiemi/quiver.sty b/Terzo anno/Elementi di Teoria degli Insiemi/quiver.sty new file mode 100644 index 0000000..7806030 --- /dev/null +++ b/Terzo anno/Elementi di Teoria degli Insiemi/quiver.sty @@ -0,0 +1,40 @@ +% *** quiver *** +% A package for drawing commutative diagrams exported from https://q.uiver.app. +% +% This package is currently a wrapper around the `tikz-cd` package, importing necessary TikZ +% libraries, and defining a new TikZ style for curves of a fixed height. +% +% Version: 1.5.2 +% Authors: +% - varkor (https://github.com/varkor) +% - AndréC (https://tex.stackexchange.com/users/138900/andr%C3%A9c) + +\NeedsTeXFormat{LaTeX2e} +\ProvidesPackage{quiver}[2021/01/11 quiver] + +% `tikz-cd` is necessary to draw commutative diagrams. +\RequirePackage{tikz-cd} +% `amssymb` is necessary for `\lrcorner` and `\ulcorner`. +\RequirePackage{amssymb} +% `calc` is necessary to draw curved arrows. +\usetikzlibrary{calc} +% `pathmorphing` is necessary to draw squiggly arrows. +\usetikzlibrary{decorations.pathmorphing} + +% A TikZ style for curved arrows of a fixed height, due to AndréC. +\tikzset{curve/.style={settings={#1},to path={(\tikztostart) + .. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$) + and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$) + .. (\tikztotarget)\tikztonodes}}, + settings/.code={\tikzset{quiver/.cd,#1} + \def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}}, + quiver/.cd,pos/.initial=0.35,height/.initial=0} + +% TikZ arrowhead/tail styles. +\tikzset{tail reversed/.code={\pgfsetarrowsstart{tikzcd to}}} +\tikzset{2tail/.code={\pgfsetarrowsstart{Implies[reversed]}}} +\tikzset{2tail reversed/.code={\pgfsetarrowsstart{Implies}}} +% TikZ arrow styles. +\tikzset{no body/.style={/tikz/dash pattern=on 0 off 1mm}} + +\endinput \ No newline at end of file