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\author{Tom Cuchta}
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\begin{frame}{}
13 April
\end{frame}

\begin{frame}{Who shaves the barber?}
Last class: imagine a town with precisely one barber. The barber obeys the \\
\textbf{Logical Barber Property:} ``the barber shaves \textit{all} the people, and only the people, who do not shave themselves." \\
\vspace{1cm}
Who shaves the barber? \\
\vspace{1cm}
\underline{Case 1: the barber does not shave himself} \\
This is impossible: in this case, the \textbf{logical barber property} implies that the barber must shave himself (contrary to ``case 1"). \\
\underline{Case 2: the barber shaves himself} \\
This is impossible: in this case, the barber shaves himself, but the \textbf{logical barber property} says that the barber shaves \textit{all} people who do not shave themselves, so he must shave himself (contrary to ``case 2"). 
\end{frame}

\begin{frame}{A ``naive" set theory}
1.) \textbf{Axiom of Extensionality} \\
$(\forall z)(z \in x \leftrightarrow z \in y) \rightarrow x=y$ \\
2.) \textbf{Axiom Schema of (Unrestricted) Comprehension} \\
For any (one-place) predicate $P$, we take the following as an axiom: \\
$(\exists z)(\forall x)(x \in z \leftrightarrow Px)$ \\
\hrule
\vspace{1cm}
\end{frame}

\begin{frame}{A ``naive" set theory}
Consider the predicate $P$ where $Px$ stands in for $\neg (x \in x)$. Consider the following proof (called ``Russell's paradox") \\
\begin{tabular}{|lll|}
\hline
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (1) $(\exists z)(\forall x)(x \in z \leftrightarrow Px)$ & Axiom 2 \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (2) $(\forall x)(x \in \alpha \leftrightarrow Px)$ & 1 ES \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (3) $\alpha \in \alpha \leftrightarrow \neg(\alpha \in \alpha)$ & 2 US \& Def of $P$ \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (4) $\alpha \in \alpha \rightarrow \neg (\alpha \in \alpha)$ & 3 Bicond, Simpl \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (5) $\neg(\alpha \in \alpha) \rightarrow \alpha \in \alpha$ & 3 Bicond, Comm, Simpl \\
$\{4\}$ & (6) $\alpha \in \alpha$ & Premise (for contradiction) \\
$\{6, \mathrm{Ax \hspace{2pt}} 2\}$ & (7) $\neg(\alpha \in \alpha)$ & 4 6 Detachment \\
$\{6, \mathrm{Ax \hspace{2pt}} 2\}$ & (8) $(\alpha \in \alpha) \wedge \neg (\alpha \in \alpha)$ &  6 7 Adjunction \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (9) $\neg (\alpha \in \alpha)$ & 6 8 RAA \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (10) $\alpha \in \alpha$ & 5 9 Detachment \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (11) $\neg (\alpha \in \alpha) \wedge (\alpha \in \alpha)$ & 9 10 Adjunction \\
\hline
\end{tabular} \\
This shows that Axiom 2 implies a contradiction... therefore we must reject it! This is what makes naive set theory ``naive". 

\end{frame}

\begin{frame}{Zermelo set theory}
Naive set theory is broken: one way to fix it is to ``build up" sets from scratch and only allow sets to contain other sets which were created ``a lower level" of construction. \\
\vspace{0.5cm} 
Axiom 1 is the same as in naive set theory. Axioms 2, 4, and 5 give us ways to make new sets. Axiom 3 is a ``restricted" form of naive set theory's Axiom 2. Axiom 7 explicitly forbids Russel's paradox. \\
\begin{enumerate}
\item two sets have same member if and only if they are the same set
\item we may always combine two sets into a new set
\item if $P$ is a predicate and $A$ is an already existing set, then the set of members $x$ of $A$, for which $Px$ is true, form a set
\item for any set, the ``union" of all elements of the set forms a set
\item for any set $x$, there is a set $y$ consisting of all subsets of $x$
\item there is an infinite set
\item sets cannot contain themselves
\end{enumerate}

\vspace{1cm}

\end{frame}

\begin{frame}{Generic set theory notation}
We use ``braces" ``$\{$" and ``$\}$" to denote sets. The empty set is precisely $\emptyset = \{ \}$. If a set contains things, it is common to write those thing between the braces; for example the set containing $a$ and $b$ is written $\{a,b\}$. \\
\vspace{1cm}
Formally, to write $y=\{a,b\}$ means $(\exists y)(\forall x)(x \in y \leftrightarrow (x=a \vee x=b))$. \\
\vspace{1cm}
Recall the subset notation: $a \subseteq b$ abbreviates $(\forall z)(z \in x \rightarrow z \in y)$. Other abbreviations: $x \cup y=\{z \colon z \in x \wedge z \in y\}$ and $x \cap y = \{z \colon z \in x \wedge z \in y\}$
\end{frame}

\begin{frame}{Zermelo set theory}
\begin{enumerate}
\setcounter{enumi}{-1}
\item \textbf{``Empty set"} $(\exists x)(x=\emptyset)$ \\
\item \textbf{``Extensionality"}: $(\forall x)(\forall y)((\forall z)(z \in x \leftrightarrow z \in y) \rightarrow x=y)$ 
\item \textbf{``Pairing"}: $(\forall w)(\forall z)(\exists y)(\forall x)(x \in y \leftrightarrow (x=w \vee x=z))$ 
\item \textbf{``Restricted Comprehension"}: For any predicate $P$ and already existing set $A$, \\
$(\exists z)(\forall x)(x \in z \leftrightarrow (x \in A \wedge Px))$ 
\item \textbf{``Union"}: $(\forall z)(\exists a)(\forall y)(\forall x)(x \in y \wedge y \in z \rightarrow x \in a)$
\item \textbf{``Power set"}: $(\forall x)(\exists y)(\forall z)(z \in y \leftrightarrow z \subseteq x)$
\item \textbf{``Infinity"}: $(\exists x)(\emptyset \in x \wedge (\forall y)(y \in x \rightarrow \{ y \cup \{y\}\} \in x))$
\item \textbf{``Foundation"}: $(\forall s)(\neg(s=\emptyset) \vee \rightarrow (\exists x)(x \in s \wedge s \cap x = \emptyset))$
\end{enumerate}
\end{frame}

\begin{frame}{A tautological equivalence to make life easier}
Let's call the following tautological equivalence ``self or". \\
\vspace{2cm}
\begin{tabular}{|l|l|l|l|}
\hline
$P$ & $P \vee P$ & $P \vee P \leftrightarrow P$ \\
\hline
$T$ & $T$ & $T$ \\
$F$ & $F$ & $T$ \\
\hline
\end{tabular}
\end{frame}

\begin{frame}{Zermelo set theory}
Prove that the set $\{\emptyset\}$ exists, i.e. prove $(\exists y)(\forall x)(x \in y \leftrightarrow x=\emptyset)$. \\
\hspace*{-0.3in}
\begin{tabular}{|lll|}
\hline
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (1) $(\forall z)(\exists y)(\forall x)(x \in y \leftrightarrow (x=\emptyset \vee x=z))$ & Axiom 2 US \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (2) $(\exists y)(\forall x)(x \in y \leftrightarrow (x=\emptyset \vee x=\emptyset))$ & 1 US \\
$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (3) $(\exists y)(\forall x)(x \in y \leftrightarrow x=\emptyset)$ & 2 TE (self or)\\
\hline
%$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (3) $\emptyset \in \alpha \leftrightarrow (\emptyset = \emptyset \vee \emptyset =\emptyset)$ & 2 US ES US \\
%$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (4) $(\emptyset = \emptyset \vee \emptyset=\emptyset) \rightarrow \emptyset \in \alpha$ & 3 Bic, Com, Sim\\
%$\{ \}$ & (5) $\emptyset=\emptyset$ & Identity law \\
%$\{ \}$ & (6) $\emptyset=\emptyset \vee \emptyset=\emptyset$ & 5 Addition \\
%$\{\mathrm{Ax \hspace{2pt}} 2\}$ & (7) $\emptyset \in \alpha$ & 
\end{tabular} \\
note: we could also express this set as $\{ \{\} \}$. 
\end{frame}

\begin{frame}{Ordered pairs}
An \textit{ordered pair} consisting of the sets $x$ (first) and $y$ (second) is described by the notation $(x,y)=\{x,\{x,y\}\}$. More formally, to say that $\{x,y\}$ exists means
$$(\exists z)(\forall a)(a \in z \leftrightarrow a=x \vee a=y).$$
And to say that $\{x,\{x,y\}\}$ exists means
$$(\exists z)(\forall a)(a \in z \leftrightarrow a=x \vee a=\{x,y\})$$
For example, 
$$(\emptyset,\emptyset) \stackrel{\mathrm{def}}{=} \{\emptyset,\{\emptyset, \emptyset\}\} \stackrel{\mathrm{Ax.1}}{=} \{\emptyset,\{\emptyset\}\}$$
\vspace{0.5cm}
A \textit{relation} is a set of ordered pairs; i.e. a relation is a set $r$ such that
$$(\forall x)(x \in r \rightarrow (\exists a)(\exists b)(x=\{a,\{a,b\}\})$$
\end{frame}


\begin{frame}{Relations and functions}
A \textit{function} is a relation $R$ with the property that if $(a,b) \in R$ and $(a,c) \in R$, then $b=c$. In other words, if $f$ is a function, then $f$ is a relation such that 
$$(\forall a)(\forall b)(\forall c)((a,b) \in f \wedge (a,c) \in f \rightarrow b=c)$$
note: we often abbreviate $(a,b) \in f$ as ``$f(a)=b$" \\
\vspace{0.5cm}
\textbf{Example:} \\
a.) Consider the relation $r=\{(a,b),(a,c),(a,a)\}$. Is it a function? \\
b.) Consider the relation $r=\{(a,b),(b,b),(c,d)\}$. Is it a function? \\
c.) Is $r=\emptyset$ a relation? Is it a function?
\end{frame}

\begin{frame}{Zermelo-Fraenkel set theory}
Zermelo-Fraenkel set theory is Zermelo set theory along with the following axiom: 
\begin{enumerate}
\setcounter{enumi}{7}
\item \textbf{``Replacement"} Let $f$ be a function. Then for any $x$, the following is a set: 
$$y=f[x] \stackrel{\mathrm{def}}{=} \{f(z) \colon z \in x \}.$$
We will not worry about a formal version of replacement here...
\end{enumerate}
\end{frame}

\begin{frame}{Well orderings}
A \textit{well-order} of a set $x$ is a total order (i.e. it obeys the axioms of total order theory) with the additional property that all subsets of $x$ have a smallest element.  \\
\vspace{0.5cm}
\textbf{Examples:} \\
a.) Is $\mathbb{N}=\{0,1,2,\ldots\}$ with the usual ``$\leq$" well ordered? \\
b.) Is the set of real numbers $\mathbb{R}$ with the usual ``$\leq$" well ordered?\\
c.) Is $\emptyset$ well-ordered?
\end{frame}

\begin{frame}{Zermelo-Fraenkel set theory (with choice) aka ZFC}
\textbf{Russell:} To select one sock from each of infinitely many pairs of socks
requires the Axiom of Choice; but for shoes the Axiom is not needed.” \\
\vspace{1cm}
The following axiom is called the axiom of choice:
\begin{enumerate}
\setcounter{enumi}{8}
\item \textbf{``Choice"} For all sets $x$, there is a relation $r$ which is a well ordering of $x$. 
\end{enumerate}
\end{frame}

\begin{frame}{Relationships between the set theories}
Denote the set theory defined by axioms 1-8 ``ZF" and denote the set theory defined by axioms 1-9 ``ZFC". \\
\vspace{1cm}
In ZFC...
\begin{itemize}
\item Banach-Tarski paradox
\item every ``connected graph" has a ``spanning tree"
\item a ``group" exists on any set
\item ``non-measurable sets" exist
\item every ``Hilbert space" has an ``orthonormal basis"
\end{itemize}

In ZF....
\begin{itemize}
\item the real numbers can be partitioned into strictly more sets than there are real numbers 
\item there is a tree $T$ with no leaves, but which has no infinite path
\item there is a field with no ``algebraic closure"...moreover there are more algebraic closures for $\mathbb{Q}$ than $\mathbb{C}$
\item there is a vector spaces without a basis
\item 
\end{itemize}
\end{frame}

\begin{frame}{Some interesting reads}
\begin{itemize}
\item {\color{blue}\href{https://www.math.upenn.edu/~ted/203S10/References/peculiar.pdf}{A Peculiar Connection Between the Axiom of Choice and Predicting the Future}}
\item {\color{blue} \href{https://xorshammer.com/2008/08/23/set-theory-and-weather-prediction/}{Set theory and weather prediction}} 
\end{itemize}
\end{frame}
\end{document}
