handouts/Advanced/Definable Sets/parts/3 sets.tex

\section{Definable Sets}

Armed with $(), \land, \lor, \lnot, \rightarrow, \forall,$ and $\exists$, we have the tools to define sets.

\definition{Set-Builder Notation}
Say we have a sentence $\varphi(x)$. \par
The set of all elements that satisfy that sentence may be written as follows:
\begin{equation*}
	\{ x ~|~ \varphi(x) \}
\end{equation*}
This is read \say{The set of $x$ where $\varphi$ is true} or \say{The set of $x$ that satisfy $\varphi$.}

\vspace{2mm}

For example, take the formula $\varphi(x) = \exists y ~ (y + y = x)$. \par
The set of all even integers can then be written as
$$
	\{ x ~|~ \exists y ~ (y + y = x) \}
$$

\definition{Definable Sets}
Let $S$ be a structure with a universe $U$. \par
We say a subset $M$ of $U$ is \textit{definable} if we can write a formula \par
that is true for some $x$ if and only if $M$ contains $x$.

\vspace{4mm}

For example, consider the structure $\bigl( \mathbb{Z} ~\big|~ \{+\} \bigr)$. \par
Only even numbers satisfy the formula $\varphi(x) \coloneqq \bigl[\exists y ~ (y + y = x)\bigr]$, \par
so we can define \say{the set of even numbers} as $\{ x ~|~ \exists y ~ (y + y = x) \}$. \par
Remember---we can only use symbols that are available in our structure!

\problem{}
The empty set is definable in any structure. How?
\begin{solution}
	Always: $\{ x ~|~ \lnot (x = x) \}$
\end{solution}

\vfill


\problem{}
Define $\{0, 1\}$ in $\Bigl( \mathbb{Z}^+_0 ~\big|~ \{<\} \Bigr)$
\hint{Define 0 and 1 as elements first, and remember that we can use logical symbols.}

\begin{solution}
	$\varphi_0(x) \coloneqq \bigl[~ \lnot \exists y ~ y < x ~\bigr]$ \par
	$\varphi_1(x) \coloneqq \bigl[~ (0 < x) ~\land~ \lnot \exists y ~ (x < y < 0) ~\bigr]$

	\vspace{2mm}

	Our final solution is $\{ x ~|~ \varphi_0(x) \lor \varphi_1(x) \}$.

	\note{A finite set of definable elements is always definable. \par
	An infinite set of definable elements might not be definable.}
\end{solution}


\vfill

\problem{}
Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\} \Bigr)$. \par
\hint{A prime number is an integer that is positive and is only divisible by 1 and itself.}

\begin{solution}
	$\psi(x) \coloneqq \bigl[~ \exists y ~ (0<y<x) ~\bigr]$ \tab \note{\say{$x$ is positive and isn't 0 or 1}} \par
	$\varphi(x) \coloneqq \bigl[~ (x<0) \land \lnot \exists ab ~ (\psi(a) \land \psi(b) \land a \times b = x) \bigr]$
\end{solution}


\vfill
\pagebreak










\problem{}
Define $\mathbb{R}^+_0$ in $\Bigl( \mathbb{R} ~\big|~ \{\times\} \Bigr)$ \par

\begin{solution}
	$\varphi(x) \coloneqq \bigl[ \exists y ~ y \times y = x \bigr]$
\end{solution}

\vfill

\problem{}
Let $\bigtriangleup$ be a relational symbol. $a \bigtriangleup b$ is only true if $a$ divides $b$. \par
Define the set of prime numbers in $\Bigl( \mathbb{Z}^+ ~\big|~ \{ \bigtriangleup \} \Bigr)$ \par

\begin{solution}
	$
		\varphi(x) \coloneqq
		\bigl[~ \lnot \exists abc ~ \bigl(
			(a \bigtriangleup x) \land
			(b \bigtriangleup x) \land
			(c \bigtriangleup x) \land
			\lnot (a = b) \land
			\lnot (a = c) \land
			\lnot (b = c)
		\bigr) ~\bigr]
	$
\end{solution}

\vfill



\theorem{Lagrange's Four Square Theorem}
Every natural number may be written as a sum of four integer squares.

\problem{}
Define $\mathbb{Z}^+_0$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$

\begin{solution}
	$\varphi(x) \coloneqq \bigl[~ \exists abcd ~ (a^2 + b^2 + c^2 + d^2 = x) ~\bigr]$,
	where $a^2 \coloneqq a \times a$.
\end{solution}

\vfill




\problem{}
Define $<$ in $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$ \par
\hint{We can't formally define a relation yet. Don't worry about that for now. \\
You can repharase this question as \say{given $x,y \in \mathbb{Z}$, write a formula $\varphi(x, y)$ that is only true if $x < y$}}

\begin{solution}
	Let $\psi(x)$ be the formula from the previous problem.

	\vspace{2mm}

	$\varphi(x, y) \coloneqq \bigl[~ \lnot (x=y) \land \exists d ~ \bigl(\psi(d) \land (x + d = y)\bigr) ~\bigr]$
\end{solution}


\vfill
\pagebreak










\problem{}
Consider the structure $S = ( \mathbb{R} ~|~ \{0, \diamond \} )$ \par
The relation $a \diamond b$ holds if $| a - b | = 1$

\problempart{}
Define $\{-1, 1\}$ in $S$.

\begin{solution}
	$\varphi(x) \coloneqq \bigl[ 0 \diamond x \bigr]$
\end{solution}

\problempart{}
Define $\{-2, 2\}$ in $S$.

\begin{solution}
	$\varphi(x) \coloneqq \bigl[~ \forall a ~ (0 \diamond x \rightarrow a \diamond x) \land \lnot (x = 0) ~\bigr]$
\end{solution}


\vfill

\problem{}
Let $\mathcal{P}$ be the set of all subsets of $\mathbb{Z}^+_0$. This is called the \textit{power set} of $\mathbb{Z}^+_0$. \par
Let $S$ be the structure $( \mathcal{P} ~|~ \{\subseteq\})$ \par

\problempart{}
Show that the empty set is definable in $S$. \par
\hint{Defining $\{\}$ with $\{x ~|~ \lnot x = x\}$ is \textbf{not} what we need here. \\
We need $\varnothing \in \mathcal{P}$, the \say{empty set} element in the power set of $\mathbb{Z}^+_0$.}

\begin{solution}
	$\varphi(x) \coloneqq \bigl[~ \forall y ~ x \subseteq y ~\bigr]$ \par
	Note that we can use the same property to define 0 in $( \mathbb{Z} ~|~ \{\leq\})$
\end{solution}

\vfill

\problempart{}
Let $x \Bumpeq y$ be a relation on $\mathcal{P}$. $x \Bumpeq y$ holds if $x \cap y \neq \{\}$. \par
Show that $\Bumpeq$ is definable in $S$.

\begin{solution}
	Let $\psi(x)$ be the formula from the previous problem.

	\vspace{2mm}

	$\varphi(x, y) \coloneqq \bigl[~ \exists x ~ (a \subseteq x) \land (a \subseteq y) \land \lnot \psi(a) ~\bigr]$
\end{solution}

\vfill

\problempart{}
Let $f$ be the function on $\mathcal{P}$ defined by $f(x) = \mathbb{Z}^+_0 - x$. This is called the \textit{complement} of $x$. \par
Show that $f$ is definable in $S$. \par
\hint{You can define a function by writing a formula $\varphi(x, y)$ that is only true when $y = f(x)$.}

\vfill
\pagebreak