\section{Definable Sets}

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

\definition{Set-Builder Notation}
Say we have a condition $c$. \par
The set of all elements that satisfy that condition can be written as follows:
$$
	\{ x ~|~ \text{$c$ is true} \}
$$
This is read \say{The set of $x$ where $c$ is true} or \say{The set of $x$ that satisfy $c$.}

\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
$$
	\{ x ~|~ \varphi(x) \} = \{ x ~|~ \exists y ~ (y + y = x) \}
$$

\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 that is true for some $x$ iff $x \in M$.

\vspace{4mm}

For example, consider the structure $\Bigl( \mathbb{Z} ~\big|~ \{+\} \Bigr)$ \par

\vspace{2mm}

Only even numbers satisfy the formula $\varphi(x) = \exists y ~ (y + y = x)$, \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{}
Is the empty set definable in any structure?

\vfill


\problem{}
Define $\{0, 1\}$ in $\Bigl( \mathbb{Z}^+_0 ~\big|~ \{<\} \Bigr)$

\begin{instructornote}
	Here's an interesting fact:

	A finite set of definable elements is always definable. Why? \par
	An infinite set of definable elements may not be definable.
\end{instructornote}

\vfill

\problem{}
Define the set of prime numbers in $\Bigl( \mathbb{Z} ~\big|~ \{\times, \div, <\} \Bigr)$
\vfill

\pagebreak


\problem{}
Define the set of nonreal numbers in $\Bigl( \mathbb{C} ~\big|~ \{\text{real}(z)\} \Bigr)$ \par

\hint{$\text{real}(z)$ gives the real part of a complex number: $\text{real}(3 + 2i) = 3$}
\hint{$z$ is nonreal if $x \in \mathbb{C}$ and $x \notin \mathbb{R}$}

\begin{solution}
	$\Bigl\{ x ~\bigl|~ \text{real}(x) \neq x \Bigr\}$
\end{solution}


\vfill


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

\vfill

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

\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 $\Bigl( \mathbb{Z} ~\big|~ \{\times, +\} \Bigr)$

\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 $a,b \in \mathbb{Z}$, can you write a sentence that is true iff $a < b$?}}

\vfill
\pagebreak

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

\problempart{}
Define 0 in $S$.

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

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

\vfill

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

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

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

\problempart{}
Let $f$ be a function on $P$ defined by $f(x) = \mathbb{Z}^+_0 - x$. This is called the \textit{complement} of the set $x$. \par
Show that $f$ is definable in $S$.

\vfill
\pagebreak