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Co-authored-by: Mark <mark@betalupi.com>
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src/Advanced/Definable Sets/main.tex Executable file
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
% for \coloneqq, a centered :=
\usepackage{mathtools}
\usepackage{units}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{Definable Sets}
\subtitle{Prepared by Mark on \today{}}
\begin{document}
\maketitle
\input{parts/0 logic.tex}
\input{parts/1 structures.tex}
\input{parts/2 quantifiers.tex}
\input{parts/3 sets.tex}
\input{parts/4 equivalence.tex}
\end{document}

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src/Advanced/Definable Sets/meta.toml Normal file
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[metadata]
title = "Definable Sets"
[publish]
handout = true
solutions = true

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src/Advanced/Definable Sets/parts/0 logic.tex Normal file
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\section{Logical Algebra}
\definition{}
\textit{Logical operators} operate on the values $\{\texttt{true}, \texttt{false}\}$, \par
just like algebraic operators operate on numbers. \par
In this handout, we'll use the following operators:
\begin{itemize}
\item $\lnot$: not
\item $\land$: and
\item $\lor$: or
\item $\rightarrow$: implies
\item $()$: parenthesis.
\end{itemize}
The function of these is defined by \textit{truth tables}:
\begin{center}
\begin{tabular}{ c | c | c }
\multicolumn{3}{ c }{and} \\
\hline
$A$ & $B$ & $A \land B$ \\
\hline
\texttt{F} & \texttt{F} & \texttt{F} \\
\texttt{F} & \texttt{T} & \texttt{F} \\
\texttt{T} & \texttt{F} & \texttt{F} \\
\texttt{T}& \texttt{T} & \texttt{T}
\end{tabular}
\hfill
\begin{tabular}{ c | c | c }
\multicolumn{3}{ c }{or} \\
\hline
$A$ & $B$ & $A \lor B$ \\
\hline
\texttt{F} & \texttt{F} & \texttt{F} \\
\texttt{F} & \texttt{T} & \texttt{T} \\
\texttt{T} & \texttt{F} & \texttt{T} \\
\texttt{T} & \texttt{T} & \texttt{T}
\end{tabular}
\hfill
\begin{tabular}{ c | c | c }
\multicolumn{3}{ c }{implies} \\
\hline
$A$ & $B$ & $A \rightarrow B$ \\
\hline
\texttt{F} & \texttt{F} & \texttt{T} \\
\texttt{F} & \texttt{T} & \texttt{T} \\
\texttt{T} & \texttt{F} & \texttt{F} \\
\texttt{T} & \texttt{T} & \texttt{T}
\end{tabular}
\hfill
\begin{tabular}{ c | c }
\multicolumn{2}{ c }{not} \\
\hline
$A$ & $\lnot A$ \\
\hline
\texttt{T} & \texttt{F} \\
\texttt{F} & \texttt{T} \\
~ & ~ \\
~ & ~ \\
\end{tabular}
\end{center}
\vspace{2mm}
$A \land B$ is \texttt{true} only if both $A$ and $B$ are \texttt{true}. $A \lor B$ is \texttt{true} if $A$ or $B$ (or both) are \texttt{true}. \par
$\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} sign. \par
\vspace{2mm}
$A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par
The only time $\rightarrow$ produces \texttt{false} is when $\texttt{true} \rightarrow \texttt{false}$.
This fact may seem counterintuitive, but will make more sense as we progress through this handout. \par
\hint{
Think about it---if event $\alpha$ implies $\beta$, it is impossible for $\alpha$ to occur without $\beta$. \par
This is the only impossibility. All other variants are valid.
}
\problem{}
Evaluate the following.
\begin{itemize}
\item $\lnot \texttt{T}$
\item $\texttt{F} \lor \texttt{T}$
\item $\texttt{T} \land \texttt{T}$
\item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
\item $(\lnot (\texttt{F} \lor \lnot \texttt{T}) ) \rightarrow \lnot \texttt{T}$
\item $(\texttt{F} \rightarrow \texttt{T}) \rightarrow (\lnot \texttt{F} \lor \lnot \texttt{T})$
\end{itemize}
\begin{solution}
\texttt{F}
\texttt{T}
\texttt{T}
\texttt{T}
\texttt{F}
\texttt{T}
\end{solution}
\vfill
\pagebreak
\begin{instructornote}
We can also think of $[x \geq 0] \rightarrow b$ as follows:
if $x$ isn't the kind of object we care about, we evaluate true and
check the next one. If $x$ \textit{is} the kind of object we care about
and $b$ is false, we have a counterexample to $[x \geq 0] \rightarrow b$,
and thus $\texttt{T} \rightarrow \texttt{F}$ must be false.
\vspace{2mm}
Say we have the sentence $\forall x ~ (a \rightarrow b)$. \par
For example, take $\varphi = \forall x ~ ([x \geq 0] \rightarrow [\exists y ~ y^2 = x])$. \par
$\varphi$ holds whenever any positive $x$ has a square root.
\vspace{2mm}
If $(\text{F} \rightarrow *)$ returned false, statements like the above would be hard to write. \par
If $x$ is negative, $\varphi$ doesn't care whether or not it has a root. In this case, $\text{F} \rightarrow *$ must be true to avoid making whole $\forall$ false.
\end{instructornote}
\problem{}
Evaluate the following.
\begin{itemize}
\item $A \rightarrow \texttt{T}$ for any $A$
\item $(\lnot (A \rightarrow B)) \rightarrow A$ for any $A, B$
\item $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ for any $A, B$
\end{itemize}
\begin{instructornote}
Note that the last formula is the contrapositive of $A \rightarrow B$.
\end{instructornote}
\begin{solution}
All are true.
\end{solution}
\vfill
% Show that A -> B ^ B -> A = T iff A = B
\problem{}
Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
That is, show that these expressions always evaluate to the same value given
the same $A$ and $B$. \par
\hint{Use a truth table}
\vfill
\problem{}
Write an expression equivalent to $A \lor B$ using only $\lnot$, $\rightarrow$, and $()$?
\begin{solution}
$((\lnot A) \rightarrow B)$
\end{solution}
\vfill
Note that both $\land$ and $\lor$ can be defined using the other logical symbols. \par
The only logical symbols we \textit{need} are $\lnot$, $\rightarrow$, and $()$. \par
We include $\land$ and $\lor$ to simplify our expressions.
\pagebreak

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\section{Structures}
\definition{}
A \textit{universe} is a set of meaningless objects. Here are a few examples:
\begin{itemize}
\item $\{a, b, ..., z\}$
\item $\{0, 1\}$
\item $\mathbb{Z}$, $\mathbb{R}$, etc.
\end{itemize}
\definition{}
A \textit{structure} consists of a universe and a set of \textit{symbols}. \par
A structure's symbols give meaning to the objects in its universe.
\vspace{2mm}
Symbols come in three types:
\begin{itemize}
\item \textit{Constant symbols}, which let us specify specific elements of our universe. \par
Examples: $0, 1, \frac{1}{2}, \pi$
\vspace{2mm}
\item \textit{Function symbols}, which let us navigate between elements of our universe. \par
Examples: $+, \times, \sin{x}, \sqrt{x}$ \par
\note{Note that symbols we usually call \say{operators} are functions under this definition. \par
The only difference between $a + b$ and $+(a, b)$ is notation.}
\vspace{2mm}
\item \textit{Relation symbols}, which let us compare elements of our universe. \par
Examples: $<, >, \leq, \geq$ \par
\vspace{2mm}
\end{itemize}
The equality check $=$ is \textit{not} a relation symbol. It is included in every structure by default. \par
By definition, $a = b$ is true if and only if $a$ and $b$ are the same element of our universe.
\vspace{3mm}
\example{}
The first structure we'll look at is the following:
$$
\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)
$$
\vspace{2mm}
This is a structure over the universe $\mathbb{Z}$ that provides the following symbols:
\begin{itemize}
\item Constants: \tab $\{0, 1\}$
\item Functions: \tab $\{+, -\}$
\item Relations: \tab $\{<\}$
\end{itemize}
\vspace{2mm}
If we look at our set of constant symbols, we see that the only integers
we can directly refer to in this structure are 0 and 1. If we want any
others, we must define them using the tools this structure offers.
\vspace{2mm}
% NOTE: this is a great example for typesetting.
% The line breaks here are ugly without a centered sentence.
To \say{define} an element of a set, we need to write a sentence that is only true for that element. \par
If we want to define 2 in the structure above,
we could use the following sentence:
\begin{center}
\say{$2$ is the $x$ that satisfies $[1 + 1 = x]$.} \par
\end{center}
This is a valid definition because $2$ is the \textit{only} element of $\mathbb{Z}$ for which $[1 + 1 = x]$
evaluates to \texttt{true}.
\problem{}
Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$.
\begin{solution}
The sentences \say{$x$ where $[x + 1 = 0]$} and \say{$x$ where $[0 - 1 = x]$} both work.
\end{solution}
\vfill
\pagebreak
Let us formalize what we found in the previous two problems. \par
\definition{Formulas}
A \textit{formula} in a structure $S$ is a well-formed string
of constants, functions, relations, \par and logical operators.
\vspace{2mm}
You already know what a \say{well-formed string} is: $1 + 1$ is fine, $\sqrt{+}$ is nonsense. \par
For the sake of time, I will not provide a formal definition --- it isn't particularly interesting.
\vspace{2mm}
As a quick example, the formula $\psi \coloneqq [\lnot (1 = 1)]$ is always false, \par
and $\varphi(x) \coloneqq [1 + 1 = x]$ evaluates to \texttt{true} only when $x$ is 2.
\definition{Free Variables}
A formula can contain one or more \textit{free variables.} These are denoted $\varphi{(a, b, ...)}$. \par
Formulas with free variables let us define \say{properties} that certain objects have.
\vspace{2mm}
For example, consider the two formulas from the previous definition, $\psi$ and $\varphi$:
\begin{itemize}
\item $\psi \coloneqq [\lnot (1 = 1)]$ \par
There are no free variables in this formula. \par
In any structure, $\psi$ is always either \texttt{true} or \texttt{false}.
\vspace{2mm}
\item $\varphi(x) \coloneqq [1 + 1 = x]$ \par
This formula has one free variable, labeled $x$. \par
The value of $\varphi(x)$ depends on the $x$ we're talking about: \par
$\varphi(72)$ is false, and $\varphi(2)$ is true.
\end{itemize}
\vspace{2mm}
\note{
This \say{free variable} notation is very similar to the function notation we are used to: \par
The values of both $\varphi(x) \coloneqq [x > 0]$ and $f(x) = x + 1$ depend on $x$.
}
\definition{Definable Elements}
Let $S$ be a structure over a universe $U$. \par
We say an element $x \in U$ is \textit{definable in $S$} if we can write a formula $\varphi(x)$ that only $x$ satisfies.
\problem{}
Define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bigr)$. \par
\hint{$\mathbb{Z}^+ = \{1, 2, 3, ...\}$. Also, $2 \times 2 = 4$.}
\begin{solution}
$2$ is the only element in $\mathbb{Z}^+$ that satisfies $\varphi(x) \coloneqq [x \times x = 4]$.
\end{solution}
\vfill
\pagebreak
\problem{}
Try to define 2 in the structure $\Bigl( \mathbb{Z} ~\big|~ \{4, \times \} \Bigr)$. \par
Why can't you do it?
\begin{solution}
We could try $\varphi(x) \coloneqq [x \times x = 4]$, but this is satisfied by both $2$ and $-2$. \par
We have no way to distinguish between negative and positive numbers. \par
\note{This problem is intentionally hand-wavy. We don't have the tools to write a proper proof.}
\begin{instructornote}
Actually, it is. Bonus problem: how? \par
Do this after understanding quantifiers.
\end{instructornote}
\end{solution}
\vfill
\problem{}
Consider the structure $\Bigl( \mathbb{R}^+_0 ~\big|~ \{1, 2, \div \} \Bigr)$
\begin{itemize}
\item Define $2^2$
\item Define $2^n$ for all positive integers $n$
\item Define $2^{-n}$ for all positive integers $n$
\item What other numbers can we define in this structure? \par
\hint{There is at least one more \say{class} of numbers we can define.}
\end{itemize}
\begin{solution}
As far as I've seen, we can define any $2^{\nicefrac{a}{b}}$ for $a, b \in \mathbb{Z}$. \par
For example, $\phi(x) \coloneqq [2 = x \div (1 \div x)]$ defines $\sqrt{2}$.
\end{solution}
\vfill
\pagebreak

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\section{Quantifiers}
Recall the logical symbols we introduced earlier: $(), \land, \lor, \lnot, \rightarrow$ \par
We will now add two more: $\forall$ (for all) and $\exists$ (exists).
\definition{}
$\forall$ and $\exists$ are \textit{quantifiers}. They allow us to make statements about arbitrary symbols. \par
\note{Quantifiers are aptly named: they tell us \textit{how many} symbols satisfy a certain sentence.}
\vspace{2mm}
Let's look at $\forall$ first. If $\varphi(x)$ is a formula, \par
the formula $\forall x ~ \varphi(x)$ is true only if $\varphi$ is true for all $x$ in our universe.
\vspace{1mm}
For example, take the formula $\forall x ~ (0 < x)$. \par
In English, this means \say{For any $x$, $x$ is bigger than zero,} or simply \say{Any $x$ is positive.}
\vspace{3mm}
$\exists$ is very similar: the formula $\exists x ~ \varphi(x)$ is true if there is at least one $x$ for which $\varphi(x)$ is true. \par
For example, $\exists ~ (0 < x)$ means \say{there is a positive number in our set.}
\vspace{4mm}
\problem{}
Which of the following are true in $\mathbb{Z}$? Which are true in $\mathbb{R}^+_0$? \par
\note{$\mathbb{R}^+_0$ is the set of positive real numbers and zero.}
\begin{itemize}[itemsep = 1mm]
\item $\forall x ~ (x \geq 0)$
\item $\lnot (\exists x ~ (x = 0))$
\item $\forall x ~ [\exists y ~ (y \times y = x)]$
\item $\forall xy ~ \exists z ~ (x < z < y)$ \tab
\note{This is a compact way to write $\forall x ~ (\forall y ~ (\exists z ~ (x < z < y)))$}
\item $\lnot \exists x ~ ( \forall y ~ (x < y) )$
\end{itemize}
\begin{solution}
\begin{itemize}
\item \say{all $x$ are positive} \tab $\mathbb{R}^+_0$
\item \say{zero doesn't exist} \tab neither
\item \say{square roots exist} \tab $\mathbb{R}^+_0$
\item \say{this set is dense} \tab\null\tab $\mathbb{R}^+_0$
\item \say{there is no minimum} \tab $\mathbb{Z}$
\end{itemize}
\end{solution}
%\begin{examplesolution}
% Here is a solution to the last part: $\lnot \exists x ~ ( \forall y ~ (x < y) )$ \par
%
% \vspace{4mm}
%
% Reading this term-by-term, we get \tab \say{not exists $x$ where (for all $y$ ($x$ smaller than $y$))} \par
% If we add some grammar, we get \tab \say{There isn't an $x$ where all $y$ are bigger than $x$} \par
% which we can rephrase as \tab~\tab \say{There isn't a minimum value} \par
%
% \vspace{4mm}
%
% Which is true in $\mathbb{Z}$ and false in $\mathbb{R}^+_0$
%\end{examplesolution}
\vfill
\pagebreak
\problem{}
Does the order of $\forall$ and $\exists$ in a formula matter? \par
What's the difference between $\exists x ~ \forall y ~ (x \leq y)$ and $\forall y ~ \exists x ~ (x \leq y)$? \par
\hint{
Consider $\mathbb{R}^+$\hspace{-1.3ex},\hspace{0.8ex} the set of positive reals. Zero is not positive. \par
Which of the above formulas is true in $\mathbb{R}^+$\hspace{-1.3ex},\hspace{0.8ex} and which is false?
}
\begin{solution}
If $\exists x$ is inside $\forall y$, $x$ depends on $y$. We may pick a different value of $x$ for every $y$. \par
If $\exists x$ is outside, $x$ is fixed \textit{before} we check all $y$.
\end{solution}
\vfill
\problem{}
Define 0 in $\Bigl( \mathbb{Z} ~\big|~ \{\times\} \Bigr)$
\begin{solution}
$\varphi(x) \coloneqq \bigl[~ \forall y ~ x \times y = x ~\bigr]$
\end{solution}
\vfill
\problem{}
Define 1 in $\Bigl( \mathbb{Z} ~\big|~ \{\times\} \Bigr)$
\begin{solution}
$\varphi(x) \coloneqq \bigl[~ \forall y ~ x \times y = y ~\bigr]$
\end{solution}
\vfill
\pagebreak
\problem{}
Define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, <\} \Bigr)$
\begin{solution}
$\varphi(x) \coloneqq \bigl[~ (x<0) \land \lnot \exists y ~ (x < y < 0) ~\bigr]$
\end{solution}
\vfill
%\problem{}
%Define $2$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, <\} \Bigr)$
%\vfill
\problem{}
Let $\varphi(x)$ be a formula. \par
Write a formula equivalent to $\forall x ~ \varphi(x)$ using only logical symbols and $\exists$.
\begin{solution}
$\forall x ~ \varphi(x)$ is true if and only if $\lnot \exists x ~ \lnot \varphi(x)$ is true.
\end{solution}
\vfill
\pagebreak

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\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

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\section{Equivalence}
\generic{Notation:}
Let $S$ be a structure and $\varphi$ a formula. \par
If $\varphi$ is true in $S$, we write $S \models \varphi$. \par
This is read \say{$S$ satisfies $\varphi$}
\definition{}
Let $S$ and $T$ be structures. \par
We say $S$ and $T$ are \textit{equivalent} (and write $S \equiv T$) if for any formula $\varphi$, $S \models \varphi \Longleftrightarrow T \models \varphi$. \par
If $S$ and $T$ are not equivalent, we write $S \not\equiv T$.
\problem{}
Show that $
\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{N} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{N} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{Z}^2 ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{Z}^2 ~\big|~ \{ +, 0 \}\Bigr)
$
\begin{solution}
All of the above are easy, but the last one can take a while. \par
The trick is to notice that $\mathbb{Z}$ has two equivalence classes mod 2, while $\mathbb{Z}^2$ has four.
\end{solution}
\vfill