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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../resources/ormc_handout}
% Typewriter tabs
\usepackage{tabto}
\TabPositions{1cm, 2cm, 3cm, 4cm, 5cm}
% for \coloneqq, a centered :=
\usepackage{mathtools}
\begin{document}
\maketitle
<Advanced 2>
<Spring 2023>
{Definable Sets}
{
Prepared by Mark on \today
}
\input{parts/0 logic.tex}
\input{parts/1 structures.tex}
\end{document}

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Advanced/Definable Sets/parts/0 logic.tex Normal file
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\section{Logical Algebra}
\definition{}
Odds are, you are familiar with \textit{logical symbols}. \par
In this handout, we'll use the following:
\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
F & F & F \\
F & T & F \\
T & F & F \\
T & T & T
\end{tabular}
\hfill
\begin{tabular}{ c | c | c }
\multicolumn{3}{ c }{or} \\
\hline
$A$ & $B$ & $A \lor B$ \\
\hline
F & F & F \\
F & T & T \\
T & F & T \\
T & T & T
\end{tabular}
\hfill
\begin{tabular}{ c | c | c }
\multicolumn{3}{ c }{implies} \\
\hline
$A$ & $B$ & $A \rightarrow B$ \\
\hline
F & F & T \\
F & T & T \\
T & F & F \\
T & T & T
\end{tabular}
\hfill
\begin{tabular}{ c | c }
\multicolumn{2}{ c }{not} \\
\hline
$A$ & $\lnot A$ \\
\hline
T & F \\
F & T \\
~ & ~ \\
~ & ~ \\
\end{tabular}
\end{center}
\vspace{2mm}
$A \land B$ is only true if both $A$ and $B$ are true. $A \lor B$ is only true if $A$ or $B$ (or both) are 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 $\rightarrow$ is false is when $T \rightarrow F$. Think about it: why does this make sense? \par
\problem{}
Evaluate the following.
\begin{itemize}
\item $(T \land F) \lor T$
\item $(\lnot (F \lor \lnot T) ) \rightarrow T$
\item $A \rightarrow T$ for any $A$
\item $(\lnot (A \rightarrow B)) \rightarrow A$ for any $A,B$
\end{itemize}
\vfill
\pagebreak
\problem{}
Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
\hint{Use a truth table}
\vfill
\problem{}
Can you express $A \lor B$ 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 logical expressions.
\pagebreak

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\section{Structures}
\definition{}<def:language>
A \textit{language} is a set of meaningless symbols. Here are a few examples:
\begin{itemize}
\item $\{a, b, ..., z\}$
\item $\{0, 1\}$
\item $\mathbb{Z}$, $\mathbb{R}$, etc.
\end{itemize}
Every language comes with the equality check $=$, which checks if two elements are the same.
\definition{}
A \textit{structure} over a language $\mathcal{L}$ consists of three sets:
\begin{itemize}
\item A set of \textit{constant symbols} $\mathcal{C}$ \par
Constant symbols let us specify specific elements of our language. \par
$\mathcal{C}$ must thus be a subset of $\mathcal{L}$.
\vspace{3mm}
\item A set of \textit{function symbols} $\mathcal{F}$ \par
Function symbols let us navigate between elements of our language. \par
$+$, $-$ are functions, as are $\sin{x}$, $\cos{x}$, and $\sqrt{x}$ \par
Functions take inputs in $\mathcal{L}$ and produce outputs in $\mathcal{L}$.
\vspace{3mm}
\item A set of \textit{relation symbols} $\mathcal{R}$ \par
Relation symbols let us compare elements of our language. \par
You are already familiar with this concept: $>$, $<$, and $\leq$ are relation symbols. \par
$=$ is \textbf{not} a relational symbol. Why? \hint{See \ref{def:language}}
\end{itemize}
\vspace{2mm}
The purpose of a structure is to give a language meaning. This is best explained by example.
\vspace{3mm}
\example{}
\def\structgeneric{\ensuremath{}}
The first structure we'll look at is the following:
$$
\Bigl(
\mathcal{L} ~\big|~ \{\mathcal{C}, ~ \mathcal{F}, ~ \mathcal{R}\}
\Bigr)
=
\Bigl( \mathbb{Z} ~\big|~ \{0, 1, ~ +, -, ~ <\} \Bigr)
$$
\vspace{2mm}
This is a structure over $\mathbb{Z}$ with the following symbols:
\begin{itemize}
\item $\mathcal{C} = \{0, 1\}$ \tab \note{(constants)}
\item $\mathcal{F} = \{+, -\}$ \tab \note{(functions)}
\item $\mathcal{R} = \{<\}$ \tab \note{(relations)}
\end{itemize}
\vspace{2mm}
Let's look at $\mathcal{C}$, our set of constant symbols. 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 the structure offers.
\vspace{1mm}
Say we want the number 2. We could use the function $+$ to define it: $2 \coloneqq [x \text{ where } 1 + 1 = x]$ \par
We would write this as $2 \coloneqq [x \text{ where } +(1, 1) = x]$ in proper \say{functional} notation.
\problem{}
Can we define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, 1, +, -, <\} \Bigr)$? If so, how?
\vfill
\problem{}
Can we define $-1$ in $\Bigl( \mathbb{Z} ~\big|~ \{0, +, -, <\} \Bigr)$? \par
\hint{In this problem, $1$ has been removed from the set of constant symbols.}
\vfill
\pagebreak
Let us formalize what we found in the previous two problems. \par
\say{Definable elements} are one of the two most important ideas in this handout.
\definition{}
A \textit{formula} in a structure $S$ is a well-formed string of constants, functions, and relations. \par
\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.
\definition{Definable Elements}
Say $S$ is a structure over a language $\mathcal{L}$. \par
We say an element $e$ of $\mathcal{L}$ is \textit{definable in $S$} if we can write a formula that only $e$ satisfies.
\problem{}
Can we define 2 in the structure $\Bigl( \mathbb{Z} ~\big|~ \{4, \times \} \Bigr)$?
\begin{solution}
No. We could try $[x \text{ where } x \times x = 4]$, but this is satisfied by both $2$ and $-2$. \\
We have no way to distinguish between negative and positive numbers.
\end{solution}
\vfill
\problem{}
Can we define 2 in the structure $\Bigl( \mathbb{Z^+} ~\big|~ \{4, \times \} \Bigr)$? \par
\hint{$\mathbb{Z}^+ = \{1, 2, 3, ...\}$}
\begin{solution}
Yes! $-2$ no longer exists, so $2$ can be defined by $[x \text{ where } x \times x = 4]$.
\end{solution}
\vfill
\problem{}
What is definable in the structure $\Bigl( \mathbb{R} ~\big|~ \{1, 2, \div \} \Bigr)$?
\begin{solution}
All powers of two, positive and negative.
\end{solution}
\vfill
\pagebreak