Enderton, note first half of axioms.

finite-set-exercises
Joshua Potter 2023-05-19 09:54:23 -06:00
parent 1da6e31581
commit aaa7052040
1 changed files with 38 additions and 11 deletions

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@ -17,6 +17,32 @@
\chapter{Reference}%
\label{chap:reference}
\section{\partial{Empty Set Axiom}}%
\label{ref:empty-set-axiom}
There is a set having no members:
$$\exists\; B, \forall\; x, x \not\in B.$$
\section{\defined{Extensionality Axiom}}%
\label{ref:extensionality-axiom}
If two sets have exactly the same members, then they are equal:
$$\forall\; A, \forall\; B,
\left[\forall\; x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$
\begin{axiom}
\lean{Mathlib/Init/Set}{Set.ext}
\end{axiom}
\section{\partial{Pairing Axiom}}%
\label{ref:pairing-axiom}
For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
$$\forall\; u, \forall\; v, \exists\; B, \forall\; x,
(x \in B \iff x = u \text{ or } x = v).$$
\section{\defined{Powerset}}%
\label{ref:powerset}
@ -28,18 +54,19 @@ The \textbf{powerset} of some set $A$ is the set of all subsets of $A$.
\end{definition}
\section{\defined{Principle of Extensionality}}%
\label{ref:principle-extensionality}
\section{\partial{Power Set Axiom}}%
\label{ref:power-set-axiom}
If $A$ and $B$ are sets such that for every object $t$,
$$t \in A \quad\text{iff}\quad t \in B,$$
then $A = B$.
For any set $a$, there is a set whose members are exactly the subsets of $a$:
$$\forall\; a, \exists\; B, \forall\; x, (x \in B \iff x \subseteq a).$$
\begin{axiom}
\section{\partial{Union Axiom, Preliminary Form}}%
\label{ref:union-axiom-preliminary-form}
\lean{Mathlib/Init/Set}{Set.ext}
\end{axiom}
For any sets $a$ and $b$, there is a set whose members are those sets belonging
either to $a$ or to $b$ (or both):
$$\forall\; a, \forall\; b, \exists\; B, \forall\; x,
(x \in B \iff x \in a \text{ or } x \in b).$$
\endgroup
@ -160,13 +187,13 @@ Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and
\lean{Bookshelf/Enderton/Set/Chapter\_1}
{Enderton.Set.Chapter\_1.exercise\_1\_2}
By the \nameref{ref:principle-extensionality}, $\emptyset$ is only equal to
By the \nameref{ref:extensionality-axiom}, $\emptyset$ is only equal to
$\emptyset$.
This immediately shows it is not equal to the other two.
Now consider object $\emptyset$.
This object is a member of $\{\emptyset\}$ but is not a member of
$\{\{\emptyset\}\}$.
Again, by the \nameref{ref:principle-extensionality}, these two sets must be
Again, by the \nameref{ref:extensionality-axiom}, these two sets must be
different.
\end{proof}