Commit e404acfc authored by Vincent Tam's avatar Vincent Tam

Post: CSB Theorem

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title: "CSB Theorem"
subtitle: "A visual argument for CSB Theorem"
date: 2018-08-29T11:25:24+02:00
- math
- set theory
<i class="fa fa-info-circle" aria-hidden></i> This _isn't_ a substitute for
books <i class="fa fa-book" aria-hidden></i>.
### Reminder
- $A \preceq B$: $A$ can be "_injected_" into $B$
- $A \sim B$: $A$ and $B$ share the _same cardinality_
- $A \prec B$: $A$ can be "_injected_" into $B$, but it's "_smaller_" than $B$.
- A **finite** set can be "_counted_" from one to some nonnegative integer.
- **Infinite** is the "_antonym_" of finite.
### Wolf's proof
When I first saw this proof in Robert S. Wolf's
[_Proof, Logic and Conjecture: The Mathematician's Toolbox_][1], I gave it up
since it _wasn't_ as intuitive as the statement.
### An informal argument
Later, I found some interesting _illustrations_ in Richard Hammack's
[_Book of Proof_][2].
1. "_Draw_" gray $A$ and white $B$.
2. "_Draw_" the given injections $f: A \to B$ ($A$ contained in $B$) and $g: B
\to A$ ($B$ contained in $A$). (Figure&nbsp;13.4)
3. "_Draw_" an _infinite chain_ of _alternating_ injections starting from $A$.
4. In "_diagram&nbsp;$A$ at step infinity $\infty$_" ($A$ containing $B$ containing $A$ …), label
the "_gray region_" as $G$.
5. Label remaining white region as $W$. (i.e. $W := A \setminus G$)
6. "_Draw_" a "_homologous_" diagram with the one in step&nbsp;4 on the right-hand side, but starting from $B$. (i.e. $B$ containing $A$ containing $B$ …) (Figure&nbsp;13.6)
7. It's natural to associate the gray regions $G \subseteq A$ with $f(G) \subseteq B$ on both sides. It remains to settle $W$.
- Applying $f$ on $W$ _won't_ lead to any useful results.
- Another given injection $g$ _can't_ be applied on $W$ due to domain mismatch.
- Reverse the "_direction_" of $g$ to that it points to the white region wrapping gray $f(A)$.
It's nice to see a constructive and _formal_ proof immediately following this
intriguing argument. The later actually guides me through the former.
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