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| author | adelon <22380201+adelon@users.noreply.github.com> | 2024-02-10 02:22:14 +0100 |
|---|---|---|
| committer | adelon <22380201+adelon@users.noreply.github.com> | 2024-02-10 02:22:14 +0100 |
| commit | 442d732696ad431b84f6e5c72b6ee785be4fd968 (patch) | |
| tree | b476f395e7e91d67bacb6758bc84914b8711593f /library/set/cantor.tex | |
Initial commit
Diffstat (limited to 'library/set/cantor.tex')
| -rw-r--r-- | library/set/cantor.tex | 18 |
1 files changed, 18 insertions, 0 deletions
diff --git a/library/set/cantor.tex b/library/set/cantor.tex new file mode 100644 index 0000000..f7c4b2b --- /dev/null +++ b/library/set/cantor.tex @@ -0,0 +1,18 @@ +\import{set/powerset.tex} +\import{function.tex} + +\subsection{Cantor's theorem} + + +\begin{theorem}[Cantor]\label{cantor} + There exists no surjection from $A$ to $\pow{A}$. +\end{theorem} +\begin{proof} + Suppose not. + Consider a surjection $f$ from $A$ to $\pow{A}$. + Let $B = \{a \in A \mid a\notin f(a)\}$. + Then $B\in\pow{A}$. + There exists $a'\in A$ such that $f(a') = B$ by \hyperref[surj]{the definition of surjectivity}. + Now $a' \in B$ iff $a' \notin f(a') = B$. + Contradiction. +\end{proof} |
