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authoradelon <22380201+adelon@users.noreply.github.com>2024-02-10 02:22:14 +0100
committeradelon <22380201+adelon@users.noreply.github.com>2024-02-10 02:22:14 +0100
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+\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}