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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}
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