Elementary Classical Analysis, Marsden, 2nd ed, Introduction, Example 1

Problem

Let $A$ be a set and let $\mathcal{P}(A)$ denote the set of all subsets of $A$. Prove that $A$ and $\mathcal{P}(A)$ do not have the same cardinality.


Answer

Suppose there is a bijection $f:A \rightarrow \mathcal{P}(A)$ and define $B=\{x \in A | x \notin f(x)\}$. Since $f$ is surjective, there exists $y \in A$ such that $f(y)=B$. If $y \in B$ then $y \notin f(y) = B$, which is contradiction. Similarly, if $y \notin B=f(y)$, then $y \in B$ by the definition of $B$. In either case, we get a contradiction.

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