Let \(S\) be a set, and let \(P(S ) \) be the power set of \(S\), which is the set of subsets of \(S \).

Suppose that \(f: S \rightarrow P(S ) \) surjects. We define \[T = \{ x\in S: x \not\in f(x) \}.\] so that \(f(x)\) is an element of \( P(S)\), hence a subset of \(S \), and \(x\) is an element of \(S\).

By assumption \(f \) surjects, so there exists \(t \in S \) such that \( f (t) = T \).

First, let \( t \in T\). Then \[ t \in \{ x \in S: x \not\in f(x) \}, \]

so \( t \in S \) such that \( t \not\in f(t) \). But since \( t \in T= f(t) \), this is a contradiction because \( t \in T = f(t) \), but \( t \not\in f(t) \).

Now let \( t \not\in T \). Then since \(t\) is not in \(T\), \(t\) cannot be an element of \(S\) such that \( t \not\in f(x) \). So if \( t \in S \), then \(t\) must also be a member of \( f(t) \). But supposing that \(f\) surjects, there is a \( t \in S \) such that \( f(t) = T \), which means that \( t \not\in T = f(t) \), but \( t \in f(t) \). This is a contradiction.

\(\therefore\) We conclude from both cases that \(f\) cannot surject, because there exists no \( t \in S \) such that \( f(t) = T \). In other words, \( T \not= f(t) \) for all \( t \in S \).

In this note, we have shown that *no* map from set \( S\) to its power set \( P(S ) \) can surject, which proves that there exists *no bijection* from S to its power set. This fundamental result means that for any set S, the set of all subsets of \(S\), \(P(S) \),has a strictly greater cardinality than \(S\). That

\[ |S| < |P(S)| \] or \[Card(S) < Card(P(S)).\]

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## Comments

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TopNewestYou need to work in something like \(\text{ZF}\) to guarantee the diagonal construction, and as mentioned below show that there be an injection \(S\longrightarrow\mathcal{P}(S)\). Your proof is otherwise standard and fine. I would write it more compactly as follows:

Claim.There is no surjection \(f:S\longrightarrow\mathcal{P}(S)\).Proof.Let \(f:S\longrightarrow P(S)\). To show: \(f\) is not surjective. By thereplacement axiomthere exists a set \(\Delta:=\{x\in S\mid x\notin f(x)\}\in\mathcal{P}(S)\). It suffices to show that \(\Delta\notin\text{ran}(f)\). Suppose this not be the case, then there exists \(\delta\in S\) such that \(f(\delta)=\Delta\). It holds that \(\delta\in f(\delta)\) \(\overset{f(\delta)=\Delta}{\Longleftrightarrow}\) \(\delta\in\Delta\) \(\overset{\text{defn. $\Delta$}}{\Longleftrightarrow}\) \(\delta\notin f(\delta)\). This is a contradiction. Hence \(\Delta\notin\text{ran}(f)\). \(\blacksquare\)Note that without suitable axioms, one cannot carry through with Cantor’s diagonal argument. In under Quine’s \(\text{NF}\) set theory this proof breaks down right at this point, and not only this, there exists sets such that \(|\mathcal{P}(X)|=|X|\).

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You may also want to show that \( f \) is injective. Otherwise, it is not obvious that \( |S| \leq |P(S)| \). (The surjective argument only shows that the cardinalities are not equal.)

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Thanks for pointing that out! Following your comment, constructing \(g: S \rightarrow P(S)\) so that every element of \(S\) is mapped to a singleton set, i.e. \(x \rightarrow \{x\}\) will do.

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