[Set Theory] Proving Cantor's Theorem by Contradicting Surjectivity on Power Sets

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

Note by Tasha Kim
1 week, 3 days ago

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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.)

Dan Wilhelm - 1 week, 1 day ago

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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.

Tasha Kim - 6 days, 23 hours ago

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