2

When we proved Cantor's theorem, we first showed that $|A| \leq |P(A)|$ by finding an injective function $A \to P(A)$. Then we showed that a surjection does not exist.

Is the injection necessary? Doesn't the non-existence of a surjection already prove the theorem?

Mixoftwo
  • 387

1 Answers1

4

Assuming the axiom of choice, the following can be shown:

  • For any sets $A,B$ either $|A| \leq |B|$ or $|B| \leq |A|$ (see this question).
  • There is a surjection $A \to B$ if and only if there is an injection $B \to A$ (see this question).

Then if you show that there is no surjection $A \to P(A)$, the second point implies that there is no injection $P(A) \to A$. Hence the first point implies $|A| \leq |P(A)|$.

So you could prove Cantor's theorem that way, but you would need to assume the axiom of choice, which people often prefer not to use if it is not necessary.

Steven
  • 3,186
  • Side note: It is also possible to show directly that there is no injection $P(A)\to A$: https://puzzling.stackexchange.com/questions/126988/leaders-and-rulers Think of the function $P(A)\to A$ as assigning an element of $A$ to be "in charge" of each subset of $A$. The fact that some element of $A$ is both a leader and a ruler, means it is hit twice (at least) so the function is not injective. – tkf Jan 19 '25 at 04:18