Cantor's Theorem:

For any set $X$, there is no onto function $f:X\to \mathcal{P}(X)$

While I was looking at the proof for this, my head decided to stop understanding it. I mean, I think I understand the proof itself, but doesn't it assume that the subset of the objects that aren't in the defined function is non-empty? Looking at it, I think that it only proves that that subset has no element. Doesn't it derive a contradiction from an assumption made earlier? Can you guys help me understand?


3 Answers 3


No. The proof assumes nothing.

If $f\colon A\to\mathcal P(A)$ is any function, then $A_f=\{a\in A\mid a\notin f(a)\}$ is not in the range of $f$.

  • If $A=\varnothing$, then the only function with the empty domain is $\varnothing$ itself, which has an empty range. Then $A_f=\varnothing$, but $\varnothing\notin\operatorname{rng}(\varnothing)$. So we're fine.
  • If $A_f$ is empty, e.g. $f(a)=\{a\}$, then it just means that no element is mapped to the empty set. This is an extension of the first case, where indeed no element is mapped to the empty set.

In either case, it's fine, since $\varnothing$ is indeed a subset of $A$ and therefore an element of $\mathcal P(A)$.

  • $\begingroup$ Ohh I see now, thanks a lot! $\endgroup$ Jan 11, 2017 at 17:49
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    $\begingroup$ +1. @J.Dionisio It may be helpful to think of this proof as constructing something: namely, we define a function $F$ from $\{$maps from $A$ to $\mathcal{P}(A)\}$ to $\mathcal{P}(A)$. We then prove that $F(f)\not\in ran(f)$. But the initial construction of $F$ makes sense without any assumptions, and our proof of $F(f)\not\in ran(f)$ does not make any assumptions either. $\endgroup$ Jan 11, 2017 at 18:18

Assuming this is your formulation of Cantor's Theorem:

For any set $S$, there is no onto mapping from $S$ to its power set.

or some equivalent formulation, let's see what happens if $S$ is empty.

If $S$ is empty, that is, of cardinality zero, then we see that its power set contains one element - the empty set. We can see no surjection exists from $\emptyset$ to $\mathcal{P}(\emptyset)$ through some work.

If this doesn't help, I suggest updating your question with the formulation of Cantor's Theorem you are looking at and details of the proof you see.


Actually your observation on the emptiness of the set $B=\{a\in A\mid a\notin f(a)\}$ is good. Following your argument one could still prove the theorem along the following lines:

Assume $f:A\to \mathcal{P}(A)$ is surjective. Let $B=\{a\in A\mid a\notin f(a)\}$ and $B^c=\{a\in A\mid a\in f(a)\}$. Then $A = B\cup B^c$ since for every element $a\in A$ we know that either $a\in f(a)$ or $a\notin f(a)$. By Cantor's argument (as you pointed out) it follows $B = \emptyset$. Thus $A=B^c$, i.e. $\forall a \in A \Rightarrow a\in f(a)$. But this means that each $f(a)\in \mathcal{P}(A)$ is not empty. Consequently $f$ cannot be surjective since the empty set is not in the range of $f$ and $\emptyset$ is always member of $\mathcal{P}(A)$.

"Never stop thinking critically :)"

  • 2
    $\begingroup$ No. Cantor's argument does not show that $B=\emptyset$. Cantor's argument shows that $B$ is not in the range of $f$, regardless of whether or not it's empty. $\endgroup$ Apr 12, 2017 at 20:55
  • $\begingroup$ Thank you for your comment. However, please read what I wrote before you post negative comments. I was referring to the line of arguments of J. Dionisio. Do also carefully read the Cantor's proof which holds for $B$ empty or not. $\endgroup$
    – JJR
    Apr 12, 2017 at 21:06
  • $\begingroup$ I know you were and you said "by Cantor's argument it follows that $B=\emptyset$", which is a false statement. This is not at all the contradiction that Cantor's theorem points to. Indeed, you cannot prove that $B$ is empty, for there are functions $f: A\to P(A)$ (obviously not surjective) for which it isn't. For instance if you take $A=\{0\}$, $P(A) = \{\emptyset, A\}$ and $f(0) = \emptyset$, then $0\notin f(0)$, so $0\in B$. The contradiction arises from assuming that $B$ is in the range of $f$, not from assuming that it is or isn't empty $\endgroup$ Apr 12, 2017 at 21:15
  • $\begingroup$ Hehe. I don't want to be pedantic but I said "By Cantor's argument (as you pointed out) it follows $B=\emptyset$". $\endgroup$
    – JJR
    Apr 12, 2017 at 21:26
  • $\begingroup$ Though your example is nice :) $\endgroup$
    – JJR
    Apr 12, 2017 at 21:33

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