Prove that for any set $A$, there is no one-one correspondence between $A$ and its power set $P(A)$.

Suppose that there is a one-one correspondence $f$ between $A$ and $P(A)$.

Let $B$ be one such set defined as follows:

$$B=({x\in A:x\notin f(x)})$$

Then $B$ is a subset of $A$ and so $B\in P(A)$. So there is an element $a\in A$ such that $f(a)=B$

case 1: $a\in f(a)$ implies $a\in B$ which is a contradiction.

case 2: $a\notin f(a)$ implies $a\in B$ which is a contradiction.

Thus we conclude that there is no one-one correspondence between $A$ and $P(A)$.

I dont quite understand the proof above like why is there a contradiction and how to arrive at it. Could anyone explain more clearly how the above proof works. Thanks

  • $\begingroup$ This proof applies to uncountable sets A while the diagonal construction does not. $\endgroup$ Mar 6, 2017 at 5:02

3 Answers 3


Before diving in to what I wrote below, it may be a good idea first to understand Russell's paradox; it's essentially the same as what's going on here, but a little less complicated.

Let's look at the first case first.

Think about what "$a\in f(a)$" means. $B$ is the set of all $x\in A$ which are not in $f(x)$. So if $a\in f(a)$, it cannot be the case that $a\in B$, by definition! But we know $f(a)=B$. So:

  • If $a\in f(a)$, then $a\in B$ (since $f(a)=B$).

  • But if $a\in B$, then $a\not\in f(a)$ (since by definition $B$ is the set of $\alpha$s which are not in $f(\alpha)$).

  • So this is a contradiction! From "$a\in f(a)$" we've concluded "$a\not\in f(a)$" - that means "$a\in f(a)$" can't be true! (If $P$ implies Not $P$, do you see why $P$ can't be true?)

The other case is similar. If $a\not\in f(a)$, then $a\in B$ (why? think about what $B$ is . . .); but that means $a\in f(a)$ (why? think about what $f(a)$ is . . .). So, again, this leads to a contradiction.

So each possibility - $a\in f(a)$ and $a\not\in f(a)$ - lead to contradictions. So neither can hold - but one of them has to!

This means that, somewhere, we must have made an incorrect assumption. And the only assumption we've made is "There is a one-one correspondence between $A$ and $P(A)$". So that assumption must be false.

  • $\begingroup$ Brilliant explanation. +1 $\endgroup$
    – Xam
    Feb 12, 2017 at 17:10

We have

$$f:A\rightarrow P(A)$$

so the image of a point in $A$ is a subset of $A$. Therefore it makes sense to define the following set

$$B=\{x\in A\mid x\notin f(x)\}$$

which contains all the points in $A$ that belong to the set $f(x)$ so, in particular, by its definition, $B\subset A$.

Now we use the assumption that $f$ es one-to-one, in particular it is surjective, so every subset of $A$ has a preimage. Let $a\in A$ be the preimage of $B$, that is


Now we try to reach a contradiction when we try to decide if $a$ belongs, or not, to $B$:

-Let's assume $a\in B=f(a)$, this is a contradiction as every element $b\in B$ does NOT belong to its image.

-If $a\notin B=f(a)$, but $B$ contains every element of $A$ that does not belong to its image, that is $a\in B$. Another contradiction.

It is obvious that one of the previous asertions had to be true, since they are not, our assumption has to be wrong, and there is no such an $f$.

  • $\begingroup$ You made a mistake in your definition of B $\endgroup$ Mar 6, 2017 at 5:01
  • $\begingroup$ Thanks for pointing it out, edited. $\endgroup$
    – Smurf
    Mar 8, 2017 at 18:35

Both cases contradict the definition of $B$.

Think about coloring each element of $A$ red or blue based on whether $a\in f(a)$ or not. $B$ is the set of blue elements. The particular $a$ that satisfies $f(a)=B$ must have a color. If it's red, then $a\in f(a)=B$, so $a\in B$. But $B$ is the blue elements, and here's a red one in there. If instead $a$ is blue, then $a\notin f(a)=B$. But $B$ contains all the blue elements, and here's one that's not in $B$.


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