# Power set of real numbers

Can someone help me with a proof that the power set of real numbers is strictly larger than the set of real numbers? I have been finding it hard using Cantor's general proof in this special case. A reference or answer will be appreciated.

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## migrated from mathoverflow.netJul 13 '13 at 13:42

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Voting to migrate to Math Stack Exchange, where such questions belong. You will probably need to describe what the problem is in more detail, because probably someone will just recite Cantor's proof for general sets $X$, which presumably you know (?) according to your post. – user43208 Jul 13 '13 at 13:27
@Mario What's the problem with using Cantor's argument for this special case? – Git Gud Jul 13 '13 at 13:45
The general argument works exactly the same on every set $X$. It's not clear why the proof would even need a "special case." – Thomas Andrews Jul 13 '13 at 13:51

This is a special case of the more general result that there is no bijection between any set $X$ and its power set. If you're going to prove it about reals then you might as well prove it about an arbitrary set.
The idea is similar to that of Cantor's diagonal argument. Suppose there is a bijection $f : X \to \mathcal{P}(X)$, and consider the set $$A = \{ x \in X : x \not \in f(x) \}$$ This is certainly a subset of $X$, and so $A = f(y)$ for some $y \in X$.
But something went wrong... is $y \in f(y)$? Is $y \not \in f(y)$?
Or, if you want a more "direct" proof, show that there is no surjection from $X$ to $\mathcal P(X)$ by taking any map $f : X \to \mathcal P(X)$ and constructing an element of $\mathcal P(X)$ not in its range. Of course, it is the set $A$ defined above. – GEdgar Jul 13 '13 at 15:56