Cardinality of set and its power set

Question

Prove that for any set $X$ we have the $|X| < |\mathcal{P}(X)|$ (power set of $X$)

How would you prove this using the definitions of bijection, surjection, and injection?

Also, does this mean when we have the empty set $X = \{\}$, then $|X| < |\mathcal{P}(X)|$ as well? Would they not be equal?

Hint for the finite case for getting an intuition for why this is true: each individual element of $X$ is an element of $P(X)$ (as a singleton set). — Cameron Williams, Nov 09 '13 at 03:25
I think this has been asked here before. It's a standard diagonal argument. — Michael Hardy, Nov 09 '13 at 03:28
To T. Bongers: It would still hold that |X| < |P(X)| when x = {} because n = 0 so |P(X)| = 2^0 = 1, correct? — stumpy, Nov 09 '13 at 03:42
In the case of the empty set $X$, there are no elements in $X$, but there is one element in $\mathcal P(X)$, namely the empty set. — Andreas Blass, Jul 31 '14 at 01:00

Michael Hardy · Answer 1 · 2013-11-09T03:42:49.847

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Suppose $f:X\to \mathcal P(X)$ is injective. Let $A=\{x\in X: x\not\in f(x)\}$. It can be shown by contradiction that $A$ is not in the image of $f$. If $A=f(a)$ then either $a\in f(a)$ or $a\not\in f(a)$, and a contradiction ensues either way.

Sometimes you see the proof written like this: Assume $f:X\to \mathcal P(X)$ is injective and surjective. Then deduce a contradiction. But phrasing the proof as a whole as a proof by contradiction, rather than just having that one part of the proof done by contradiction, just makes it look more complicated than it needs to be.

edited Nov 09 '13 at 03:42

answered Nov 09 '13 at 03:31

Michael Hardy

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It is not needed that $f$ is injective. – Andrés E. Caicedo Jul 31 '14 at 01:24
True, but I think it may be a bit easier to understand that way. – Michael Hardy Jul 31 '14 at 03:24

Cardinality of set and its power set

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