$|X|\lt|\power(X)|$

How can you argue with a cookie? Especially one which has such a magnificent theorem written all over it! Baking this was a wonderful way to spend the afternoon. And because I love this theorem so much, here is the proof:

Theorem. (Cantor) If $f\colon X\to\power(X)$ then $f$ is not surjective.

Proof. Let $D=\{x\in X\mid x\notin f(x)\}$. If $D\in\rng f$ then there is $x\in X$ such that $D=f(x)$. If $x\in D$ then $x\in f(x)$, but $x\in D$ if and only if $x\notin f(X)$. Therefore $x\notin D$, but then $x\notin f(x)$ and therefore $x\in f(x)$. Either way we have a contradiction and therefore $D\notin\rng f$ and $f$ is not surjective. $\square$

Corollary. (The Cookie Theorem) For every $X$, $|X|<|\power(X)|$.

Proof. There is an obvious injection, $x\mapsto\{x\}$ from $X$ into $\power(X)$, so $|X|\leq|\power(X)|$, but equality is impossible because Cantor’s theorem tells us that there is no surjection from $X$ onto $\power(X)$, let alone a bijection. Therefore $|X|<|\power(X)|$ as wanted. $\square$

(Oh yeah, it tastes great too!)