König's theorem on the continuum

We first prove the general König inequality: if $\kappa_i < \lambda_i$ for every $i$ in an index set $I$, then $\sum_{i\in I}\kappa_i < \prod_{i\in I}\lambda_i$. Fix sets $B_i$ with $|B_i|=\lambda_i$ and subsets $A_i \subsetneq B_i$ with $|A_i|=\kappa_i$. The inequality $\sum_i \kappa_i \le \prod_i \lambda_i$ is routine (send each $a\in A_i$ to the tuple that is $a$ in coordinate $i$ and some fixed default value elsewhere), so the content is showing they are not equal.

Suppose, for contradiction, some function $h : \bigsqcup_{i\in I} A_i \to \prod_{i\in I} B_i$ is surjective. For each $i$, let $h_i:A_i\to B_i$ send $a \mapsto h(a)(i)$ (the $i$-th coordinate of $h(a)$). Since $|A_i|=\kappa_i<\lambda_i=|B_i|$, the map $h_i$ cannot be surjective onto $B_i$ (if it were, picking a preimage for each element of $B_i$ would inject $B_i$ into $A_i$, forcing $\lambda_i\le\kappa_i$ — contradiction). So choose $d_i \in B_i \setminus \operatorname{ran}(h_i)$ for every $i$, and let $g \in \prod_i B_i$ be the tuple with $g(i)=d_i$.

Since $h$ is surjective, $g=h(a)$ for some $a$, say $a \in A_j$. Then $g(j)=h(a)(j)=h_j(a) \in \operatorname{ran}(h_j)$. But $g(j)=d_j \notin \operatorname{ran}(h_j)$ by construction — a direct contradiction. So no surjection $h$ exists, giving $\sum_i\kappa_i < \prod_i\lambda_i$. (Setting $I=X$, $\kappa_i=1$, $\lambda_i=2$ recovers Cantor's classical diagonal argument $|X|<2^{|X|}$ as a special case.)

Now suppose, for contradiction, $\operatorname{cf}(2^{\aleph_0})=\aleph_0$. Then $2^{\aleph_0}$ is the sum of a strictly increasing $\omega$-sequence of smaller cardinals $\kappa_0<\kappa_1<\kappa_2<\cdots$, i.e. $2^{\aleph_0}=\sum_{n<\omega}\kappa_n$ with every $\kappa_n < 2^{\aleph_0}$. Apply König's inequality with $\lambda_n := 2^{\aleph_0}$ held constant for every $n$ (valid since $\kappa_n < 2^{\aleph_0} = \lambda_n$ for every $n$):

$$2^{\aleph_0} = \sum_{n<\omega}\kappa_n \;<\; \prod_{n<\omega}\lambda_n = \left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0}.$$

This says $2^{\aleph_0}<2^{\aleph_0}$, absurd. So $\operatorname{cf}(2^{\aleph_0})\neq\aleph_0$; since cofinality is never smaller than that (it is at least $\aleph_0$ for any infinite cardinal, and it cannot be finite), we conclude $\operatorname{cf}(2^{\aleph_0})>\aleph_0$. $\blacksquare$