For any infinite binary string $x = x_1x_2x_3\dots$, let $f( \,x) \, = g(x_1)g(x_2)g(x_3)\dots$.

If for each element $x$ of a countable set $X$ of infinite binary strings, $f( \,x) \,$ is also in $X$, then let $X$ be considered â€˜balanced.â€™

Let V be a collection of countable sets of infinite binary strings that partition the set of infinite binary strings where each element of V is balanced.

Given a countable list of infinite binary strings, Cantorâ€™s diagonal argument can be used to create an infinite binary string that is not in the list (an anti-diagonal).

Let each element of V be well ordered in a fashion such that the anti-diagonal of the ordering is equal to $a = 111\dots$.

By definition, no element of V can contain $a$, so the elements of V cannot partition the set of infinite binary strings.

Now, either no such collection V can exist where each element of V is balanced or Cantorâ€™s diagonal argument is faulty.

On a side note, December 20th is my birthday. Yay!