Hex Theorem/Corollary

Theorem

Every game of Hex has exactly one winner.

Aiming for a contradiction, suppose there is a sequence of moves:

$P_1, Q_1, P_2, Q_2, \dotsc$

such that neither player wins.

By the Pigeonhole Principle, every tile is marked after $n^2$ moves.

But, by the Hex Theorem, that board has exactly one winner.

Thus, by Proof by Contradiction, every game of Hex has at least one winner.

$\Box$

Aiming for a contradiction, suppose there is a game of Hex in which both players win.

Let $B$ be the board at the end of the game.

Let every unmarked tile of $B$ be marked with an arbitrary player symbol, by the Principle of Finite Choice.

Then, the resulting board $B'$ has every tile marked.

But, by definition of winning, each player has a finite sequence of adjacent tiles marked with their own symbol, which spans their opposite sides.

In particular, every tile in the sequence is marked in $B$, and so is marked the same in $B'$.

Thus, both players win in $B'$.

Thus, by Proof by Contradiction, every game of Hex has no more than one winner.

The result follows.

$\blacksquare$