This proof of the Baire Category Theorem works in the interval [0,1], or in {0,1} However, it carries most of the force of the full proof, and tastes a lot better. Also related to Banach-Mazur games.

We wish to prove that a countable intersection of open dense sets is not empty (this is another equivalent formulation). To that end, we define a 2-player game: First, set some target set G. Alice's aim is to end up inside G, Bob's is to end up outside it. The idea of the proof is to pick sets G with the appropriate set-theoretic properties.

To play the game, the players write down "0.". Then Alice writes down any finite string of digits; then Bob does the same, and they continue, alternating turns. The number written down after each player's turn converges to some limit x. If x is in G, Alice wins, else Bob does. We say that Alice is guaranteed a win on the set G if she has a strategy which always ensures x will be in G (no matter what Bob does), and that Bob is guaranteed a win on the set G if he has a strategy which always ensures x won't be in G (no matter what Alice does). Of course, for "most" sets G we won't be able to show either player is guaranteed a win. Also note that if G is a subset of G', then if Alice is guaranteed a win on G she's also guaranteed a win on G', and if Bob is guaranteed a win on G' he's also guaranteed a win on G.

To complete the proof, just prove the following statements (or follow links to read them):

  1. If G is an open set then Alice is guaranteed a win. (GpBCT: proof that Alice wins on an open set).
  2. If the complement of G is open and dense, then Bob is guaranteed a win (GpBCT: proof that Bob wins on the complement of an open dense set).
  3. If Bob is guaranteed a win of each of the sets G1, G2, ..., then Bob is also guaranteed a win on their union. (GpBCT: proof that Bob wins on a countable union of sets if he's guaranteed a win on each one of them)
The conclusion is that Bob is guaranteed a win if the complement of G is a countable intersection of open dense sets. Since Bob cannot win if G is the entire space, it follows that the countable intersection of open dense sets is non-empty.
QED.

Log in or registerto write something here or to contact authors.