# Homework Help: Baire category theorem

1. Jan 11, 2008

### quasar987

1. The problem statement, all variables and given/known data
My professor stated (but did not prove) the Baire category theorem as follows: "If (X,d) is a complete metric space, then for any sequence F_n of closed sets whose reunion is X, there is a k such that the interior of F_k is non-empty.

How is this equivalent to the more traditional statement " If (X,d) is a complete metric space, then for any sequence F_n of closed sets with empty interior, their reunion has empty interior as well."

???

Last edited: Jan 11, 2008
2. Jan 11, 2008

### morphism

Assuming that X is nonempty, yes they are equivalent.

That your version implies your professor's is easy: X doesn't have empty interior, and so cannot be the union of closed sets each of which has empty interior.

Going the other way seems to be trickier. Your professor's version immediately implies the following: In a complete metric space X, the intersection U of a sequence of dense open subsets is nonempty. But now take any closed ball in your metric space - this is going to be a complete metric space in its own right. So the intersection of U with this closed ball will also be nonempty (just shift everything from the space X to the space that is the closed ball). This in turn implies that U is dense in X (take any open ball B, then take the concentric closed ball who radius is half that of B - U will nontrivially intersect the closed ball and hence B). Now we're ready to prove that what you're professor said implies what you said: Take a sequence (F_n) of closed sets with empty interior in a complete metric space X, and let G_n=X\F_n; (G_n) is thus a sequence of dense open sets, and consequently their intersection G is dense in X. But then X\G, which is the union of the F_n's, must have empty interior.

Phew. Maybe I'm overcomplicating things though.

Last edited: Jan 11, 2008
3. Jan 12, 2008

### quasar987

Indeed, I have found a much swifter way!

We want to show that my professor's version imply my version.

Let Fn be a sequence of closed sets of empty interior. Suppose absurdly that the interior of their intersection is not empty. Call it X'. Now consider X' as a topological subspace of X. Call Fn' the trace of Fn on X'. It is a closed set in the subspace topology and the sequence of Fn' exhaust X'. Therefor, by my version of the thm, there exists a Fn' whose interior is not empty. That is to say, there is an open O in X such that O n X' is contained in Fn'. But O n X' is also contained in Fn, which is a contradiction because O n X' is also open in X, X' being itself open.

4. Jan 13, 2008

### morphism

Why are you taking their intersection?

5. Jan 13, 2008

### quasar987

typo; make that reunion

Last edited: Jan 13, 2008
6. Jan 13, 2008

### morphism

But then we have no way of knowing that it's a complete metric space.

7. Jan 13, 2008

### quasar987

darn :grumpy:

8. Jan 17, 2008

### quasar987

Why can't there be a closed ball that does not intersect U?

9. Jan 19, 2008

### morphism

Let's start over.

Your professor's version implies: If X is a complete metric space, then the intersection of a sequence of dense open subsets of X is nonempty.
Proof: If {U_n} is a sequence of dense open sets, then {X\U_n} is a sequence of closed sets and the identity int(X\U_n)=X\cl(U_n) tells us that each X\U_n has empty interior. So $\cup_n$X\U_n cannot be the whole of X, and consequently $\cap_n$U_n must be nonempty.

Next, let's enhance the above result and state that the intersection is in fact dense in X.
Proof: As before, let {U_n} be a sequence of dense open subsets of X. Take an arbitrary closed ball B in X - this is a complete space in the inherited metric topology. In particular, {U_n$\cap$B} is a sequence of dense open subsets of B. So their intersection $\cap_n$(U_n$\cap$B) = ($\cap_n$U_n)$\cap$B is nonempty, by the above result. But now if we take any open ball B' in X, then $\cap_n$U_n will intersect any closed ball that sits in B'. So $\cap_n$U_n is dense in X.

Finally let's wrap up and prove your statement. Let {F_n} be a sequence of closed subsets of X with empty interior. Then {X\F_n} is a sequence of dense open subsets of X (again, this follows from the identity cl(X\F_n)=X\int(F_n)). So the intersection $\cap_n$X\F_n is dense in X, and thus int($\cup_n$F_n) = X\cl(X\$\cup_n$F_n) = X\cl($\cap_n$X\F_n) = {}. Done!

Hopefully it's clearer now. (And hopefully without mistakes!)