# [SOLVED] How do I prove R is a baire space?

• Jun 23rd 2008, 02:36 PM
aurora
[SOLVED] How do I prove R is a baire space?
I need to use the fact that R is a baire space somewhere, but I just don't seem to be able to prove it. I'm not even sure if it's trivial or not (meaning, if it's even required to prove it), but I guess I can't take the risk. Moreover, it's disturbing I don't know how too.
I know that R is a complete metric space.
I've read that complete spaces are baire spaces, but I haven't got the proof for that.
My definition for baire space is one of the following (which are equivalent, I've proved that):
1) every countable intersection of dense open sets is a dense set.
2) every countable unification of non-dense (not sure of the term) sets has an empty interior.

Can't see why R satisfies that, nor how to prove the more general claim, that complete space => Baire space.

Bless you!
Tomer. (Clapping)
• Jun 23rd 2008, 03:43 PM
Plato
Have you looked at a site such as this: Baire Space -- from Wolfram MathWorld?
It seems that you have a problem with definitions.
• Jun 23rd 2008, 09:14 PM
aurora
I don't know what's the deal with my lecturer, but all the definitions I get are for some reason theorems, and the theorems I need to prove are definitions :)
I've got an excercise "teaching" me what a baire space is, using the definition I wrote down. I need to prove some things.
Later on, I need to prove a more "practical" question, regarding R, but I need to use the fact that it's a baire space. I cannot just say "R is complete (or homeomorphic to itself = a complete space) and therefore it's a baire space", because we have not studied that. All we studied in class is Baire's theorem, but it mentions nothing of Baire spaces (though I can see the connection).

In other words, I'm still stuck (Nerd).

Thank you very much for responding!
• Jun 23rd 2008, 11:12 PM
Opalg
Quote:

Originally Posted by aurora
I need to use the fact that R is a baire space somewhere, but I just don't seem to be able to prove it. I'm not even sure if it's trivial or not (meaning, if it's even required to prove it), but I guess I can't take the risk. Moreover, it's disturbing I don't know how too.
I know that R is a complete metric space.
I've read that complete spaces are baire spaces, but I haven't got the proof for that.
My definition for baire space is one of the following (which are equivalent, I've proved that):
1) every countable intersection of dense open sets is a dense set.
2) every countable unification of non-dense (not sure of the term) sets has an empty interior.

Here is the proof of Baire's theorem, as given in Gert Pedersen's excellent book Analysis Now.

Theorem If {A_n} is a sequence of open, dense subsets of a complete metric space (X,d), then the intersection \$\displaystyle \bigcap A_n\$ is dense in X.

Proof. Let \$\displaystyle B_0\$ be an closed ball in X with radius r>0, and let \$\displaystyle B_0^o\$ denote its interior. Since A_1 is dense in X and open, the set \$\displaystyle A_1\cap B_0^o\$ contains a closed ball B_1 with radius < r/2. Since A_2 is also dense and open, \$\displaystyle A_2\cap B_1^o\$ contains a closed ball B_2 with radius < r/4. By induction we find a sequence (B_n) of closed balls in X, such that \$\displaystyle B_n\subseteq A_n\cap B_{n-1}^o\$ and radius\$\displaystyle (B_n)<2^{-n}r\$ for every n. Since X is complete, there is evidently a point x in X such that \$\displaystyle \{x\} = \bigcap B_n\subseteq B_0\cap\bigl(\bigcap A_n\bigr).\$ This shows that \$\displaystyle \bigcap A_n\$ intersects every nontrivial ball in X, ensuring its density.
• Jun 24th 2008, 06:33 AM
aurora
Thank you very much! That's exactly what I needed! (Clapping)