Showing Dense G_δ-Subspace of Baire Space is a Baire Space

[Lie]

Hello!

Does anyone have any idea how to show that every dense G_\delta-subspace of a Baire space is a Baire space?

Grateful.

 

[Lie]

Hello!

Does anyone have any idea how to show that every dense G_\delta-subspace of a Baire space is a Baire space?

Grateful.

I did it! ;)

 

[Greg Bernhardt]

Hi there! Yes, I can help with that. First, let's define some terms for clarity. A Baire space is a topological space in which the intersection of countably many dense open sets is still dense. A dense G_\delta-subspace is a subspace that is both dense and a countable intersection of open sets (G_\delta stands for countable intersection of open sets in Greek).

Now, to show that every dense G_\delta-subspace of a Baire space is a Baire space, we can use the fact that Baire spaces are also completely regular. This means that for any closed set C and any point x not in C, there exists a continuous function f such that f(x) = 0 and f(C) = {1}.

So, let's take a dense G_\delta-subspace, call it Y, of a Baire space X. Since Y is dense, it intersects every open set in X. Now, let C be a closed set in Y and x be a point not in C. By the definition of a Baire space, there exists a countable intersection of open sets in X, call it U, such that U ∩ Y is still dense in Y. Since Y is a G_\delta-subspace, U ∩ Y is also a G_\delta-subspace of Y.

Now, using the completely regular property of Baire spaces, there exists a continuous function f: Y → [0,1] such that f(x) = 0 and f(U ∩ Y) = {1}. Since U ∩ Y is dense in Y, f(Y) = [0,1].

But since Y is a G_\delta-subspace, it is also closed in itself. Therefore, f(Y) is a closed subset of [0,1]. This means that f(Y) is also a Baire space, since it is a closed subset of a Baire space.

In conclusion, we have shown that every dense G_\delta-subspace of a Baire space is a Baire space by using the properties of Baire spaces and completely regular spaces. I hope this helps! Let me know if you have any further questions.