Are the Irrationals and Product Spaces Non-Homeomorphic?

[hedipaldi]

Why are the irrationals R-Q and the product space (R-Q)XQ not homeomorphic?
The first space i Baire space.may be the second space is not?

 

[Office_Shredder]

Their completions aren't homeomorphic, I think that does the trick but maybe there's some weird counterexample

 

[micromass]

The completions of $\mathbb{Q}$ and $\{x\in \mathbb{Q}~\vert~0<x<1\}$ aren't homeomorphic either, even though the two spaces are homeomorphic. The problem is that completion is a metric concept and not a topological concept.

I think looking at Baire spaces is the way to go

 

[Bacle2]

Why are the irrationals R-Q and the product space (R-Q)XQ not homeomorphic?
The first space i Baire space.may be the second space is not?

Well, how about from the perspective that R-Q can be embedded in R, but , at least that I can

tell, (R-Q)xQ cannot?

 

[hedipaldi]

why not?
it seems that a homeomorphism actually can be defind by using continuous fractions.Isn't it?

 

[ForMyThunder]

(R-Q)XQ isn't Baire. To prove that look at "slices".

 

[hedipaldi]

what is "slices" ? can you give me a link for the proof?
Thank's a lot.

 

[ForMyThunder]

Take the sets (R-Q)x{q} for rational q. This is a countable set of closed sets with nonempty interior but their union is the entire space. Hence, it isn't Baire.

 

[hedipaldi]

You mean with EMPTY interior right?