Are the Irrationals and Product Spaces Non-Homeomorphic?

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SUMMARY

The discussion centers on the non-homeomorphic nature of the irrationals R-Q and the product space (R-Q)×Q. It is established that R-Q is a Baire space, while (R-Q)×Q is not, primarily due to the properties of their completions and the concept of metric versus topological spaces. The conversation highlights that the completion of the sets ##\mathbb{Q}## and ##\{x\in \mathbb{Q}~\vert~0 PREREQUISITES

  • Understanding of Baire spaces
  • Familiarity with metric and topological concepts
  • Knowledge of homeomorphism in topology
  • Basic understanding of continuous fractions
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  • Research the properties of Baire spaces and their implications in topology
  • Study the concept of metric completions and their relationship to homeomorphism
  • Explore the use of continuous fractions in defining homeomorphisms
  • Investigate the concept of "slices" in topological spaces and their significance
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Mathematicians, topologists, and students studying advanced concepts in topology, particularly those interested in the properties of Baire spaces and homeomorphism.

hedipaldi
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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?
 
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Their completions aren't homeomorphic, I think that does the trick but maybe there's some weird counterexample
 
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
 
hedipaldi said:
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?
 
why not?
it seems that a homeomorphism actually can be defind by using continuous fractions.Isn't it?
 
Last edited:
(R-Q)XQ isn't Baire. To prove that look at "slices".
 
what is "slices" ? can you give me a link for the proof?
Thank's a lot.
 
Last edited:
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.
 
You mean with EMPTY interior right?
 

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