Are Q and N Homeomorphic in Terms of Topology?

  • Context: Graduate 
  • Thread starter Thread starter Scousergirl
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around whether the set of rational numbers (Q) is homeomorphic to the set of natural numbers (N) in terms of topology. Participants explore different topological structures and their implications for homeomorphism, including the subspace topology from the reals and the discrete topology.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant questions the continuity of a bijection between Q and N, suggesting that a continuous function would need to preserve limits, which seems problematic.
  • Another participant points out that the topology being considered is crucial, noting that Q and N are not homeomorphic under the normal topology of the reals but may be under the discrete topology.
  • A claim is made that sequences in N converge constantly, leading to a proposed proof that they cannot be homeomorphic, although this claim is challenged by another participant.
  • There is a discussion about the nature of convergence in N, with a participant clarifying that only eventually constant sequences converge, raising questions about the implications for a potential homeomorphism.
  • One participant suggests focusing on the topologies themselves, indicating that N has the trivial topology while Q does not, which could be sufficient to show they are not homeomorphic.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are competing views regarding the topologies and the implications for homeomorphism between Q and N. The discussion remains unresolved with differing interpretations of convergence and continuity.

Contextual Notes

Limitations include the dependence on the choice of topology and the assumptions made about convergence in the context of homeomorphisms. The discussion highlights the need for clarity on what topological properties are being considered.

Scousergirl
Messages
15
Reaction score
0
Is Q homeomorphic to N?

I understand that there exists a bijection from Q to N but I cannot figure out how this function is continuous and it's inverse is also continuous.
 
Physics news on Phys.org
With what topology? The subspace topology from being a subspace of the reals with their normal topology, or the discrete topology. They're not homeomorphic in the former, but are in the latter.
 
The normal topology...this is what i came up with...not sure if its right:

Claim: they are not homeomorphic
Proof: Assume they are. Then There exists a continuous function f from Q to N. Therefore all of the sequences in Q are mapped to a sequence in N preserving limits. But since sequences in N converge constantly, this cannot be a bijection. therefore they are not homeomorphic.

Does this make sense?
 
Why do sequences in N converge constantly? The sequence (1, 2, 3, ...) certainly doesn't converge. Did you mean to say that the only sequences in N that converge are the ones that are eventually constant? But then why is this a contradiction? Maybe your homeomorphism f maps convergent sequences to sequences that are eventually constant. Can this happen?

But honestly I wouldn't bother thinking about it this way. Just think of what the topologies are. N will get the trivial topology, i.e. every set is going to be open (why?). Will Q get the trivial topology? No, it won't. Try to find a set that's not open in Q. This will be enough to show that they're not homeomorphic.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K