Are Q and N Homeomorphic in Terms of Topology?

[Scousergirl]

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.

 

[matt grime]

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.

 

[Scousergirl]

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?

 

[morphism]

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.