Is Q Homeomorphic to N?

1. Oct 9, 2007

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.

2. Oct 10, 2007

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.

3. Oct 10, 2007

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?

4. Oct 10, 2007

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.