# Is the metric space Q of rationals homeomorphic to N, the natural numbers?

1. Oct 1, 2008

### gmn

I don't know if this is more appropriate for the topology forum, but I am learning this in analysis. I am asked to say whether or not Q and N are homeomorphic to eachother and to justify why. I am confused as to how to prove precisely that two spaces are homeomorphic, for there are no formal proofs of this in my text, only that (0,1] is not homeomorphic to the unit circle. everything else is just a visualisation of a donut and a coffee cup or something like that, but that doesn't really help me when thinking about N and Q. I know there exists a bijection, but I feel like it might not qualify for a homeomorphism because it doesn't necessarily send points that are close together or far away from eachother in N to points that are close together or far away in Q. I'm confused. Please help!

thanks

2. Oct 1, 2008

### morphism

Try to find a topological property that N possesses but Q does not. Hint: what can you say (topologically) about an arbitrary subset of N?

3. Oct 2, 2008

### mathman

It depends on the topology you put on the sets. If you use a discrete topology for the rationals, then they will be homeomorphic. If you use the "natural" topology for the rationals, they will not be.

4. Oct 2, 2008

### Hurkyl

Staff Emeritus
And he specified he's using the metric topology.

5. Oct 2, 2008

### ice109

doesn't the metric function of the space have to continuous?

6. Oct 2, 2008

### jostpuur

I don't see this being made clear anywhere.

edit: Whoops. It was in the title. Well, that is a hidden place...

Yes. One never has to worry about it, though, because the metric function is always continuous in the topology it itself defines.

7. Oct 3, 2008

### mathman

The metric function does not have to be continuous!
Examples: Integers d(x,y) = |x-y|.
Rational numbers: Express all rationals in lowest terms, d(a/b,c/f)=|a-c|+|b-f|.

These will both lead to discrete topologies.

8. Oct 3, 2008

### Hurkyl

Staff Emeritus
And in both cases, that makes the metric a function from a discrete space into the reals.

Exercise: Prove that any function with a discrete domain is continuous.

9. Oct 3, 2008

### xaos

the topology on Q is probably the subspace topology inherited from R, a metric space.

10. Oct 4, 2008

### mathman

The distances in the topology I described are all integers (which is of course a subset of the reals). The point I was trying to make is that with the topologies I defined, the integers and the rationals are homeomorphic.

11. Oct 8, 2008

### ice109

c'mon that's easy

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?