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

[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 each other 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 each other in N to points that are close together or far away in Q. I'm confused. Please help!

thanks

 

[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?

 

[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.

 

[Hurkyl]

It depends on the topology you put on the sets.

And he specified he's using the metric topology.

 

[ice109]

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

 

[jostpuur]

And he specified he's using the metric topology.

I don't see this being made clear anywhere.

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

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

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

 

[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.

 

[Hurkyl]

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.

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.

 

[xaos]

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

 

[mathman]

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.

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.

 

[ice109]

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.

c'mon that's easy