# Homeomorphism between Rationals and Natural numbers

Is $\mathbb{Q}$ homeomorphic to $\mathbb{N}$? Please explain.

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I just edited the question. Sorry it was homeomorphic not isomorphic. Thanks – chandu1729 Mar 15 '13 at 2:17
Please edit also the title. – Berci Mar 15 '13 at 2:17
It depends on what topology you're using for each space. – Christopher A. Wong Mar 15 '13 at 2:18
I disagree. A space already has a fixed topology (in contrast to a set). And in the question we are given spaces because otherwise it would be nonsense to ask if they are homeomorphic. The topologies on rational, natural numbers etc. are always the euclidean ones unless otherwise stated. This abuse of notation is necessary because otherwise we would get lost with brackets, tuples, etc. – Martin Brandenburg Mar 15 '13 at 2:30
I don't agree with you, @MartinBrandenburg: even if you're given the sets under the name spaces it is a valid, logical and even necessary question to ask under what topology. Perhaps in some courses/universities it is always assumed the topology is the inherited from euclidean one, but that doesn't need to be so everywhere. and there is no doubt the answer depends on the topology chosen for each set (or space, if you will) – DonAntonio Mar 15 '13 at 2:45

In $\mathbb{N}$, every point is isolated. This is not the case for $\mathbb{Q}$. Here no point is isolated.

By the way, there is a nice theorem due to Sierpinski which states that $\mathbb{Q}$ is in fact the only countable metrizable topological space without isolated points. This has some curious applications, for example $\mathbb{Q}$ is homeomorphic to $\mathbb{Q} \times \mathbb{Q}$.

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On top of what you mentioned, $\mathbb{Q}$ has properties that $\mathbb{N}$ doesn't like between every two rationals, there is another. But, as a beginner, I'm not quite sure how this directly implies that there is no homeomorphism between the two spaces. Could you explain? – agrasin Mar 27 '15 at 15:52
@agrasin: You say "between" - this refers to the order structure. This doesn't imply a priori that the topologies can be distinguished. But of course we can just say that any non-empty open subset of $\mathbb{Q}$ is infinite. – Martin Brandenburg Mar 27 '15 at 21:04
@agrasin: You say "between" - this refers to the order structure. This doesn't imply a priori that the topologies can be distinguished. But of course we can just say that any non-empty open subset of $\mathbb{Q}$ is infinite. This property is stable under homeomorphisms. – Martin Brandenburg Mar 27 '15 at 21:13
Okay, so there are two ways to show $\mathbb{Q}$ and $\mathbb{N}$ aren't homeomorphic: the isolated point argument in your post, and the one you just mentioned about non-empty open subsets of $\mathbb{Q}$ being infinite. How would you complete the proof? That is, how would you show that the isolated point property of $\mathbb{N}$ implies there is no homeomorphism? And how would you show that the other property of $\mathbb{Q}$ is stable under homeomorphisms? – agrasin Mar 27 '15 at 21:44
I won't write down the proofs, because they are very easy and just use the definitions. If $f : X \to Y$ is a homeomorphism and $x \in X$ is isolated, then $f(x) \in Y$ is isolated - you can prove this. By the way, there are probably hundreds of different proofs, that $\mathbb{N}$, $\mathbb{Q}$ are not homeomorphic, not just two. – Martin Brandenburg Mar 27 '15 at 22:46

No. The points (one point element sets) are open in $\Bbb N$, but not in $\Bbb Q$ (with its standard topology, inherited from $\Bbb R$).

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