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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
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It depends on what topology you're using for each space. –  Christopher A. Wong Mar 15 '13 at 2:18
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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
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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

2 Answers 2

up vote 11 down vote accepted

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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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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