# The integers and the rationals are homeomorphic?

We know since $\mathbb{Q}$ is countable that there exist a bijection $f : \mathbb{Z} \to \mathbb{Q}$. If we view $\mathbb{Q}$ and $\mathbb{Z}$ are topological subspaces of $\mathbb{R}$, are theo homeomorphic??

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Which topology do you consider? Subsets of $\mathbb{R}$ or another? – Duronman Oct 29 '13 at 10:24
the induced topology – ILoveMath Oct 29 '13 at 10:26
Is $\{17\}$ an open subset of $\mathbb{Q}$, under the induced topology? – André Nicolas Oct 29 '13 at 10:28
It’s true that it’s not open, but that has nothing to do with its being closed: a set can be both open and closed. For example, $\{q\in\Bbb Q:q^2<2\}$ is clopen (both open and closed) in $\Bbb Q$. – Brian M. Scott Oct 29 '13 at 10:34
In the induced topology on $\mathbb{Q}$, they are not open. Under the induced topology on $\mathbb{Z}$, they are open. – André Nicolas Oct 29 '13 at 10:36

Solution 1: Is $\mathbb{Q}$ a discrete topological space?

Solution 2: When does a sequence in $\mathbb{Z}$ converge?

Solution 3: Is $\mathbb{Q}$ locally compact?

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First pick a topology. So in this case I assume its the induced topology. Now any topological invariant will give an obstruction to a homeomorphism. For example, $\mathbb{Q}$ is everywhere dense in $\mathbb{R}$, but $\mathbb{Z}$ is not. So the answer is no, they are not homeomorphic.

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See Seirios below for more examples of topological obstructions to constructing a homeomorphism. – Don Shanil Oct 29 '13 at 10:39
“$\mathbb{Q}$ is everywhere dense in $\mathbb{R}$” does not give an obstruction to a homeomorphism! It’s not an intrinsic topological property of $\mathbb{Q}$, but rather a property of the embedding of $\mathbb{Q}$ into $\mathbb{R}$. For instance, $\mathbb{Q} \setminus [0,1]$, with its induced topology from $\mathbb{R}$, is not dense in $\mathbb{R}$ but is homeomorphic to $\mathbb{Q}$. – Peter LeFanu Lumsdaine Oct 29 '13 at 16:03
Yes, you are correct! Sorry my mistake. Thanks for pointing this out. – Don Shanil Oct 30 '13 at 3:28