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Prove that $\mathbb{Q}$ isn't homeomorphic to $\mathbb{Z}$

I know that a homeomorphism is a bijection function from two sets where both the function and its inverse are continuos.

So I figure that if I show that there is no continuos invective mapping from $\mathbb{Q}\rightarrow\mathbb{Z}$ then I have done the proof. But I do not know how to do this (if his is the right way). How can I start to show this?

Burgundy
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    They can't be homeomorphic since $\mathbb{Z}$ has the discrete topology but $\mathbb{Q}$ does not. This can easily seen since every interval $(a,b)$ contains an infinite number of elements from $\mathbb{Q}$, while for $n\in \mathbb{Z}$ the intervals $(n-\frac{1}{2},n+\frac{1}{2})$ contains only one element from $\mathbb{Z}$. – Nex Nov 10 '15 at 19:11
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    "Homeomorphic" depends on the topology, and different topologies can be given to $\mathbb{Q}$. Thinking about the answer below, we know there is a bijection between $\mathbb{Q}$ and $\mathbb{Z}$, and so the spaces will be homeomorphic if $\mathbb{Q}$ has the discrete topology... But yes, with the usual standard topology, the answer is a nice easy proof. – charlestoncrabb Nov 10 '15 at 19:17
  • Possible duplicate of $\mathbb{Z}$ and $\mathbb{Q}$ are not homeomorphic – Dietrich Burde Nov 15 '17 at 10:09

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Assume that there is a continuous bijective map $f \colon \mathbb{Q} \rightarrow \mathbb{Z}$. Then since singletons are open in $\mathbb{Z}$, you see that each singleton $\{ q \} = f^{-1}(\{ f(q) \})$ should be open in $\mathbb{Q}$ which means that $\mathbb{Q}$ has the discrete topology. This is obviously false (since the open sets in $\mathbb{Q}$ are of the form $U \cap \mathbb{Q}$ where $U$ is open in $\mathbb{R}$ - $U$ contains an open interval and every open interval contains infinitely many rationals).

levap
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