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?
Best Answer
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).