R with usual topology and Z with discrete topology are homeomorphic? I can't decide if the reals with the usual topology is homeomorphic to Z with the discrete topology. I know know that there must be uncountable sets in both topologies because the power set of z has uncountably many sets within itself and there are an uncountable number of reals but this doesn't get me very far. Is this something to do with compactness although I don't know how to use it? Are there standard ways of doing this? An explanation of what's going on here would be appreciated! Thanks
 A: If $(X,\tau_X)$ and $(Y,\tau_Y)$ are homeomorphic, then there is a bijection between the two spaces. Namely, there is a function $f\colon X\to Y$ which is a bijection, and satisfies that $f[U]$ is open if and only if $U$ is open (for $U\subseteq X$).
Now ask yourself, is there a bijection $f\colon\Bbb R\to\Bbb Z$?
A: Homeomorphic spaces share their topological properties.


*

*A discrete space with more than one element is not connected; the real line with the usual topology is connected.

*The compact subsets of a discrete space are finite, the real line has plenty of infinite compact subsets.

*Sequences with pairwise distinct terms are not convergent in a discrete space, they can be convergent in the real numbers.

*…
You seem to have unclear the concept of homeomorphism. If $(X,\tau)$ and $(Y,\sigma)$ are topological spaces, a homeomorphism between them is a map $f\colon X\to Y$ so that


*

*$f$ is bijective

*$f$ is continuous

*$f^{-1}$ is continuous


(continuity with respect to the considered topologies, of course). As a consequence, the map $f$ also induces a bijection between $\tau$ and $\sigma$: this bijection sends $U\in \tau$ to $f[U]\in \sigma$; it is well defined because, by hypothesis, $f^{-1}$ is continuous; the inverse is $V\mapsto f^{-1}[U]$, which is also well defined, because $f$ is continuous.
A: They are not homeomorphic. The topology of $Z$ is "bigger" than the usual topology of $\Bbb R$. Here "bigger" means $Z$ has more open sets than $\Bbb R$.
Now if they are homeomorphic, i.e. there exists: $f: \Bbb R\to Z$ which is a bijiction and continuous and $f^{-1}: Z \to \Bbb R$ is also continuous.
We pick $z\in Z$, as we know $\{z\}$ is open in $Z$, however, $f^{-1}(\{z\})$ is not open in the usual topology of $\Bbb R$.
