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
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$\begingroup$ You seem to have unclear what a homeomorphism is: it is a bijective map from one space to the other, continuous and with continuous inverse. It's not a map between the topologies (that is, the families of open sets) of the spaces. $\endgroup$– egregOct 2, 2014 at 23:32
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3 Answers
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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 $f\colon\Bbb R\to\Bbb Z$?
answered Oct 2, 2014 at 23:18
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$\begingroup$ I don't see how that's relevant as we are mapping the sets, not the elements, I'm missing something here sorry $\endgroup$– MmmOct 2, 2014 at 23:23
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$\begingroup$ The homeomorphism is a function between the spaces which preserves properties of certain sets. It's not a function between power sets (although as a bijection, it can be seen as a function between power sets; but one which maps singletons to singletons bijectively, and therefore can be reduced back to a function between the original spaces). $\endgroup$– Asaf Karagila ♦Oct 2, 2014 at 23:27
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$\begingroup$ I think I am misunderstanding what the homeomorphism is mapping. Does it map an interval in the usual topology to a set of integers in the second? $\endgroup$– MmmOct 2, 2014 at 23:35
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$\begingroup$ Max, what is the definition of a homeomorphism? Please write it in a comment here. $\endgroup$– Asaf Karagila ♦Oct 2, 2014 at 23:35
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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.
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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.
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$\begingroup$ @AsafKaragila I added some other examples. But, of course, the argument about countability works perfectly. $\endgroup$– egregOct 2, 2014 at 23:31
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$\begingroup$ Yes, I noticed. I'm sorry, but I can only upvote this once. $\endgroup$– Asaf Karagila ♦Oct 2, 2014 at 23:31
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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 $\Bbb R$.
