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Would anyone be able to explain why not?

I feel like you can show it using sequences, but I'm not sure how.

Usual metric of $\mathbb{R}$.

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    $\begingroup$ One has a limit point, the other doesn't. $\endgroup$ – GEdgar Apr 8 at 1:20
  • $\begingroup$ Just edited. They should both be metric spaces. $\endgroup$ – ssvnormandysr2 Apr 8 at 1:24
  • $\begingroup$ It depends on the topology. If they both have discrete topology, then any bijection will do, since any function from a set with the discrete topology to the another set also with discrete topology is continuous. $\endgroup$ – C Squared Apr 8 at 1:24
  • $\begingroup$ sorry for deleting. just wanted to be sure i was correct before saying that :) $\endgroup$ – C Squared Apr 8 at 1:25
  • $\begingroup$ $\mathbb{Z}$ is not compact and $\{1/n:n\in\mathbb{N}\}\cup\{0\}$ is... $\endgroup$ – C Squared Apr 8 at 1:33
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The intuition is in the comments:

One has a limit point, the other doesn't

So let's run with this and write a proof by contradiction: We'll show that any bijection $$h: \bigg\{\frac{1}{n}\ \bigg|\ n=1,2,\ldots\bigg\}\cup\{0\} \to \mathbb{Z}$$ cannot be continuous. We will apply the intuition by focusing on the limit point $0$ and its image under $h$.

For convenience denote $X = \bigg\{\frac{1}{n}\ \bigg|\ n=1,2,\ldots\bigg\}$.

Let $p = h(0)$. The set $\{p\}$ is open in $\mathbb{Z}$. But its preimage $h^{-1}(\{p\}) = \{0\}$ is not open in $X$ because every open set of $X$ that contains $0$ also contains at least one other point.

Since a homeomorphism is, in particular, a continuous bijection, there is no homeomorphism between $X$ and $\mathbb{Z}$.


Bonus material

An accumulation point of a topological space $X$ is a point $p\in X$ such that every open set $U$ containing $p$ also contains at least one other point of $X$, that is, for every open $U\ni p$ there exists $q\in X$ with $q\neq p$ and $q\in U$.

Exercise 1: Let $X,Y$ be topological spaces and let $f:X\to Y$ be continuous and injective. Show that if $p\in X$ is an accumulation point of $X$, then $f(p)$ is an accumulation point of $Y$.

Exercise 2: Apply Exercise 1 to your question :)

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    $\begingroup$ Oh, now I see it. I was a bit confused by the intuition, didn't get the relationship. Great answer, thank you. $\endgroup$ – ssvnormandysr2 Apr 8 at 1:39
  • $\begingroup$ The Bonus material is deep and excellent!!! @Neal $\endgroup$ – srm99 Apr 8 at 1:49
  • $\begingroup$ Exercise 1 is false in general, beware! $\endgroup$ – Henno Brandsma Apr 8 at 9:03
  • $\begingroup$ @HennoBrandsma how bout now? $\endgroup$ – Neal Apr 10 at 14:14
  • $\begingroup$ @Neal with injective added it’s OK. $\endgroup$ – Henno Brandsma Apr 10 at 14:16
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$0$ is a limit point in the first set. But $\mathbb Z $ has none.

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  • $\begingroup$ Would you be able to expand on that? I'm not sure I see the relationship between limit points and homeomorphisms as bijective, continuous functions with continuous inverse. $\endgroup$ – ssvnormandysr2 Apr 8 at 1:36
  • $\begingroup$ Well, using say the limit point definition of continuity, we easily get a contradiction. $\endgroup$ – Chris Custer Apr 8 at 1:39
  • $\begingroup$ Another way: one space is discrete, the other isn't. The point $\{0\} $ isn't open in the first space. $\endgroup$ – Chris Custer Apr 8 at 1:41

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