# Are $\{\frac{1}{n}|n∈\mathbb{N}\}\cup\{0\}$ and $\mathbb{Z}$ homeomorphic? [closed]

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 $$\mathbb{R}$$.

• One has a limit point, the other doesn't. – GEdgar Apr 8 at 1:20
• Just edited. They should both be metric spaces. – ssvnormandysr2 Apr 8 at 1:24
• 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. – C Squared Apr 8 at 1:24
• sorry for deleting. just wanted to be sure i was correct before saying that :) – C Squared Apr 8 at 1:25
• $\mathbb{Z}$ is not compact and $\{1/n:n\in\mathbb{N}\}\cup\{0\}$ is... – C Squared Apr 8 at 1:33

## 2 Answers

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 :)

• Oh, now I see it. I was a bit confused by the intuition, didn't get the relationship. Great answer, thank you. – ssvnormandysr2 Apr 8 at 1:39
• The Bonus material is deep and excellent!!! @Neal – srm99 Apr 8 at 1:49
• Exercise 1 is false in general, beware! – Henno Brandsma Apr 8 at 9:03
• @HennoBrandsma how bout now? – Neal Apr 10 at 14:14
• @Neal with injective added it’s OK. – Henno Brandsma Apr 10 at 14:16

$$0$$ is a limit point in the first set. But $$\mathbb Z$$ has none.

• 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. – ssvnormandysr2 Apr 8 at 1:36
• Well, using say the limit point definition of continuity, we easily get a contradiction. – Chris Custer Apr 8 at 1:39
• Another way: one space is discrete, the other isn't. The point $\{0\}$ isn't open in the first space. – Chris Custer Apr 8 at 1:41