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My question is to consider $\mathbb{Z} \subseteq \mathbb{R}$, with the subspace metric. What are the open subsets of $\mathbb{Z}$? The answer to this is that any subset of $\mathbb{Z}$ is open. I am struggling to prove this. Can anyone help me out? I want to prove this using the metric space characterization of open sets:

Let $(X,d)$ be a metric space. A subset $U \subseteq X$ is an open set if for each $x \in U$ there exists $r > 0$ such that $B(x,r) \subseteq U$. Where $B(x,r)$ is the open ball with radius $r$ and centre $x$.

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    How is one expected to prove a topology statement without any topology? – Lee Mosher Feb 14 '18 at 17:08
  • How do you define an open set if not via topology? How do you define the subspace metric? Write down the definitions, then see what you can do with them. – Xander Henderson Feb 14 '18 at 17:15
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    Sorry! I'm currently doing a metric spaces and basic topology course and we haven't quite gotten to the topology section yet. Everything i found online to help was using the word topology. I didn't realise that was a topological statement. –  Feb 14 '18 at 17:15
  • Let (X,d) be a metric space. A subset U ⊂ X is an open set if for each x ∈ U there exists r > 0 such that B(x,r) ⊂ U. Where B(x,r) is the open ball with radius r and centre x. –  Feb 14 '18 at 17:16
  • @pullofthemoon I've edited your question to remove the "without any topology" line. Is this what you meant to ask? –  Feb 14 '18 at 20:47

2 Answers2

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Hint:

Since every set can be written as a union of singletons, you should try to prove that singletons are open sets.

Hint 2:

For a singleton $\{n\}\subseteq\mathbb Z$, try to find an open set $O$ in $\mathbb R$ such that $O\cap\mathbb Z=\{n\}$

Hint 3:

Think of simple candidates for $O$.

5xum
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  • Is the union of open sets open? And so if you prove each {$n$} where $n \in \mathbb{Z}$ is open then that implies every subset of $\mathbb{Z}$ is open? –  Feb 14 '18 at 17:19
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    "Is the union of open sets open?"... you should already know the answer to this question if you are solving the problem you posted. – 5xum Feb 14 '18 at 17:23
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Hint: remember that the open ball $B(x, r)$ of radius $r$ about $x$ depends on the subspace: for example, in the subspace $\Bbb{Z}$ of the metric space $\Bbb{R}$, $B(0, 2)$ means something different from what it means in $\Bbb{R}$: in $\Bbb{Z}$ it means all the integers at distance less than $2$ from $0$, not all the real numbers with that property. So in $\Bbb{Z}$, $B(0, 2)$ has just three elements. Now can you think of a $d > 0$ such that $B(i, d)$ (in $\Bbb{Z}$) is contained in any subset that contains the integer $i$?

Rob Arthan
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  • Okay, so say we have the open ball $B(n,1/2)$ such that $n$ is any integer. Then we have the open ball $B(n,1/2}={n}$ in $\mathbb{Z}$. Since open balls are open we know that {$n$} is therefore open. Any subset of $Z$ will be a union of singletons {$n$} and since every union of open sets is open then every subset of $\mathbb{Z} is open. –  Feb 15 '18 at 11:28
  • If you have now learnt that unions of open sets are open, then, yes. My answer was intended to give you a proof that does not require that knowledge. – Rob Arthan Feb 16 '18 at 23:55