My proof is:

Let $x$ be any point in $\mathbb{N}$, and let $\beta$ be the closest number (such as a factor or irrational number) to $x$, such that $\beta \notin \mathbb{N}$ and $\beta > x$.

Let $B_\epsilon(x)$ be the epsilon ball around $x$ in $\mathbb{N}$. Let $\epsilon = \beta - x$.

Clearly $\forall x \in \mathbb{N}: \exists \epsilon >0, \epsilon \in \mathbb{R}: B_\epsilon \left({x}\right) \subseteq \mathbb{N}$. So we have that all singleton sets in $\mathbb{N}$ are open.

All subsets of $\mathbb{N}$, except the empty set, consist of a union of many such $x \in \mathbb{N}$ and since all unions of open sets are open, we have that all subsets of $\mathbb{N}$ are open.

This means that the complement to every subset in $\mathbb{N}$ is closed. However, since the complement to any subset of $\mathbb{N}$ is itself a subset of the space, we have that all subsets of $\mathbb{N}$ are closed.

Therefore, all subsets of $\mathbb{N}$ are clopen.

Is this correct?

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    $\begingroup$ Surely $x + \frac{\beta - x}{2}$ is closer to $x$ than $\beta$ is, showing that no such $\beta$ exists. If $x$ is $3$ what is $\beta$ supposed to be? $\endgroup$ – user4894 Feb 7 '17 at 5:47
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    $\begingroup$ I don't really understand your definition of $\beta$. Why not just take $\varepsilon=\frac{1}{2}$? $\endgroup$ – carmichael561 Feb 7 '17 at 5:47
  • $\begingroup$ I see this now, very dumb mistake. Thanks. $\endgroup$ – jackson5 Feb 7 '17 at 5:51
  • $\begingroup$ @carmichael561 if you'd like to post this as an answer, I will gladly accept it. $\endgroup$ – jackson5 Feb 7 '17 at 5:52

There is no closest number to $x$, there is always another number between $x$ and any $\beta$ you come up with.

However you have the right idea about showing the singletons are open, just use $\epsilon = \frac{1}{2}$.


Just show all singletons are open, which is easy, as all natural numbers are at least 1 apart: $\{n\} = B_\frac{1}{2}(n)$ where the ball is taken in the restricted metric. If all singletons are open, all subsets $A \subseteq \mathbb{N}$ are open, as $A = \cup\{\{n\}: n \in A\}$ is a union of open sets, so open.

This also makes all subsets closed, as all complements are open as well.

  • $\begingroup$ Just to test my understanding: since we have $\{ y \in \mathbb{N} : d(x,y) \leq 1 \} = \mathbb{N}$, doesn’t that mean $\mathbb{N}$ is clopen? $\endgroup$ – Taylor Rendon Dec 1 '20 at 16:08
  • $\begingroup$ I was just thinking about this since for the closed ball of radius $1$, the ball itself is equal to the whole space. $\endgroup$ – Taylor Rendon Dec 1 '20 at 19:15
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    $\begingroup$ @TaylorRendon yes and the whole space is always clopen. $\endgroup$ – Henno Brandsma Dec 1 '20 at 19:39
  • $\begingroup$ So when we say the whole space is always clopen, we mean its clopen in itself? For example, $\mathbb{Q}$ is neither closed nor open in $\mathbb{R}$ but it is clopen in itself? $\endgroup$ – Taylor Rendon Dec 1 '20 at 20:08
  • $\begingroup$ @TaylorRendon yes it’s clopen in the induced subspace topology. The axioms of topology state that $X$ and $\emptyset$ are always open ( and thus closed too). $\endgroup$ – Henno Brandsma Dec 1 '20 at 21:41

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