Is there a distance $d$ on $\mathbb{R}$ so that a non-empty subset $\mathcal{U}$ of $\mathbb{R}$ is open wrt to $d$ if and only if its complement $\mathbb{R}\setminus\mathcal{U}$ is finite?

My observations so far: Let $\emptyset\neq\mathcal{U}\subset \mathbb{R}$. Denote the metric topology by $T(d)$. Then $$\mathcal{U} \in T(d) \iff \mathbb{R}\setminus\mathcal{U} \text{ finite} \implies\mathbb{R}\setminus{\mathcal{U}} \text{ compact}.$$ Since every finite set in a metric space is compact.

So the question becomes, is there a $d$ on $\mathbb{R}$ such that every closed set is compact?

  • 2
    $\begingroup$ There is certainly a metric for which every closed set is compact. E.g. take some bijection $\phi:\Bbb R\leftrightarrow [0,1]$. Then $d(x,y):=|\phi(x)-\phi(y)|$ has this property. A closed subset of a compact space is compact. And here we using the metric of $[0,1]$ (which is compact) on $\Bbb R$. But this says nothing about your initial question. $\endgroup$
    – M. Winter
    Jan 11, 2018 at 20:36
  • $\begingroup$ Ok, yes I see. I was hoping that would not be true and I would have a contradiction.. $\endgroup$ Jan 11, 2018 at 20:38
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    $\begingroup$ math.stackexchange.com/questions/145516/… $\endgroup$ Jan 11, 2018 at 20:40
  • $\begingroup$ The language in that question is unfamiliar to me, but i will do some searching for definitions now to try and understand it $\endgroup$ Jan 11, 2018 at 20:45
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    $\begingroup$ You are asking whether the cofinite topology is metrizable. The answer is no, because it is not Hausdorff. $\endgroup$ Jan 11, 2018 at 20:59

1 Answer 1


Let me try to answer the question with fewer (unexplained) technical terminology than in the linked question.

No. There is no such metric (i.e. "distance" as you say) on $\Bbb R$ so that all cofinite sets (sets with finite complement) are open.


We know that your kind of (non-empty) open sets must have infinitely many points. Therefore any two (non-empty) open sets $U$ and $V$ must intersect, because otherwise $V\subseteq \Bbb R\setminus U$ would be finite.

Now choose two distinct points $x,y\in\Bbb R$. If $\delta:=d(x,y)$ is their distance, then the both open sets

$$U_{\delta/2}(x)=\{z\in\Bbb R\mid d(x,z)<\delta/2\},\qquad U_{\delta/2}(y)=\{z\in\Bbb R\mid d(y,z)<\delta/2\}$$

would be disjoint. Contradiction. $\square$

  • $\begingroup$ Okay I understand this. After saying that $U$ and $V$ must intersect I understand that you can say that would not be a Haussdorf Space, thank you $\endgroup$ Jan 11, 2018 at 21:17

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