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Let $X$ be an infinite set.

Then, is it possible to construct a metric space $(X,d)$ such that every closed set except the whole space $X$ is finite?

If possible, what would be the example of such $X$ and $d$? If not possible, why? Can you give me a proof?

User
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    The whole space is closed, so we need to change the wording somewhat. – André Nicolas Aug 16 '14 at 23:45
  • Even if you exclude $X$, this can't be, because the topology of a metric space is necessarily a Hausdorff (or T2) topology. – Thomas Andrews Aug 16 '14 at 23:51
  • @AndréNicolas Oh. I forgot to exclude $X$. Thank you. I editted the wording. – User Aug 16 '14 at 23:52
  • @ThomasAndrews im not familiar with the Hausdorff topology. Can you explain in other words? – User Aug 16 '14 at 23:53
  • A topology is said to be "Hausdorff" if for any two distinct points $x,y$ there are disjoint open sets $U,V$ with $x \in U, y \in V$. In plainer terms they can be separated by open sets. Any metric does this by simply letting $U,V$ be the balls of radius $d(x,y)/2$ about $x,y$ respectively. This terminology will make less sense if you've only ever seen metric topology. – jxnh Aug 18 '14 at 19:17

2 Answers2

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The whole space $X$ is closed, so the answer is no.

Now let us change the problem to ask whether it is possible for every closed set apart from $X$ to be finite.

Let $p$ and $q$ be distinct points. Let $\epsilon=d(p,q)/4$. Let $B_p$ be the open ball with centre $p$ and radius $\epsilon$, and define $B_q$ similarly.

If every closed set apart from $X$ is finite, then the complement of $B_p$ and the complement of $B_q$ are both finite. But the union of these complements is $X$, contradicting the fact that $X$ is infinite.

André Nicolas
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If $N_r(x)=\{y\in X:d(x,y)<r\}$ is the neighborhood of $x$ of radius $r$, then given $x\neq y$, let $r=\frac{d(x,y)}{2}$. Then $N_r(x)\cap N_r(y)=\emptyset$.

But $N_r(x)$ and $N_r(y)$ are non-empty and open, so they must be the complements of closed sets. If only finite sets are closed sets, can $N_r(x)^c\cup N_r(y)^c=X$?

Thomas Andrews
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