If for metric space $(X,d)$ we have $d(A,B)>0$ for any pair of non-empty disjoint closed subsets $A$ and $B$. Show that $(X,d)$ is complete.

I am confused. If I let $A=N\subseteq R$ and $B=\{n+ \frac1n | n\in N, n\geq2\}$ then both $A$ and $B$ are disjoint and closed subsets of $(R, |\cdot|)$, which is complete but $d(A,B)=0$

So does this disproves the above claim?

EDIT: Looking carefully, the condition is not if and only if so this does not disproves the above claim.

Please check my proof:

Suppose $X$ is not complete $\rightarrow \exists (x_n)\in X $ that is Cauchy but not convergent. If the set $F=\{x_n \mid n\in N\} $ is finite, then $(x_n)$ has a constant subsequence and thus $(x_n)$ converges to that constant. So $F$ has to be infinite. Hence, we can extract a subsequence from $(x_n)$, say $(y_n)$ with all its terms distinct.

Let $G=\{y_{2n}\mid n\in N\}$ and $H=\{y_{2n+1}\mid n\in N\}$

Then $G$ and $H$ are disjoint, closed subsets of $X$ but $d(G,H)=0$ as $(y_n)$ is also Cauchy.

Is this proof okay?

  • 1
    $\begingroup$ Your proof is great. I think everything is accounted for. $\endgroup$
    – D. Brogan
    Oct 28, 2019 at 18:45
  • $\begingroup$ @D.Brogan thank you very much for verifying. It really means a lot to me $\endgroup$
    – chesslad
    Oct 28, 2019 at 19:17
  • 1
    $\begingroup$ The proof is nice, but you might want to point out why $G,H$ are actually closed. $\endgroup$
    – PhoemueX
    Oct 28, 2019 at 20:14
  • $\begingroup$ It is correct, and before I read it, I thought that i show one could answer your question $\endgroup$
    – Mirko
    Oct 29, 2019 at 2:22

1 Answer 1


This community wiki solution is intended to clear the question from the unanswered queue.

Your proof is correct. Perhaps you should add the observation that if a Cauchy-sequence has a cluster point $\xi$, then it converges to $\xi$. See If a Cauchy Sequence has an accumulation point, then it converges to said accumulation point.

This shows that $G,H$ are closed.


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