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

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.

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?

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

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.