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?
Best Answer
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.