I am trying to solve the following question:
Question: Let $(X,d)$ be a metric space such that $d(A,B)>0$ for every pair of disjoint closed subsets $A$ and $B$. Show that $(X,d)$ is complete.
My Attempt :
Let $(x_n)_n$ be a Cauchy sequence with distinct terms in $X$ which is not convergent in $X$. Now I am trying to arrive a contradiction…and then our result will be done…!!!
To do this I consider the sets
$A=\{x_{2k}| k \in \mathbb{N} \}$ and $B=\{ x_{2k+1}|k \in \mathbb{N} \}$
then Cauchyness of $(x_n)_n$ imply $d(A, B)=0$ and also $A$ and $B$ are disjoint.
Now If A and B are closed sets then we are done…!!!!
But I cannot prove these sets to be closed in $X$.
I know the range set of a sequence is not closed, in general, in a metric space.
Am I on the right track to prove the theorem? If $A$ and $B$ are closed, how can I prove it here…? If this procedure is not in the right way, suggest me any other solution.
And sorry if my question is so classic….!!
Thank you.!!
Best Answer
Cauchy sequences are convergent iff it has a convergent subsequence.
Now the two sets you defined, has no limit points, for if $A$ has limit point, say $y$, there is a subsequence of $A$ that converges to $y$, which is also a convergent subsequence of $(x_n)$, but since $(x_n)$ is a Cauchy sequence which is not convergent in $X$, then no subsequence of it can converge, hence a contradiction.
So $A$ and $B$ have no limit points, so the closure of $A$ is $A$ and closure of $B$ is $B$, and closures are closed sets, and hence we got two disjoint closed sets whose distance is $0$.