Today in class we proved the following:
Let $(E,d)$ be a metric space.
E is Complete $\Rightarrow$ For every decreasing sequence $(A_n)_{n∈\Bbb N}$ of non empty subsets of E such that $\lim_{n\to\infty}diam(A_n)=0$, the set $\bigcap_{n\in\Bbb N} \bar A_n$ has one and only one element.
Supposedly this is Cantor's nested set theorem (or something like that, the name might be different depending on the source material). The $\Leftarrow$ proof was left as homework and the professor gave us a hint. Now, we defined a Complete Metric Space as one were every Cauchy Sequence converges so the hint was that for every $k∈\Bbb N$, given a Cauchy Sequence $(x_n)_{n∈\Bbb N} ⊆ E$, we should consider the set $A_k$ = {$x_n$: $n≥k$}, i.e., the set of the k-th tail of the sequence.
Now that I have this sequence the professor suggested, I guess I have to show that $(i)$ all of it's elements are non empty, $(ii)$ it's decreasing, (i.e. $A_n ⊆ A_{n-1}$ for all $n∈\Bbb N$) and $(iii)$ ${diam(A_n)\to 0}$ as ${n\to \infty}$. My question is: I'm not sure if I'm also supposed to prove that the intersection of the closures contains only one element or that if $(i), (ii)$ and $(iii)$ combined guarantee that.
Thanks in advance :)