I could not prove the following question could you please help me?
Best Regards
Let $X, d(x, y)$ be a metric space. By definition, diameter of a
bounded set $A ⊂ X$ is the number $diam(A)$ = $sup${$d(a, b) : a, b ∈ A$}.
a) Suppose that $X$ is a complete and $A1 ⊃ A2 ⊃ A3 ⊃ · · · An ⊃ · · ·$
is a nested sequence of closed subsets of $X$. Prove, that if $diam (An )$ → $0$ then there is a unique point $a$ such that $a ∈ An$ for every $n$.
b) Give an example of a sequence $A1 ⊃ A2 ⊃ A3 ⊃ · · · An ⊃ · · ·$ of
closed subsets of $\mathbb{R}$ such that $diam (An ) ⇸ 0$ and $\bigcap{_n}$ $An = ∅$
c) b) Give an example of a sequence $A1 ⊃ A2 ⊃ A3 ⊃ · · · An ⊃ · · ·$ of
open subsets of $\mathbb{R}$ such that $diam (An ) → 0$ but $\bigcap{_n}$ $An = ∅$
Best Answer
Since this is a homework, I will give you some hints to lead you on the right track.
(a) For each $n\in\mathbb{N}$ choose $x_{n}\in A_{n}$. Can you show that this sequence converges, and that the limit is in $A_{n}$ for all $n\in\mathbb{N}$? Also, if there would be another such point $y$ in every $A_{n}$, then how would this contradict diam$(A_{n})\to 0$?
(b) Try to look for a nested sequence of unbounded closed sets. Bounded closed sets will not work, because Heine-Borel says that compact sets in $\mathbb{R}$ are those that are closed and bounded, and if each $A_{n}$ is compact, then it can be shown that $\bigcap A_{n}\neq\emptyset$ even if diam$(A_{n})\not\to 0$.
(c) Try to look for a nested sequence of open intervals where the other end stays fixed in each set, and choose the second ends so that diam$(A_{n})\to 0$.