# Proof about diameter of a set

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 $\bigcap{_n}$ $An = ∅$

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For (a), I would try picking an element from each set and using completeness to justify taking a limit and finding $a$. Then you'd have to worry about uniqueness. For (b), you can find an example where the diameter of each set is infinite if you think about it. You can do (c) using open intervals and a few minutes' thought. – user108903 Dec 16 '12 at 15:38
I guess this should be enough for a) Choose $x_n ∈ A_n$ , and show that the sequence $(x_n)$ satisfies the Cauchy condition. – Amadeus Bachmann Dec 16 '12 at 15:45
@user108903 I would really appreciate if you can show me how to prove (a) explicitly? – Xentius Dec 16 '12 at 16:03
Do you see why each set $A_n$ is non-empty? If so, start by choosing $x_n\in A_n$ for each $n$. Then you have to show that $(x_n)_{n\ge1}$ is Cauchy; this should come out of the condition $diam(A_n)\to 0$. Then apply completelness. I think it's better for you to work this out yourself if possible. If you understand the definitions of these concepts, each step shouldn't be difficult. – user108903 Dec 16 '12 at 16:04
@user108903 Could you please check my reply to ThomasE.'s answer. – Xentius Dec 17 '12 at 0:57

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$(A_{n})\to 0$.

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@Xentius. Since the sequence of sets is nested, then for any $n,m\in\mathbb{N}$, $n<m$, you have $d(X_{n},X_{m})\leq diam(A_{m})$, so if $n,m\to\infty$, then $d(X_{n},X_{m})\to 0$. Hence $(X_{n})$ is a Cauchy sequence and conveges to some $x\in X$ because $X$ is complete. You still have to show that $x\in\bigcap A_{n}$, and if $y\in\bigcap A_{n}$ then $y=x$. First use that each $A_{n}$ is closed and thus $X\setminus A_{n}$ are open. So if $x\notin A_{k}$ for some $k$‚ then you find $r>0$ s.t. $B(x,r)\subset X\setminus A_{k}$. Since $x_{n}\in A_{k}$ for all $n\geq k$, then what does this imply? – T. Eskin Dec 17 '12 at 8:05
@Xentius The trick is that the limit is outside $A_{k}$, while rest of the sequence continues inside $A_{k}$. I.e. since $x_{n}\to x$ you find $j$ s.t. $d(x_{n},x)<r$ for all $n\geq j$. Now choose $m=\max\{k,j\}$. Then $d(x_{n},x)>r$ for all $n\geq m$ (since $B(x,r)\subset X\setminus A_{k}$ and $x_{n}\in A_{k}$ for all $n\geq m$) and simultaneously $d(x_{n},x)<r$ for all $n\geq m$ since $x_{n}\to x$. This is an obvious contradiction, so we conclude that $x\in A_{n}$ for all $n$. Now what happens if there exists another $y\in A_{n}$ for all $n$? Since $d(x,y)>0$ and diam$(A_{n})\to 0$, then... – T. Eskin Dec 18 '12 at 6:20
I guess for b we can choose $A_n=[n,\infty)$ right? Their intersection will be empty but they are all nonempty closed sets. – Xentius Dec 18 '12 at 8:26
and for c) I should take $A_n=(0,1/n)$ ? – Xentius Dec 18 '12 at 8:28
@Xentius Yes, both are correct. – T. Eskin Dec 18 '12 at 8:43