Diameter

Definition

Let $A\subset X$. Then

$$\text{diam}(A):=\sup \{d(x,y)\mid x,y\in A\}$$

Remark

This generalizes the diagonal from Heine-Borel Theorem.

Remark (Boundedness)

We have that $\forall x,y\in A,d(x,y)\in \mathbb{R}$. So,

$$B=\{d(x,y\mid x,y\in A\}\subset \mathbb{R}$$

If it is unbounded, then

$$\text{diam}(A)=\sup \{d(x,y)\mid x,y\in A\}=+\infty$$

Remark (Cauchy Sequences)

Let $(p_n)$ be a Cauchy Sequence. Let $\epsilon >0$. Then $\exists N\in \mathbb{N}$ such that $\forall n\ge N$, $d(p_n,p_m)<\epsilon$. That is, if

$$A_N:=\{p_n\mid n\ge N\}=\{p_{N+1},p_{N+2},\cdots \}$$

If $x,y\in A_N$, then $d(x,y)<\epsilon$, such that it is an upper bound. Then

$$\text{diam}(A_N)\le \epsilon$$

We note that

$$A_1\supset A_2\supset A_3\supset \cdots$$

and $\forall \epsilon >0,\exists N\in \mathbb{N}$ such that $\text{diam}(A_N)\le \epsilon$. Indeed, this implies

$$\underset{N\in \mathbb{N}}{\inf }\text{diam}(A_N)=0$$

Lemma (Cauchy - Diameter)

$(p_n)$ is Cauchy $⟺$ ${\inf }_N\text{diam}(A_N)=0$. This is proven from before.

Lemma

Let $A\subset X$ be nonempty.

1. $\text{diam}(A)=\text{diam}(\overline{A})$
2. Let $K_1\supset K_2\supset \cdots$, be nonempty nested Compact Sets. Assume $\text{diam}(K_n){\to }_{n\to \infty }0$. Then

$$\bigcap _{n=1}^{\infty }{K}_n=\{p\}$$

Proof: Note that since $A\subset \overline{A}$ we have

$$\{d(x,y)\mid x,y\in A\}\subset \{d(x,y)\mid x,y\in \overline{A}\}$$

which implies $\text{diam}(A)\le \text{diam}(\overline{A})$. More generally, $A\subset B⟹\text{diam}(A)\le \text{diam}(B)$. In particular, $\text{diam}(A)=\infty ⟹\text{diam}(\overline{A})=\infty$ and so suppose $\text{diam}(A)<\infty$.

Then we want to show that $\forall \epsilon >0,$

$$\text{diam}(\overline{A})\le \epsilon +\text{diam}(A)$$

To see this, let $\epsilon >0$. By definition of $\overline{A}$, we have that $\forall p,q\in \overline{A},\exists p^{\prime },q^{\prime }\in A$ such that

$$d(p,p^{\prime })<\epsilon /2d(q,q^{\prime })<\epsilon /2$$

then that implies

$$\begin{array}{rl}d(p,q)& \le d(p,p^{\prime })+d(p^{\prime },q^{\prime })+d(q^{\prime },q)\\ & \le \epsilon +d(p^{\prime },q^{\prime })\\ & \le \epsilon +\text{diam}(A)\end{array}$$

Then by the Supremum, we get

$$\text{diam}(\overline{A})\le \epsilon +\text{diam}(A)$$

Therefore $\text{diam}(A)=\text{diam}(\overline{A})$.

For the second part, suppose $(k_n)$ are decreasing nonempty Compact Sets where

$$\bigcap _{n\in \mathbb{N}}{k}_N\ne \varnothing$$

For contradiction, suppose there are at least two points. That is, we have $q,p$ and $q\ne p$. Then $p,q\in k_n$ for some $n\in \mathbb{N}$. We see that

$$\text{diam}(k_n)\ge d(p,q)>0$$

We use the fact that $\text{diam}(k_N)\to 0$ since each $k$ is a subset, so the Diameter must be decreasing. Let $\epsilon =d(p,q)>0$ since $p\ne q$. Then $\exists N\in \mathbb{N}$ where

$$d(p,q)=\epsilon >\text{diam}(k_N)\ge d(p,q)$$

and so we have a contradiction.