I have the following question but without any correction so I will really appreciate a feedback on my answers.
Question:
Let be $K$ a non-empty compact metric space. We define $ \text{diam} (K) = \sup \left \{ d(a, b): a,b \in K \right \}$. Show that $\text{diam} (K) < \infty$ and $\exists a,b \in K$ such that $\text{diam} (K)= d(a, b)$.
My Answer:
Prove that $\text{diam} (K) < \infty$
By absurd we suppose that $\text{diam} (K) = \infty$. It means that there exists at least one strictly increasing sequence defined WLOG, as follow: $(u_n=d(a,b_n))_n$ verifying $u_n \underset{n \to \infty}{\rightarrow} \infty$ and this for any given fixe $a \in K$.
Hence we can not extract a convergent (in $K$) subsequence from $u_n$. And this contradict the definition of $K$ being a compact set.
Prove that $\exists a,b \in K$ such that $\text{diam}(K)= d(a,b)$
We writte $\text{diam}(K)=C$.
Now we define by reccurence the following sequence:
- We take $a_0,b_0$ as we want in $K$ and $u_0=d(a_0,b_0) \leq C$
- We choose $a_1, b_1 \in K$ s.t. $0 \leq C- d(a_0,b_0) < u_1=d(a_1,b_1) \leq C$. Such $a_1, b_1$ has to exists by the definition of $C$. Indeed it is not the case we get $d(a_0;b_0)=C$ (and this completes the proof).
- … same for all $u_n$…
So we have $u_n \underset{n \to \infty}{\rightarrow} C $ and the subsequence $(a_{nk} ; b_{nk})$ that must cvge to $(a;b)$ by the compactness of $K$ verifies too $u_{nk} \underset{k \to \infty}{\rightarrow} K $ by the uniqueness of the limit. $ \Rightarrow d(a_{nk},b_{nk}) \underset{n \to \infty}{\rightarrow} d(a,b)=C$
Is this correct? thank you.
Best Answer
We first prove that $\text{diam}(K) < \infty$.
Proof 1
Suppose not. Then, $\exists (a_n), (b_n) \in K$ such that $d(a_n,b_n) \uparrow \infty$.
Now observe that, for any $x \in K$, we have: $$d(a_n, b_n) \le d(a_n, x) + d(b_n,x)$$
In particular, LHS $\uparrow \infty \implies \sup \{d(a_n,x), d(b_n, x)\} = \infty$. Therefore, $K$ is not bounded. But every compact metric space is totally bounded, so this is a contradiction. $\square$
Proof 2
We continue from work done in the first proof.
As we showed, $\sup \{d(a_n,x), d(b_n, x)\} = \infty$ for any $x \in K$.
Therefore, if we fix an arbitrary $x \in K$, we may pick $u_k \in \{a_n\} \cup \{b_n\}$ such that $d(x, u_k) \to \infty$.
Suppose that $(u_k)$ has a convergent subsequence $u_{k_l} \to u \in K$. Then, by the continuity of the metric $d$,
$$d(x, u_{k_l}) \to d(x,u)$$
But $\lim_{l} d(x, u_{k_l}) = \infty$, so this is a contradiction. $\square$
Remark: I think a little more justification than you gave was also needed for the last part of this proof.
We now prove that the supremum $s := \text{diam}(K)$ may be attained.
By definition of the supremum, $\exists (a_n), (b_n) \in K$ such that $d(a_n,b_n) \uparrow s$.
Further, as $K$ is sequentially compact, we may find a subsequence $a_{n_k}$ which converges (say, to $a$). Now consider the sequence $(b_{n_k})$. By appeal to sequential compactness again, this further has a convergent subsequence(!) $b_{{n_{k_l}}} \to b$.
The good news is that we still have $a_{{n_{k_l}}} \to a$. Therefore, by the continuity of $d$, we may deduce that:
$$d(a, b) = \lim_l d(a_{{n_{k_l}}}, b_{{n_{k_l}}}) = s$$
so the supremum is attained.
Remarks: I changed my mind again; continuity of the metric is required to prove this! Also, note that it is necessary to apply sequential compactness twice here. It is not entirely obvious that there exists $(n_k)$ such that both $(a_{n_k}), (b_{n_k})$ simultaneously converge, as you asserted in your proof.
Proof suggested by Infinity Hunter
Let $K$ be compact. Then, by Tychonoff, $K \times K$ is also compact. Hence, the image of the continuous function $d : K \times K \to \mathbb{R}$ is bounded and attains its bounds. This implies exactly that $\text{diam}$ has a finite value which is achieved by some pair of points.
(If you are not familiar with any of the results I quoted, you should be able to find proofs for all of them in any textbook, or even on this site.)