Show that $x,y \in A$ exist such that $d(A) = d(x,y)$.

Question

Let $(X,d)$ be a metric space and let $A\subseteq X$ be a non empty compact subset.
Define $d(A) = \sup \{d(x,y):x,y\in A\}$.

Exercise: Show that $x,y\in A$ exist such that $d(A) = d(x,y).$

I know that:

Since $A$ is a non empty compact subset, $A$ is closed.
$A$ is closed implies that if $B_\epsilon(x) \cap A \neq \emptyset$ for every $\epsilon >0$, then $x\in A$.

My solution:

Showing that $x,y \in A$ exist such that $d(A) = d(x,y)$, is equivalent to showing that $\inf(A) \in A$ and $\sup(A) \in A$, right? Take $a = \inf(A)$ and suppose that $B_\epsilon(a) \cap A = \emptyset$. We can find $\bar{a} = a + \epsilon/2$ such that $\bar{a} > a$ and $\bar{a} \notin A$, since $\bar{a} \in B_\epsilon(a)$. This is a contradiction because this would mean that $a$ is not the infimum of $A$. Therefore $B_\epsilon(a) \cap A \neq \emptyset$, and $a\in A$. The same reasoning can be used for the supremum of $A$.

Question: Is my solution correct? Or should I be more elaborate/careful when concluding that proving the exercise is essentially showing that $\inf(A) \in A$ and $\sup(A) \in A$?

$\inf$ and $\sup$ only make sense if there is an order. – drhab Nov 04 '17 at 14:37 — drhab, Nov 04 '17 at 14:37

score 1 · Answer 1 · answered Nov 04 '17 at 14:12

1

It is not correct. It makes no sense to assert that $\inf(A),\sup(A)\in A$, since $\inf(A)$ and $\sup(A)$ are real numbers and $A$ is a subset of an arbitrary metric space.

answered Nov 04 '17 at 14:12

José Carlos Santos

427,504

So do infimums and supremums only exist for sets that are subsets of $\mathbb{R}$? – titusAdam Nov 04 '17 at 14:17
@titusAdam More generally, they only exist on ordered sets. – José Carlos Santos Nov 04 '17 at 14:18
This is the best answer on "Is my solution correct?" But...downvoted?! I really don't understand. – drhab Nov 04 '17 at 14:41
@drhab Me neither. It's discouraging. – José Carlos Santos Nov 04 '17 at 14:43
@JoséCarlosSantos What's discouraging? I didn't downvote your answer! – titusAdam Nov 04 '17 at 15:37
@titusAdam I don't know who downvoted my answer, and I surely never thought that it was you. What's discouraging is that someone downvoted my answer. – José Carlos Santos Nov 04 '17 at 15:38

score 0 · Answer 2 · answered Nov 04 '17 at 14:25

If $A \subseteq X$ is compact then $A \times A \subseteq X \times X$ is also compact in the product topology: see here for example.

The metric $d : X \times X \to\mathbb{R}$ is a continuous function so it attains its maximum on the compact subset $A \times A$.

score 0 · Answer 3 · edited Nov 04 '17 at 14:28

0

Suppose $y= d(A) = \sup \{d(x,y):x,y\in A\}$ so, for every $n\in N$ there exists $x_n, z_n \in A$ such $|y-d(x_n,z_n)| \leq \frac1n$. $A$ is compact so there is $x\in A$ such $x_n \to x$ and $z\in A$ such $z_n\to z$. You can easily see that $y= d(x,z)$.

edited Nov 04 '17 at 14:28

mechanodroid

46,490

answered Nov 04 '17 at 14:25

GhD

926

score 0 · Answer 4 · answered Nov 05 '17 at 19:38

Let's use compactness directly. For every $n \geq 1$ there are $x_n,y_n \in A$ such that $d(x_n,y_n) >d(A) + 1/n$. By compactness, there is a convergent subsequence $(x_{n_k})_k$ of $(x_n)_n$, converging to $x^* \in A$. Again by compactness, there is a convergent subsequence $(y_{n_{k_\ell}})_\ell$ of $(y_{n_k})_k$ converging to $y^* \in A$. Taking the limit in the first estimative, we get $d(x^*, y^*) \geq d(A)$. However, $d(x^*,y^*) \leq d(A)$ by definition of $d(A)$. We conclude that $d(x^*,y^*) = d(A)$.

Show that $x,y \in A$ exist such that $d(A) = d(x,y)$.

4 Answers4