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Let $(X,d)$ be a metric space and $K$ a compact set in $X$. If $x\in K$, let $\mathcal{O}_x$ be a neighborhood of $x$ with finite diameter. Since $K$ is compact, there are finitely many points $x_1,\dotsc,x_n$ in $K$ such that $$K\subseteq \bigcup_{\ell\leq n}\mathcal{O}_{x_\ell}=U.$$ Since all $\mathcal{O}_{x_\ell}$ have finite diameter, it follows that there exists a constant $M>0$ sufficiently large such that $\mathrm{diam}(\mathcal{O}_{x_\ell})\leq \frac{M}{2}$ for all $\ell\leq n$. Therefore if $x,y\in K$, then $x\in \mathcal{O}_{x_i},y\in\mathcal{O}_{x_j}$ for some $i,j\leq n$ and thus \begin{align*} d(x,y)&\leq d(x,x_i)+d(x_i,x_j)+d(x_j,y)\\ &\leq M+\sup_{k,\ell\leq n}\{d(x_k,x_\ell)\}. \end{align*}

My question is how can I show that $\sup_{k,\ell\leq n}\{d(x_k,x_\ell)\}$ exists without using sequences or the fact that a compact set is bounded? This seems obvious since there are finitely many $x_\ell$ but I don't know how to proceed.

Nerhú
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  • Bounded? It's a constant. It is fixed finite set of numbers - there are $n^2$ of them, and one of them is the largest. Maybe you just need to realize that even though $x$ and $y$ can change, the same value of $n$ works for all possible $x$ and $y$? – JonathanZ May 16 '23 at 21:00
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    The set $K \times K$ is compact and the map $d: K \times K \to \mathbb{R}$ is continuous hence has a $\max$. – copper.hat May 16 '23 at 23:05
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    A metric never assigns an infinite value to the distance between two points, it's part of the definition of a metric. – JonathanZ May 16 '23 at 23:09
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    As for what this constant is, we don't know, we just know that it has some value independent of the points $x$ and $y$. I noticed you said you don't know how to proceed - you proceed by observing that that quantity is independent of $x$ and $y$, and you are done. – JonathanZ May 16 '23 at 23:15
  • It's kind of like there is a limit to how high a shelf someone in your family can get something down from. We don't know what it is, but you've only got a finite number of people in your family, and none of them has an infinitely high reach. Without knowing more details we can't get any more specific. – JonathanZ May 16 '23 at 23:19
  • @JonathanZsupportsMonicaC oh I see, I was thinking that perhaps a metric space could be "infinitely large" in size, even though the notion of infinite distance is not well defined in this context. This is the answer I was looking for, thanks! – Nerhú May 17 '23 at 00:30
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    A metric space can be "infinitely large", but this means that for any given number you can always find a pair of points whose distance is larger than that number. But for any specific pair of points, the distance between them is a finite number. Does that make sense? – JonathanZ May 17 '23 at 00:39
  • @JonathanZsupportsMonicaC Yes, although I wonder if there is any way to determine how big a metric space is. For example, to show that a compact set $K$ is closed in $X$ we take a point $p\in X\setminus K$ and open balls $\mathcal{O}x$ centered at $x$ with radius $r_x=\frac{1}{2}d(x,p)$ for each $x\in K$. Then by compactness there is a finite number of elements $x_1,\dotsc,x_n$ in $K$ such that $$K\subseteq \bigcup{\ell\leq n} \mathcal{O}{x\ell}=U.$$ Then we define $V$ as the ball centered on $p$ by taking the smaller of the radii $r_{x_\ell}$ and noticing that $U\cap V=\emptyset$ – Nerhú May 17 '23 at 01:18
  • @JonathanZsupportsMonicaC we conclude that $p$ is an interior point of $X\setminus K$. The question is, if our metric space is small, what happens if $p$ is far enough from the set $K$ that some ball of radius $r_x$ falls out of $X$? Visualizing what is happening in this proof, it seems that a metric space must be large enough to contain all these balls, is this correct? – Nerhú May 17 '23 at 01:20
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    @copper.hat Yes. Or, if the fact that the product of two compact spaces is compact is not available, just choose a point $x_0$, and $diam(K)\le2\cdot\max{d(x,x_0):x\in K}$. – bof May 17 '23 at 03:50
  • @copper.hat It seems to me that this argument is circular, since in order to prove that a continuous function on a compact set attains its maximum one needs to prove that a compact set is closed and bounded. In this question what is being proved is basically that $K$ is bounded. – Nerhú May 17 '23 at 04:27
  • @Nerhú There is no circularity in the argument. I have no idea where you get that from. The question assumes that $K$ is compact. – copper.hat May 17 '23 at 04:35
  • @copper.hat https://math.stackexchange.com/questions/109548/x-compact-metric-space-fx-rightarrow-mathbbr-continuous-attains-max-min – Nerhú May 17 '23 at 04:51
  • @Nerhú Why did you send me that link? That is a standard result. – copper.hat May 17 '23 at 05:40
  • @copper.hat what I mean is that to use what you say (that a continuous function defined on a compact set attains its maximum) we need to prove it using the definition of compact set and continuous function. We can't use the fact that $K$ is bounded to prove it (like in the post I sent) since that's exactly what I'm trying to prove by saying that the diameter is finite. I know that what you say can also be proved with sequences and subsequences, but I can't use it either since it is a later result. If you can prove what you say using only the definitions it would be very helpful. – Nerhú May 17 '23 at 06:17
  • You have a finite list of finite numbers, so one of them is greater than the others. – Didier May 17 '23 at 12:04
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    @Nerhú I see what you are saying. Try this, pick some $k_0 \in K$ then the collection $B(k_0,n)$ is an open cover hence there is a finite subcover and so for some $n_0$ we have $K \subset B(k_0,n_0)$. Then the triangle inequality shows that $d(x,y) \le 2n_0$ for all $x,y \in K$. – copper.hat May 17 '23 at 16:47

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The set $\{d(x_k, x_l)\}$ is a finite set of real numbers, say $\{y_1, \dots, y_r\}$. It has supremum $$\max \{y_i\} = \max(y_1, \max(y_2, \max(\dots \max(y_{r-1}, y_r)\dots ))).$$

ronno
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