2
$\begingroup$

Let $(X,d)$ be a compact metric space. Fix some $\epsilon >0$. Then it is clear that any set $S\subset X$ such that for all $x,y \in S$ one has that $d(x,y) > \epsilon$ is finite. In fact, an infinite one would contain a sequence with no convergent subsequence contradicting the compactness of $X$.

But now one can ask, whether there is a constant $N$, such that any subset $S$ of $X$ with that property has at most $N$ elements.

Unfortunately I have no clue how to prove this or find a counterexample. I'm thinking of the following (somewhat analogous) question where there is no such $N$:

Let $X$ be a noetherian scheme. Then its topological space is noetherian meaning that every descending chain of closed subsets of $X$ stabilizes. However this does not imply that dim$(X)$ is finite. (see for example here: https://mathoverflow.net/questions/21067/noetherian-rings-of-infinite-krull-dimension)

So my question is, if there exists such a bound $N$ in general (i.e. for arbitrary compact metric spaces) or if there are nice conditions under which such a bound exist.

Moreover I would be interested if there is a way to build up a dimension theory using the minimal bound. (It seems to be similar to the following: https://en.wikipedia.org/wiki/Equilateral_dimension, but without the assumption that all distances coincide. Here one could also ask, whether the best possible choice is always given by equilateral points, I'm also not sure about that.)

To finish I want to add that I'm not familiar with the notions of nets, filters and ultrafilters (and so on), so I would appreciate, if a solution would be more elementary if possible (if not, I'm willing to accept that I have to learn about those things first..)

  • 0
    Certainly not for all compact metric spaces - you may at least bound the diameter. But even then, an $n$-dimensional hypercube with supremum norm contains a set of size $2^n$ (the vertices) such that any two points have distance equal to the diameter. – 2017-01-11
  • 0
    I don't understand how a compact metric space can have unbounded diameter. The map $X \times X \to \mathbb{R}$ given by $(x,y) \mapsto d(x,y)$ is continuous and as $X \times X$ is compact, it is bounded. – 2017-01-12
  • 0
    Seems I misinterpreted the order of quantifiers you intended, i..e, that you wanted $N=N(\epsilon)$ [which does not work] and not $N=N(X,\epsilon)$ [which works, see accepted answer]. My previous comment shows that the "intermediate" interpretation $N=N(\operatorname{diam}(X),\epsilon)$ won't work – 2017-01-12

2 Answers 2

2

I think you are asking whether for a given compact metric space $(X, d)$ and a given $\epsilon > 0$ there is an $N$ such that subsets $S$ of $X$ with more than $N$ elements are guaranteed to contain elements that are within $\epsilon$ of each other. If so, then this is true: by compactness, there is a finite set of points $x_1, \ldots, x_N \in X$ such that the open balls $B_i = B(x_i, \epsilon/2)$ cover $X$. Then, if $S$ is any subset of $X$ with more than $N$ elements, the pigeon hole principle implies that there are two distinct elements $x, y \in S$ that belong to the same open ball $B_i$ and hence have $d(x, y) < \epsilon$.

2

There is always such an upper bound. Suppose not. Then for each $n\in\Bbb Z^+$ there is an $S_n\subseteq X$ of cardinality $n$ such that $d(x,y)>\epsilon$ whenever $x,y\in S_n$ and $x\ne y$. It follows (with a little work) that there are points $x_{m,n}$ for $1\le m\le n$ such that $d(x_{k,n},x_{\ell,n})>\epsilon$ whenever $1\le k<\ell\le n$.

There is an infinite $M_1\subseteq\Bbb Z^+$ such that the sequence $\langle x_{1,n}:n\in M_1\rangle$ converges to some $y_1\in X$. Given an infinite $M_m\subseteq\{k\in\Bbb Z^+:k\ge m\}$, there is an infinite

$$M_{m+1}\subseteq M_m\cap\{k\in\Bbb Z^+:k\ge m+1\}$$

such that $\langle x_{m+1,n}:n\in M_{m+1}\rangle$ converges to some $y_{m+1}\in X$; in this fashion we construct an infinite decreasing nest of sets $M_m$ for $m\in\Bbb Z^+$.

There is an $M\subseteq\Bbb Z^+$ such that $M\setminus M_m$ is finite for each $m\in\Bbb Z^+$. (Such a set can easily be constructed recursively.) Clearly $\langle x_{m,n}:n\in M\rangle\to y_m$ for each $m\in\Bbb Z^+$. Suppose that $k,m\in\Bbb Z^+$ with $k\epsilon$ for each $n\ge m$, so $d(y_k,y_m)\ge\epsilon$. But then $\{y_m:m\in\Bbb Z^+\}$ is an infinite, discrete subset of $X$, which is impossible.