21
$\begingroup$

Back in college I tried to find characterizations of compact (metric) spaces that make use of Lebesgue's Covering Lemma. During the course of this, I defined the following (pretty arbitrarily named) property of subsets of topological spaces:

If $X=(X,\mathcal{O})$ is a topological space and $Y\subseteq X$, we say that $Y$ is relatively finite (in $X$) if and only if for every neighborhood $N$ of the boundary $\partial Y$ the set $Y\setminus N$ is finite.

My professor found it quite interesting, but couldn't recall seeing something similar before.

My assumption is that this property is either not very meaningful, or that it has previously been used elsewhere under a different name. Can anyone enlighten me?

Edit: Some context, in case it helps: As I said I was looking for a converse of the Covering Lemma – i.e. what properties do I have to require of the metric space $X$ in addition to $X$ fulfilling the Lebesgue condition (every open cover has a Lebesgue number $\delta>0$ such that every $\delta$-ball is contained in some cover element) in order for $X$ to be compact?

A property I found was the “relative finiteness” of the set $X^\star$ of the isolated points of $X$. Since I did this out of mere curiosity and never actually used it for anything, the only additional properties of “relatively finite” sets I can give you are these small ones that I needed for the proof:

  • If, in addition to the above definition, $Y$ is open, then every sequence of points from $Y$ has a cluster point in $X$.
  • If, in addition to the above definition, $X$ is first-countable, then the interior of $Y$ is at most countable.
  • 0
    So if the topological space had no boundary...2011-03-06
  • 1
    @badp: If you mean "if $Y$ had no boundary" -- in that case, $N=\emptyset$ is a neighborhood of $\partial Y$, hence "relatively finite" and "finite" are equivalent.2011-03-06
  • 1
    So dense sets with empty interior are "relatively finite?"2011-03-06
  • 1
    @HennoBrandsma: Yes; indeed I used that as an example to show that in general, "relatively finite" sets need not be countable.2011-03-06
  • 0
    @HennoBrandsma: Actually, "empty interior" is sufficient -- $Y\setminus N$ needs to be finite, not necessarily $X\setminus N$.2011-03-06
  • 0
    @balpha: Can anything interesting about such subspaces be said? Normally one doesn't just define stuff for the fun of it...2011-03-08
  • 0
    @S.L.: Requiring this property of the set of $X$'s isolated points in addition to $X$ satisifying the Lebesgue condition is necessary and sufficient for compactness; that was the reason for defining it. I'm just not sure if this fact is interesting at all; hence the question.2011-03-08
  • 0
    @S.L. I expanded my question with some context.2011-03-12
  • 1
    @balpha: First off, I'm no expert by any means. But: I think your notion might give a nice(ish) characterization of compact subsets of the rationals for example. I also guess for it to be really interesting, you would have to see whether it has any useful applications.2011-03-12

2 Answers 2