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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.
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So if the topological space had no boundary... –  badp Mar 6 '11 at 9:26
@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. –  balpha Mar 6 '11 at 9:28
So dense sets with empty interior are "relatively finite?" –  Henno Brandsma Mar 6 '11 at 12:30
@HennoBrandsma: Yes; indeed I used that as an example to show that in general, "relatively finite" sets need not be countable. –  balpha Mar 6 '11 at 12:35
@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. –  Sam Mar 12 '11 at 21:38
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2 Answers

I know this is an old question, but I just ran into it and had this thought.

In a metric space, I believe your property is equivalent to $Y$ having discrete interior with compact closure.

To see this, first assume your property. For any point $p$ in the interior, let $d$ be such that the ball centered at $p$ of radius $d$ is contained in the interior. Then the ball of radius $\frac{d}{2}$ lies outside the $\frac{d}{2}$-neighborhood of the boundary, and thus is finite. So every point in the interior has a finite neighborhood, thus the interior is discrete.

Further, take any sequence of distinct points $\{a_n\}$ in the interior of $Y$. There is a subsequence of $\{a_n\}$ which converges to a point in $\partial Y$, because otherwise you could take the closure of the union of balls of radius $r_n$ around each $a_n$, where $r_n$ is half the distance from $a_n$ to the boundary. The complement of this closure is an open set which excludes all (infinitely many) the $a_n$, and thus cannot contain all of $\partial Y$. So a point of $\partial Y$ is contained in the closure of these balls, and one can then easily see that the corresponding $a_i$ must converge to that point. Thus, any sequence of points in the interior has a convergent subsequence, giving us compact closure.

Next, assume $Y$ has discrete interior with compact closure, and let $N$ be a neighborhood of the boundary $\partial Y$. Suppose $Y - N$ is infinite. Then $Y - N$ either has cluster points in the interior of $Y$, which violates discreteness, or it has no cluster points in $Y$ at all (such points cannot be in $\partial Y$ since they are separated by $N$), which would imply an infinite discrete subset of $Y$ that doesn't have any cluster points, violating compactness. Either way you get a contradiction, so $Y - N$ must be finite.

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This is now a very old question, but I want to correct an error in the original post. It is not the case that a relatively finite subset of a first countable space must have countable interior: the space $\omega_1$ with the order topology is a counterexample. Let $A = \{0\} \cup \{\alpha+1:\alpha \in \omega_1 \}$, the set of isolated points in $\omega_1$; $\operatorname{bdry}A = \omega_1 \setminus A$, the set of countable limit ordinals. Every infinite subset of $A$ has a cluster point in bdry $A$, so every nbhd of $\operatorname{bdry}A$ must be cofinite in $\omega_1$. That is, $A$ is an uncountable set of isolated points that is relatively finite in $\omega_1$

As for the original question, since a subset $A$ of a space $X$ is relatively finite in $X$ iff its interior is, it seems fair to say that relative finiteness is ‘really’ a property of open sets, and it's in that form (if at all) that I’d expect to find prior uses. I've not actually seen any, however.

As MartianInvader pointed out, in metric spaces it’s equivalent to having discrete interior and compact closure. More generally, in countably paracompact $T_3$ spaces it’s equivalent to having discrete interior and countably compact closure.

Proposition: Let $X$ be a $T_3$ space, and let $Y \subseteq X$ have discrete interior and countably compact closure; then $Y$ is relatively finite in $X$.

Proof: Let $W$ be an open nhbd of $\operatorname{bdry}Y$, and suppose that $Y \setminus W$ is infinite. Let $\{W_n:n \in \omega \}$ be a partition of $Y \setminus W$; points of $Y \setminus W$ are isolated in $X$, so $\{W\} \cup \{X \setminus \operatorname{cl}Y \cup \{W_n:n \in \omega \}$ is a countable open cover of $X$ that clearly has no finite subcover.

Theorem: Let $X$ be a countably paracompact $T_3$ space. If $Y$ is a relatively finite subset of $X$, then $\operatorname{cl}Y$ is countably compact. (In fact, $Y$ need only satisfy the weaker property that if $W$ is an open nbhd of $\operatorname{bdry}Y$, then $Y \setminus W$ is compact.)

Proof: Let $H = \operatorname{cl}Y$, and suppose that $H$ is not countably compact. Then $H$ has an increasing open cover $\mathcal U = \{U_n:n \in \omega\}$ with no finite subcover. Countable paracompactness of $X$ implies that $\mathcal U$ has an open refinement $\mathcal V = \{V_n:n \in \omega\}$ such that $\operatorname{cl}V_n \subseteq U_n$ for each $n \in \omega$. By passing to a (faithfully indexed) refinement if necessary, we may further assume that $\mathcal V$ is locally finite.

Let $V = \operatorname{int}Y$. Let $n(0) = 0$, and choose $x_0 \in V \cap V_0$. Suppose that for some $m > 0$ and all non-negative integers $k < m$ we have chosen $n(k) \in \omega$ and $x_k \in V$ so that (a) $x_{m-1} \in V \cap V_{n(m-1)}$, and (b) $n(i) < n(k)$ and $x_i \in V \cap V_{n(i)} \setminus V_{n(k)}$ whenever $0 \le i < k < m$. $\mathcal V$ is point-finite and has no finite subfamily covering $H$, so there is an $n(m) \in \omega$ such that $n(m) > n(m-1)$ and $\{x_k:0 \le k < m\} \cap V_i = \emptyset$ for all $i \ge n(m)$. Clearly we may now choose $x_m \in V \cap V_{n(m)}$, and the recursion goes through to yield a set $A = \{x_n:n \in \omega\} \subseteq V$ such that $x_k \in V \cap V_{n(k)} \setminus V_i$ whenever $i,k \in \omega$ and $i \ge n(k+1)$.

Now let $p$ be any point of $H$. $\mathcal V$ is locally finite, so $p$ has an open nbhd $W$ that meets only finitely many members of $\mathcal V$. Clearly $W \cap A$ is finite, so $p$ is not a cluster point of $A$. Thus, $A$ is a closed, discrete subset of $V$, and $X \setminus A$ is an open nbhd of $\operatorname{bdry}Y$ whose relative complement in $Y$ is not compact (and hence certainly not finite).

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Interesting, in that case I'll have to check my original proof for errors. My topology has become quite rusty, so give my some time :) –  balpha Jul 12 '11 at 17:18
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