1
$\begingroup$

I'm confused about the fact that in a compact space $X$ we cannot find a locally finite collection of subsets. This is said and proved in the wikipedia article : https://en.wikipedia.org/wiki/Locally_finite_collection

But in this post Compact space, locally finite subcover

They are saying that If every open cover of $X$ has a locally finite subcover, then $X$ is compact. I don't understand which one is true.

Also if the first one is true can we use it to show that the space $\mathbb R$ is not compact by saying that $\mathbb R$ contains a locally finite collection of subsets which is given by the intervals $(n,n+2)$ for each $n\in \mathbb Z$.

$\endgroup$
2
$\begingroup$

If $\mathcal{F}$ is a locally finite collection of sets of $X$ where $X$ is compact, then $\mathcal{F}$ is finite (and thus trivially locally finite).

Proof: let $x \in X$. Then $x$ has an open neighbourhood $U_x$ such that $$\mathcal{F}_x = \{F \in \mathcal{F}: F \cap U_x \neq \emptyset\}$$ is finite.

Then $\mathcal{U} = \{U_x: x \in X\}$ is an open cover of the compact $X$, so there is a finite subcover $\{U_{x_1}, \ldots, U_{x_N}\}$ of this $\mathcal{U}$.

Claim: $$\mathcal{F}\setminus\{\emptyset\} = \bigcup_{i=1}^N \mathcal{F}_{x_i}$$

The right to left inclusion is clear, as, for all $x$, $\mathcal{F}_x \subseteq \mathcal{F}$ and all members are non-empty, and if on the other hand $F \in \mathcal{F}\setminus\{\emptyset\}$, pick $p \in F$ and pick $x_i$ for some $i \in \{1,\ldots,N\}$ such that $p \in U_{x_i}$ (we have a subcover), and then by definition, $F \in \mathcal{F}_{x_i}$ for this $x_i$.

It follows that $\mathcal{F} \subseteq \bigcup_{i=1}^N \mathcal{F}_{x_i} \cup \{\emptyset\}$ and is thus finite, as a finite union of finite sets.

We could say, if we consider finite collections of sets to be trivially locally finite, a compact space has no non-trivial (i.e. infinite) locally finite families. Wikipedia correctly identifies the infinite condition.

An immediate corollary is

If $X$ is a space such that every open cover $\mathcal{U}$ has a locally finite subcover $\mathcal{U}'$, then $X$ is compact. This condition trivially holds for compact spaces, as there we even have a finite (so trivially locally finite) subcover.

This holds as the promised locally finite subcover must be finite by the first fact and we are done straight away.

Such considerations serve to motivate the definition of so-called "paracompact" spaces where every open cover must have a locally finite refinement. The above shows we actually need "refinement" instead of subcover to get a weaker notion than compactness. We also need "locally finite" because if a space $X$ satisfies the condition that every cover has a finite refinement, then $X$ is also compact, and we don't get a new notion. So only the two combined weakenings of "finite" to "locally finite" and "subcover" to "refinement" gives us a new interesting property. It turns out that all metrisable spaces are paracompact (not trivial!) and this is a very convenient class of spaces. It also gave rise to many generalised derived notions (what if we demand point-finite instead of locally finite, or we restrict to countable covers etc..). See the Handbook of Set Theoretic topology (the chapter on covering properties), or Engelking's chapters on paracompactness and metrisability.

$\endgroup$
2
$\begingroup$

Both statements are true, except that you are misquoting the first one - the Wikipedia article says that

No infinite collection of a compact space can be locally finite.

(emphasis mine). Thus, if $X$ is compact, then it is certainly true that every open cover of $X$ has a locally finite subcover, because every open cover has a finite subcover (by the definition of compactness).

$\endgroup$
1
  • $\begingroup$ Thank you Zev, I'm aware of infinite condition, I just know that finite collection of subsets is easily seen to be locally finite. So talking about locally finite collection is interesting only if the collection is infinite. Hence I think we better say that if $X$ is compact then every open cover of $X$ has a finite subcover. Otherwise it is confusing I think.. $\endgroup$ – palio Dec 25 '17 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.