# Is there an analogue for a compact set using closed sets?

Compact sets have the property that given any open cover, you can extract a finite subcover. I was wondering if there was an analogue to compactness that could be used for closed sets. Since the union of infinitely many closed sets is not necessarily closed, you'd probably have to use the infinite intersection of closed sets. But infinite intersections of closed sets can easily be the empty set or just a point. Is there any interesting way to think of closed sets that is similar to how compact sets can be understood in terms of their open covers?

• I'm not an expert in this area, but my understanding is no, the whole point of this concept is that open sets do not contain boundary points and so you are analyzing how that works with covering a given set. – Morgan Rodgers Nov 8 '18 at 23:39
• Well, I suppose we could Invent a type of property that if $A = \cap_{\alpha} C_{\alpha}$ then there would exist a finite class of $C_i$ so that $A = C_i$. Of maybe if $\cap_{\alpha} C_{\alpha}\subseteq A$ then there is a finite $\cap_{i} C_i \subseteq A$. No harm in making definitions. What significance sets with that property (if any exist) have is ... another question. – fleablood Nov 9 '18 at 1:19

## 2 Answers

I can add some things that may be of interest. Writing this more to spark your interest and may lead you to find a suitable answer to your question.

First, note that if $$X$$ is a compact topological space then any closed subset of it is also compact, and so can be characterised as above.

Moreover, if $$X$$ is Hausdorff, not compact, but locally compact we have its Alexandroff one-point compactification; this is essentially $$X$$ adjoined to a new point $$*$$, with the topology that if $$U \subset X$$ is open, it is also open in this new space, and if $$* \in U$$ we require $$X - U$$ to be compact in $$X$$ for $$U$$ to be open. It turns out that this new space is compact (it has other interesting properties). Then if $$C \subset X$$ is closed, then $$\{*\} \cup C$$ is closed in the compactification and is thus compact, so you can think about the relation between covers of $$C$$ and covers of this new subset.

An example of the above is that the one-point compactification of $$\mathbb{R}$$ is just the unit circle $$S^1$$.

Another interesting example is for a space $$X$$ which is countably compact; every countable open cover has a finite subcover. This property is equivalent to the property that every nested sequence of closed, non empty, sets is non-empty.

Hope this assists you in some ways..

You can just take the definition of compact using open sets and throw in a bunch of complements:

A topological space is compact if given an arbitrary collection of closed sets $$\{C_i\}$$ with empty intersection there's some finite sub-collection with empty intersection.

As an example of how this works in practice, a proof that $$\mathbb{R}$$ isn't compact is to take the collection of closed subsets $$\{C_n=[n, \infty)| n \in \mathbb{N}\}$$ which has empty intersection. Any finite sub-collection $$\{C_{n_i}\}$$ has intersection $$C_{max(n_i)}$$, which isn't empty. Therefore $$\mathbb{R}$$ isn't compact by this definition.

• Or, taking a contrapositive, this reduces to the finite intersection property characterization: A topological space is compact if and only if whenever a family of closed sets $\{ C_i \}$ has the property that any intersection of a finite subfamily is nonempty, then $\bigcap_i C_i$ is nonempty. – Daniel Schepler Nov 9 '18 at 1:19