Definition request: explicit definition of covering compactness in terms of set notation

Question

Part of my confusion with covering compactness stems from the fact that it is a definition given almost completely in a high level manner (in English no less).

When I look at:

A set $A \subset (M,d)$ is compact if every open cover of $A$ has a finite subcover

I have to remember what $1.$ open cover is, $2.$ subcover is, and $3.$ what finite subcover is, $4$. are the subcovers themselves open.... So every single proof regarding covering compactness requires me to translate/remember those things before everything else and it gets tedious.

Can someone please offer a workable definition of covering compactness in terms of set notation?

I could construct one myself but I am not completely certain if it is correct and whether if the notations/style etc is widely used.

I will start:

Let $C = \{C_n\}$ be a cover of $A$, where each $C_n$ is open, and let $A \subseteq \bigcup_{n=1}^\infty C_n $, then $A$ is compact if....

Answer 1

It’s not possible to complete the definition that you’ve started. Compactness of $A$ says something about every cover of $A$ by open sets, so you cannot phrase the definition in terms of a particular cover of $A$ by open sets.

In the comments Daniel Fischer has given a correct version expressed entirely in set-theoretic notation. Most people find that substituting words for at least some of the ‘connective tissue’ in the symbolic expression makes it easier to understand. Doing so, we might end up with something like this:

Let $\langle X,\tau\rangle$ be a topological space, and let $A\subseteq X$. Then $A$ is compact if and only if the following statement is true:

• If $\mathscr{U}\subseteq\tau$, and $A\subseteq\bigcup\mathscr{U}$, then there is a finite $\mathscr{V}\subseteq\mathscr{U}$ such that $A\subseteq\bigcup\mathscr{V}$.

We could use even more words without getting bogged down in terminology:

• If $\mathscr{U}$ is any cover of $A$ by open sets, then there is a finite subfamily $\mathscr{V}\subseteq\mathscr{U}$ that is also a cover of $A$.