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....

  • 1
    $\begingroup$ Let $\mathscr{T}$ be the collection of open sets in $M$. Then $$A\text{ is compact} \iff \bigl(\forall \mathscr{U}\subset \mathscr{T}\bigr) \bigl( A \subset \bigcup \mathscr{U} \implies (\exists \mathscr{V} \subset \mathscr{U})(\mathscr{V} \text{ is finite} \land A \subset \bigcup \mathscr{V})\bigr).$$ We can modify that to use more or fewer words. $\endgroup$ Mar 11 '16 at 21:36

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 $A$.
  • $\begingroup$ In the above definition, how can we restrict to a metric space instead of a topological space? Second would it be too pedantic to specify what it is meant by "finite"? Also the first union $\bigcup$ is different in a sense compared to the second union. The second one must be countable finite union. The first one not necessarily. Just want the definition clear as day that's all $\endgroup$
    – Fraïssé
    Mar 12 '16 at 4:28
  • $\begingroup$ @Lookbehindyou There's no difference between taking the union of a "small" (e.g. finite) family of sets and of a "large" (e.g. infinite) family of sets - both are defined the same way: the union of $\mathcal{A}$ is the set $\{x: \exists B\in\mathcal{A}(x\in B)\}$. As for defining "finite," this is a place where looking at a set theory text might help: for example, we can define the notion of an inductive set, and then a set is finite iff (a set equinumerous to) it is an element of every inductive set. (In the presence of Choice, "finite" also means "no nonsurjective self-injection".) $\endgroup$ Mar 12 '16 at 5:43
  • $\begingroup$ @Lookbehindyou: There’s no good reason to restrict it to metric spaces, and doing so adds another step: you have to define $\tau$ from $d$ instead of taking it as a given. \\ As Noah said, unions are unions: the definition doesn’t change depending on the size of the collection whose union is being taken. If you really want to emphasize the finiteness of $\mathscr{V}$, you could say: then there are a positive integer $n$ and sets $U_1,\ldots,U_n\in\mathscr{U}$ such that $A\subseteq U_1\cup\ldots\cup U_n$. $\endgroup$ Mar 12 '16 at 5:56

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