Why don't we use closed covers to define compactness of metric space?

Question

I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?

Answer 1

It is important to understand that, although definitions often look arbitrary, they never are. Mathematical objects are intended to model something, and you can't understand why the definition is the way it is until you understand what it is trying to model. The question you asked is exactly the right one: why is it defined this way and not some other way? What is it trying to model?

(For example, why does a topology say that arbitrary unions of open sets are open, but infinite intersections of open sets might not be? It's because topology is intended to be an abstraction of certain properties of the line and the plane, and open sets are intended to be a more general version of open intervals of the line and open discs in plane, and that is how the intervals and discs behave.)

This case is similar. Mathematicians noticed that there are certain sorts of “well-behaved” subsets of the line and of metric spaces in general. For example:

1. A continuous function is always uniformly continuous — if and only if its domain is well-behaved in this way
2. A continuous real-valued function is always bounded — if and only if its domain is well-behaved in this way
3. If $f$ is a continuous real-valued function on some domain, there may be some $m$ at which $f$ is maximized: $f(x) ≤ f(m)$ for all $x$. This is true of all such $f$ if and only if the domain is well-behaved in this way
4. Every sequence of points from a subset of $\Bbb R^n$ contains a convergent subsequence — if and only if the subset is well-behaved in this way

and so on. It took mathematicians quite a long time to understand this properly, but the answer turned out to be that the "well-behaved" property is compactness. There are several equivalent formulations of it, including the open cover formulation you mentioned.

In contrast, the alternative property you propose, with closed covers, turns out not to model anything interesting, and actually to be trivial, as the comments point out. It ends nowhere. But even if it ended somewhere nontrivial, it would be a curiosity, of not much interest, unless it had started from a desire to better understand of something we already wanted to understand. It's quite easy to make up new mathematical properties at random, and to prove theorems about those properties, and sometimes it might seem like that is what we are doing. But we never are.

Properly formulated, compactness turns out to be surprisingly deep. Before compactness, mathematics already had an idea of what a finite set was. Finite sets are always discrete, but not all discrete sets are finite.
Compactness is the missing ingredient: a finite set is one that is both discrete and compact. With the discovery of compactness, we were able to understand finiteness as a conjunction of two properties that are more fundamental! Some of the properties we associate with finiteness actually come from discreteness; others come from compactness. (Some come from both.) Isn't that interesting?

And formulating compactness correctly helps us better understand the original space, $\Bbb R^n$ and metric spaces in general. Once we get compactness right, we see that the properties of "well-behaved" sets I mentioned above are not true of all compact spaces; metric spaces are special in several ways, which we didn't formerly appreciate.

Keep asking these questions. Every definition is made for a reason.

Answer 2

Just a not-so-trivial answer from a not-so-fluent student, I think the open cover definition can be understood in terms of their unions.

Suppose $X$ is a topological space and $\mathcal{O}$ is an open cover. Give $\mathcal{O}$ a well-ordering, and consider the chain of unions $S_\kappa:=\bigcup_{i=1}^{\kappa}U_i$, where $U_i\in\mathcal{O}$ and $\kappa$ is an ordinal. Then $\kappa\mapsto S_\kappa$ is a non-strictly increasing map. But note that, we can give different well-orderings on $\mathcal{O}$ and that same map would always remain non-strictly increasing. The important thing is, for every well-ordering, the set $\{S_\kappa\}$ is an open cover. And $\mathcal{O}$ has a finite subcover, iff there exists a specific well-ordering s.t. $S_N=X$.

Now recall the generalized version of sequential compactness in $\mathbb{R}$ to general topological spaces is that every net admits an accumulating point. If space $X$ admits an open cover that has no finite subcover, then, using the above notations, no $S_N$ equals $X$. WLOG, we can assume, for every $\kappa_1<\kappa_2$, $S_{\kappa_1}\subsetneq S_{\kappa_2}$ and $\bigcup_{\kappa<\kappa_1}S_\kappa\subsetneq S_{\kappa_1}$. Then, consider the nested closed sets $X- S_\kappa$, we have $\bigcap_{\kappa}(X- S_\kappa)=X-\bigcup_\kappa S_\kappa=\emptyset$. Now, we can construct a net $(x_\kappa)$ by arbitrarily choosing an $x_\kappa\in S_\kappa -\bigcup_{\eta<\kappa}S_\eta$ for every $\kappa$. By this construction, the net $(x_\kappa)$ eventually lies in the closed set $X-S_\kappa$ for every $\kappa$. Note since they are closed sets, every one of these contains all its own limit points. Therefore, if the net admits an accumulating point $y\in X$, this point must also lies in every closed set $X-S_\kappa$. Hence $y\in \bigcap_{\kappa}(X- S_\kappa)=\emptyset$. Contradiction. Thus, we have a net with no accumulating points.

To summarize the above, if an open cover admits no finite subcover, then it's always possible to construct a net accumulating at a non-existing point. Hence breaking the generalized version of compactness definition. Using open covers, not closed covers, is because we care about their complements, which only contains all the accumulating points if they are closed. Hence it can link with the convergence of subnets.

PS: The above discussion can be seen as a generalization of an open interval in $\mathbb{R}$, where you could have an open cover without a finite subcover. Then you could have a sequence approaching one of the end points of the interval, which is absent. It's just here the topology is first-countable, and so sequences are enough.