How to Motivate Open-Cover Formulation of Compactness in a Metric Space?

Question

The open cover formulation of compactness always seemed to come out of nowhere for me. I've consulted many Analysis textbooks, but all of them have been like - 'Here's the open cover formulation, now we prove this and the sequential formulation are equivalent.' None of them actually go on to explain where this open cover formulation comes from.

So, my question is this - suppose I was a researcher trying to come up with an open set formulation of compactness for the first time. All I know is Real Analysis, and I have defined a compact set as one in which a sequence has a convergent subsequence. How would I go about doing so?

Answer 1

An imaginary history of a discovery: You consider a metric space $(X,d)$ that is $not$ sequentially compact, like $\Bbb R.$ You take a sequence $(x_n)_{n\in \Bbb N}$ in $X$ with no convergent sub-sequence. As it is a metric space, you see that each $x\in X$ has a nbhd $U$ such that $\{n: x_n\in U\}$ is finite. So the $set$ $S=\{x_n: n\in \Bbb N\}$ must be infinite. And each $x\in X$ has an open nbhd $V_x$ such that (i): $V_x\cap S=\emptyset$ if $x\not\in S,$ (ii): $V_x\cap S=\{x\}$ if $x\in S.$ So $S$ is a countably infinite closed discrete sub-space of $X.$ Now you look at $C=\{V_x: x\in S\}\cup \{X\setminus S\}$ and realize that $\cup C=X$ but $\cup D\ne X$ for any finite $D\subset C.$

You now ask "What about open covers of $X$ if $(X,d)$ $is$ sequentially compact?" and discover the converse.

Along the way, you also found that a metric space is not sequentially compact iff it has an infinite closed discrete subspace iff it has a countable open cover with no finite sub-cover.

[ Unlike the $\in$-order topology on the ordinal $\omega_1,$ which is not compact, but is countably-compact and has no infinite closed discrete sub-space.]

Answer 2

If you start with the notion of sequences having convergent subsequences and call that compactness then you'll never find the open-cover definition (realistically) because you've not started out with intuition: you've just applied a label to a concept you've come up.

Let's go back a bit further: what are we trying to convey when we say compact? We're trying to explain that what we're looking at is somehow all together in one place, not too spread out, that any point of what we're looking at is "not too far" from any other.

Ok, so how do we make that more mathematical? We could try considering the distances between points... but that requires a metric and we know that a general set doesn't have to have that. In fact, when we think about general sets we run into the standard problem: there's not much structure there to work with. Typically we have open sets and neighbourhoods and... well, that's about it.

But that's actually all we need! We have counting measure available to us and that gives us a way to describe how spread out (or not) our set is: we see if we can cover our set with finitely many open sets. If we can never do that, then we can't possibly be compact: our set must be spread out quite significantly. If we can do it sometimes but not others... that's probably not compact then, as it shouldn't really depend on how we're choosing our sets. But if every time we cover our set we can find a finite set of neighbourhoods that still covers it, we can call that compact.

This way of looking at it already alerts you to the idea that sequential compactness might not always be good: we quickly see that these sequences might run off arbitrarily far in any direction while having convergent subsequences, and that those might be quite messy (we might start thinking about Besicovitch sets and how strange those can be).

Note that compact doesn't have to mean small and that some non-compact sets, with this definition, can be enclosed in compact sets (consider your favourite bounded non-compact set and then any origin-centred closed call that contains it).

Answer 3

Here's one attempt at motivation. I'll assume we're in $\mathbb{R}$, but this makes sense in any metric space.

Suppose we're interested in the relationship between continuity and uniform continuity, and we want a sufficient condition on a set $E$ so that if a function $f$ is continuous on $E$, then $f$ is uniformly continuous on $E$.

Continuity on $E$ gives: for every $\varepsilon > 0$, for every $x \in E$, there exists $\delta_x > 0$ such that $t \in (x-\delta_x, x+ \delta_x) \cap E$ implies $f(t) \in (f(x) - \varepsilon, f(x) + \varepsilon)$. For uniform continuity, we need a $\delta>0$ that works for all $x \in E$. But the problem is $\delta:=\inf \{ \delta_x \}_{x \in E}$ might be zero. The existence of a finite subcover of the open cover $\{ (x-\delta_x, x+\delta_x) \}_{x \in E}$ would ensure we can find a strictly positive $\delta$.

Answer 4

For me, the first thing that motivated it was continuous functions. Most people really like continuous functions, which are just functions where at every point we can fix the difference of the output to within some threshold, and that guarantees we have an open set around that point. Similarly we might wonder about what if we fix the size of the open sets to not be dependent on any specific point, which is of course what uniform continuity is.

So now what kinds of spaces allow us to easily say continuous functions are also uniformly continuous? Because continuity at every point gives us an open set at every point, and so long as there are infinitely many points, we have an infinite open cover naturally. Now by magically saying that we'd like every open cover to have a finite subcover, we now have enough to move forward with a tiny bit more legwork and say our continuous function is also uniformly continuous.