What should be the intuition when working with compactness?

Question

I have a question that may be regarded by many as duplicate since there's a similar one at MathOverflow.

1. In $\mathbb{R}^n$ the compact sets are those that are closed and bounded, however the guy who answered this question and had his answer accepted says that compactness is some analogue of finiteness. In my intuitive view of finiteness, only boundedness would suffice to say that a certain subset of $\mathbb{R}^n$ is in some sense "finite". On the other hand there's the other definition of compactness (in terms of covers) which is the one I really need to work with and I cannot see how that definition implies this intuition on finiteness.

2. To prove a set is compact I know they must show that for every open cover there's a finite subcover;the problem is that I can't see intuitively how one could show this for every cover. Also when trying to disprove compactness the books I've read start presenting strange covers that I would have never thought about.

I think my real problem is that I didn't yet get the intuition on compactness. So, what intuition should we have about compact sets in general and how should we really put this definition to use?

Can someone provide some reference that shows how to understand the process of proving (and disproving) compactness?

Answer 1

The following story may or may not be helpful. Suppose you live in a world where there are two types of animals: Foos, which are red and short, and Bars, which are blue and tall. Naturally, in your language, the word for Foo over time has come to refer to things which are red and short, and the word for Bar over time has come to refer to things which are blue and tall. (Your language doesn't have separate words for red, short, blue, and tall.)

One day a friend of yours tells you excitedly that he has discovered a new animal. "What is it like?" you ask him. He says, "well, it's sort of Foo, but..."

The reason he says it's sort of Foo is that it's short. However, it's not red. But your language doesn't yet have a word for "short," so he has to introduce a new word - maybe "compact"...

---

The situation with compactness is sort of like the above. It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.

But in some sense you've never encountered the notion of compactness by itself before, isolated from the notion of discreteness (in the same way that above you've never encountered the notion of shortness by itself before, isolated from the notion of redness). This is just a new concept and you will to some extent just have to deal with it on its own terms until you get comfortable with it.

Answer 2

You may read various descriptions and consequences of compactness here. But be aware that compactness is a very subtle finiteness concept. The definitive codification of this concept is a fundamental achievement of $20^{\,\rm th}$ century mathematics.

On the intuitive level, a space is a large set $X$ where some notion of nearness or neighborhood is established. A space $X$ is compact, if you cannot slip away within $X$ without being caught. To be a little more precise: Assume that for each point $x\in X$ a guard placed at $x$ could survey a certain, maybe small, neighborhood of $x$. If $X$ is compact then you can do with finitely many (suitably chosen) guards.

Answer 3

I answered a very similar question here.

Recapitulating:

Compactness does for continuous functions what finiteness does for functions in general.

If a set $A$ is finite then every function $f:A\to \mathbb R$ has a max and a min, and every function $f:A\to\mathbb R^n$ is bounded. If $A$ is compact, the every continuous function from $A$ to $\mathbb R$ has a max and a min and every continuous function from $A$ to $\mathbb R^n$ is bounded.

If $A$ is finite then every sequence of members of $A$ has a subsequence that is eventually constant, and "eventually constant" is the only kind of convergence you can talk about without talking about a topology on the set. If $A$ is compact, then every sequence[ net ] of members of $A$ has a convergent subsequence.

Answer 4

The definition of compactness that reads: "every cover has a finite subcover" is most directly related to the idea that being compact is, in some sense, like being finite: compact sets share with finite sets the property that every cover has a finite subcover. It is a concept that takes some time getting used to, so at first proofs of/with it may look weird.

Compactness in $\mathbb R^n$ is equivalent to being closed and bounded. This again is a property shared with finite sets: any finite set in $\mathbb R^n$ is closed and bounded. Also, in a metric space, a set is compact if, and only if, every sequence in it has a convergent subsequences. Again, a property that holds for finite sets: Every sequence in a finite set contains a convergent subsequence.

Finally, there is a less straightforward analogy using nonstandard analysis. A set is compact if, and only if, every point in its enlargement is near-standard. Intuitively, an enlargement of a set is obtained by adding new points generated from the set. Being near-standard means a new point is infinitesimally close to an already existing point in the set. For a finite set, the enlargement is equal to the set, so every point in the enlargement is simply equal to some point from the set. So again, this is a property shared with compact sets: a set is compact if every point in the enlargement is infinitesimally close to some point from the set.

Answer 5

Also, I feel it's pretty strange that when people come to prove some set is compact, they must show that for every open cover there's a finite subcover.

That’s a bit like finding it pretty strange that in order to prove that an integer $n$ is even, one must show that $n=2k$ for some integer $k$: in both cases we’re simply verifying that something satisfies the definition of a particular property. An integer $n$ is by definition even if and only if there is an integer $k$ such that $n=2k$, so the most straightforward way to show that $n$ is even is to show that such a $k$ exists. Similarly, a subset $K$ of a space $X$ is by definition compact if and only if every open cover of $K$ has a finite subcover, so the most straightforward way to show that $K$ is compact is to prove that every open cover of $K$ has a finite subcover.

For the intuition, note first that every finite set is clearly compact. If $F$ is finite, and $\mathscr{U}$ is an open cover of $F$, then for each $x\in F$ we can choose a $U_x\in\mathscr{U}$ such that $x\in U_x$, and $\{U_x:x\in F\}$ will then be a finite subfamily of $\mathscr{U}$ that still covers $F$. In fact, finite sets are the only ones that are guaranteed to be compact no matter what the topology on the space might be: if the topology is the discrete topology, in which every set is open, the compact sets are precisely the finite sets. Thus, compactness generalizes a property of finite sets. What probably isn’t at all clear at this point is why this particular property of finite sets is so important. In fact, as you can see from this brief historical survey, it took topologists a number of years to realize its central importance.

More generally, one might say that compact sets are in a certain important sense small sets. Suppose that every point $x$ of some compact set $K$ has an open neighborhood $U_x$ with some property $P$. Then $\mathscr{U}=\{U_x:x\in K\}$ is an open cover of $K$, so it has a finite subcover $\{U_{x_1},\dots,U_{x_n}\}$: $K\subseteq U_{x_1}\cup\ldots\cup U_{x_n}$. If $P$ is one of a great many ‘nice’ properties, the union of finitely many open sets having $P$ also has $P$, and $U_{x_1}\cup\ldots\cup U_{x_n}$ is then an open neighborhood of $K$ with the property $P$. The compact set $K$ behaves just like a single point in terms of having an open neighborhood with $P$. The importance of this characteristic of compact sets probably won’t really become clear until you’ve actually used this property of compact sets in a variety of contexts with a variety of properties $P$, but at least it should offer further evidence that compactness is in some sense a kind of smallness.

Answer 6

I like to think of compactness as generalised finiteness.

When studying the topology of a space, the information we care about is in the structure of open sets. In particular we might associate with some open sets a mathematical object, like a number (such as the radius of a open ball) or something more sophisticated like a chart on a manifold.

Compactness is useful because some properties, like taking minimums of lists or making sure an arbitrary sum converges, only work for finite sets. Consider the following sketch:

Given a function $f : [a,b] \rightarrow \mathbb{R}$, can it attain arbitrarily large values?

The key is to notice that since $f$ is continuous, if we cover its image with open sets in any way we like, the preimages of these open sets will be open in $[a,b]$. Let's say we pick balls of radius $1$ around each point in the image. There are uncountably many of them, so there's not much we can say, but surely, the preimages of these balls will cover the whole domain.

Now, since $[a,b]$ is compact, we can refine this cover (of preimages) to a finite one, and the corresponding refinement still covers Im $f$. But this means $f$ must be bounded, since its image is covered by finitely many balls of radius $1$!

Most applications of compactness follow this same pattern. We start we some arbitrary cover that has nice properties, but which don't serve us much if the cover is infinite. We then invoke compactness to say there's a finite subcover which is just as good.

Answer 7

Regarding the definition in terms of open covers, this turns out to be technically very useful, but I agree that it can be hard to understand at first. You might want to consider one of the equivalent definitions in terms of closed sets.

The contrapositive of the standard open cover condition says that if $U_i$ is a collection of open sets so that no finite union of $U_i$'s covers the compact set $X$, then the entire set of $U_i$'s doesn't cover $X$. Passing to complements, we find that if $F_i$ is a collection of closed subsets having non-empty intersection with $X$, then the intersection of the entire collection has non-empty intersection with $X$.

Now this can be rephrased in a slightly simpler way: given a chain of closed subsets (i.e. a collection of non-empty closed subsets such that for any two members of the collection, one is contained in the other) each having non-empty intersection with $X$, their intersection has non-empty intersection with $X$. (For a discussion of this, see this answer, or more-or-less any topology text book.)

So this says that we can't find a shrinking collection of closed subsets each having a point in common with $X$, but whose intersection doesn't have a point in common with $X$.

Contrast with say the interval $X := (0,1).$ The subsets $[0,1/n]$ are closed intervals having non-empty intersection with $X$, but their intersection is the singleton $\{0\}$, which doesn't have non-empty intersection with $X$.

Consider also the whole real line. The subsets $[n,\infty)$ are closed and non-empty, but their intersection is empty.

So these two examples capture the basic intuition for how compactness fails: a set may be missing some kind of "boundary point" or "point at infinity". Conversely, the intuition is that a compact set is not missing any such points.

The definition in terms of subsequences necessarily containing a convergent subsequence (valid in a metric space) captures the same idea.

As in the above linked answer, one can also rephrase the definition in terms of chains of open sets. Then it says that if $U_i$ is any chain of open sets whose union covers $X$, then already one of the of open sets $U_i$ must contain $X$.

So you can think of this as saying that it is not possible to fill out $X$ by taking an increasing union of proper open sets (in the induced topology).

Compare this with the open interval $(0,1)$, which is the increasing union of the proper open subsets $(1/n,1-1/n)$.

---

Also, here is an aside on index sets, which can sometimes be confusing when you first encounter them: note that while one can't reduce to countable open covers, or decreasing sequences of closed subsets, in the general definition of compactness (and one can't use sequences to test for compactness in general --- although there is a generalization of sequences (in fact more than one --- nets or filters) which can be used), when building up intuition it is fairly safe to imagine that the open covers in question are countable, or that the chain of closed subsets is just a decreasing sequence of closed subsets.

While it is technically important to learn to argue with non-countable index sets as in the general definition of compactness, I wouldn't let the possible non-countability of those index sets be the thing I focus on when building up intuition; this truly is a technical point which will just get in the way from an intuitive view-point. As you become more confident in manipulating index sets, the distinction between the countable and general case won't seem like that big of a deal anyway.

Answer 8

On $\mathbb{R}^n$, compactness is equivalent to a set being closed and bounded. In a compact set every sequence has convergent subsequences and every continuous function is uniformly continuous, attains a maximum and a minimum, and can be integrated without any trouble.

Compactness in a general topological space is basically a way to capture the nice properties of closed and bounded subsets of $\mathbb{R}^n$.

Answer 9

You know pigeonhole or Dirichlet principle? After you have 6 baseballs put into 5 holes you ended up having in at least one hole at least two balls. Now if you have infinite number of baseballs putting into 5 holes(assume the holes have no bottom) you ended up having in at least one hole the infinite number of baseballs. ( proof by contradiction assume there were finite number of baseballs in all 5 holes you would have finite number of balls in total, contradiction). It works for any finite number of holes. In continuous R^3 case rather than a finite number of holes you have a bounded set (or totally bounded to be precise, you have a set inside a finite number of arbitrarily small balls) and instead of baseballs you have points. You put infinite number of points into this totally bounded set. in at least one of the arbitrarily small balls the infinite number of points must have accumulated around a certain point. (accumulate around a point means in arbitrarily small ball covering the point you get infinite number of points) The points can accumulate around a point inside the set or around the point outside the set (for example if the point is at the boundary which does not belong to set, open set). We are interested only in class of sets which always accumulate around the points inside the set. The property: "Any collection of the infinite number of the points from an (infinite) set always accumulate around at least one point of the set" is called the compact property of the set. This property must imply boundness because for unbounded set you can find one specific collection of infinite points that do not accumulate at all.(you space points equidistantly into direction where it is not bounded). It must also imply completeness because Any/Every Cauchy sequence (as any sequence in fact) is made of infinite number of points so it does accumulate around at least one point of the set (from the compact property) , but from the Cauchy sequence property it has never more than one accumulation point. So in this case it has exactly one accumulation point which is its limit, so it converges.(Definition of completeness.) This property also implies that any infinite open cover {O_i} of the set you can simplify to finite cover. Because otherwise if you can not simplify the cover to finite one you can create an infinite sequence of points x_i where every point is in none of the sets O_1...O_i. (escaping sequence of points). Obviously this sequence would not converge to any point of the set x because x would be inside (at least) one open set O_k, but every open set in cover would have at most finite number of points (escaping sequence of points) which prevents convergence and accumulation point inside any open set in cover and hence in the set itself. This is contrary to our compact property.

Answer 10

What should be the intuition when working with compactness?

ANS: PACMAN

In the picture you see Pacman about to eat two compact topological spaces one after the other, but imagine Pacman being allowed to grab a bite from anywhere it wants to. As it eats the whole thing, it might bite into air (reworking the same areas of the lunch plate), only eating one-tenth of some dot. It can take an infinite number of bites, but it only really needs a finite number of chomps to take care of lunch.

If we look at Pacman eating up the closed unit interval $[0,1]$ (a compact, complete and connected space), taking 'open bites' you can ask how far it can go in the form $[0,k)$ using only a finite number of 'interval' bites. With a little thought (no supremum $k \lt 1$ can stop Pacman), you come to the conclusion that Pacman can consume at least up to $[0,1)$. If that took $n$ bites, since Pacman is relentless it will also eat the singleton $\{1\}$ (that supremum thing again), so the interval $[0,1]$ is covered in $n+1$ bites.

Using only open interval bites works since they form a basis for the topology.

Answer 11

Boundedness is a metric concept, not a topological concept. Boundedness means that you can be "before" (< M) or "beyond" (> M) the bound M.

On general topological spaces, there is no ">" or "<" so this doesn't make sense. Hence the need for a more general concept : compactness.

Answer 12

Let $ (X, d) $ be a metric space, and say $ f : X \longrightarrow \mathbb{R} $ is a continuous function. Can we impose an interesting condition on $ X $ (and $ X $ alone) to ensure that $ f $ is bounded ?

Continuity makes $ f $ "locally bounded", i.e. for every $ x \in X $ there is a $ \delta_x > 0 $ such that $ f $ is bounded on $ B(x, \delta_x) $. Also $ X $ is covered by these $ \{ B(x, \delta_x) : x \in X \} $.

Had we had the provision to pick finitely many points $ x_1, \ldots, x_n \in X $ such that $ X $ is covered by $ \{ B(x_1, \delta_{x_1}), \ldots, B(x_n, \delta_{x_n}) \} $ itself, $ f $ would've been bounded. So if we impose that every cover of $ X $ by open balls has a finite subcover, that will do. This condition is precisely compactness of $ X $.

[This is just one way to motivate compactness. Compactness of a metric space has lot many consequences, especially on how continuous functions behave on that space.]

Answer 13

In my opinion, I think it's okay that you treat compactness as a clever trick as you said. But as it is an abstract concept, I think it's recommended to take the concept as itself, not comparing to real examples.

Maybe the concept of compactness is motivated from Heine-Borel property, but even without a motivation, I think the definition of compactness itself is very natural.

There are many branches of mathematics and each branch has its main vocabularies. For example, in number theory, one's main vocabulary is an "integer". Just like this, "openness" is a main vocabulary in general topology. (The question of asking why openness is a main vocabulary is a completely different question and this is well-answered in Wikipedia.) So when you do general topology, you play with open sets. And definitely it's hard to make an argument on infinite open objects, but compactness ease this so that when you play with compactness you can only consider finite objects (for each compact set).

If you take concepts in this way, definitions of Lindelöf space and paracompact space and etc are very natural just like compactness. I'm quite sure no one has a clear "real life example" of paracompactness but almost all of people here think the definition is very natural.

Answer 14

Finite Union Property

Let $X$ be any set and property $p$ identify a family of subsets of $X$, i.e $P = \{x \in \mathcal{P}(X): p(x) \}$. Property $p$ (or family $P$) is said to have Finite Union Property (FUP) if $$ U_1, U_2, ..., U_n \in P \implies \bigcup_{i=1}^n U_i \in P $$ where $n \in \mathbb{N}$

Then, we can define compactness as

Compactness

A topological space $(X, \mathcal{T})$ is compact if and only if for every property $p$ with FUP, for each point $x \in X$, there is an open neighbourhood $O_x$ (not necessarily distinct) that satisfies $p$, then $X$ satisfies $p$.

In short, the definition gives sufficient conditions to generalize any property with FUP on a compact set, i.e. if a FUP property is true for a neighbourhood for each point in a compact space, then the property is true for the compact space.

The idea is from https://www.math.ucla.edu/~tao/preprints/compactness.pdf

Answer 15

Although this question already received a lot of nice answers, in mine I would to present a different accent. I have a lot of experience dealing with compact sets, so I can call them my good friends. :-) But I have to confess that despite my familiarity with compacts, I don’t have in my mind a clear idea of them, for instance, thinking them as “small and complete” spaces. On the other hand, from a professional point of view this is not very important, because usual ways to think about compacts are methods to prove something about them. To be acquainted with these methods, you may look, for instance, a chapter on compact spaces in Engelking’s “General topology”. I remark that compactness proofs are not always direct, but often use that compactness is a very nice property, preserved by taking closed subspaces, continuous images, and Tychonoff products. For an illustration, here is a big list of ways to prove that every closed interval in $\Bbb R$ is compact and my most upvoted answer at MSE. :-)

Answer 16

I had a lot of difficulty understanding any of these answers because they have no pictures and I consider myself some what a visual learner, so this is an attempt at a Visual answer. Technical note: I'll be talking about discussing about metric spaces induced from vector space with norm (coz different norm give equivalent metric space structure). I will

Compactness through covers

I think the easiest to understand definition of compactness is through the open set definition. We begin by defining an open cover:

Open cover(deftn):A collection of open sets of a topological space whose union contains a given subset.

The picture for is that, you can imagine set as a region in space (here I take plane), and you want to cover it through some blankets(=subset). Now, what the blankets actually look like depends on the topology you choose. Here is a drawing:

You can see that the points that the red boundary and all points inside (call the union as A), is contained inside the union of the sets $P,G,B,O$. And if our topology were that all these sets were actually open, then we would say that these sets cover $A$.

Now, when we specialize to metric space (specifically 2-norm) and go by the induced topology,the shapes these covering blankets can be sewn in is restricted to that of circles (generally n-spheres) and these circles can not have edges (no boundary).

Compactness (deftn):Formally, a topological space X is called compact if each of its open covers has a finite subcover.

What the compactness tells is that if we had an infinite collection of blankets covering our set ( I can't draw this for obvious reasons), then it would be possible to sieve out some of these blankets and still cover the set with a finite number of blankets. In the picture, if it were actually that the sets I choose were compact, then it could be pretty easily concluded that it is compact. I mean, even if we were in standard topology it should be clear that we can take a ball large enough such that $A$ is in it.

You can imagine a blanket like the above but in a circular shape and the edges removed. Note that the actual "shape" of the border depends on the metric.

Compactness as being closed and bounded

In the above, I have drawn a typical compact set in $\mathbb{R}^2$ , and we can see that it is clearly bounded because we can draw a circle containing it (see purple) , and the set also has all it's boundary points hence it is closed.

Specific remarks on question by OP

OP asks, why boundedness is not enough for compactness. I think the best way to understand is through behaviour of sequence. If we really have a sense of finiteness, then it intuitive that the converging sequence converge inside the set only... and that is infact a theorem.

---

Bonus (?):

From page-234 of Road to Reality by Road to Penrose, the following picture is given: