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A compact topological space is defined as a space, $C$, such that for any set $\mathcal{A}$ of open sets such that $C \subseteq \bigcup_{U\in \mathcal{A}} U$, there is finite set $\mathcal{A'} \subseteq \mathcal{A}$ such that $C \subseteq \bigcup_{U'\in \mathcal{A'}} U'$.

Now, this definition leads to many interesting results, but if I were teaching someone about compact sets, how would I motivate this? Concepts like sequential compactness, open and closedness, and even connectedness are reasonably easy to motivate. I can not see how to motivate this definition. Compact spaces are often seen as generalizations of finite spaces. They are also seen as a generalization of boundedness and closedness. I can't see how to connect the definition with these concepts.

Alternatively, is there a definition of a compact set which is easier to motivate?

Euler88 ...
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    I have an answer in mind about emphasizing that a set being compact is kind of like a set being small or finite, but a question: these students have presumably taken Real Analysis, so are familiar with the Heine-Borel or something? – Moya Aug 26 '15 at 02:41
  • @Moya I actually didn't have any specific students in mind (I was actually thinking of a blog post.) I was planning on making it pretty basic actually (starting from set theory.) – Christopher King Aug 26 '15 at 02:44
  • Well that's slightly more difficult. To me, the prototypical discussion of compactness as a generalization of finiteness is the separation of disjoint compact sets in a Hausdorff space. In this way, compact sets kind of behave like finite point sets. However, that takes a bit of work to get to. – Moya Aug 26 '15 at 02:47
  • @Moya A more advanced answer is fine. It just a simpler one would be better. – Christopher King Aug 26 '15 at 02:49
  • If sequential compactness is easy to motivate, so is (quasi)compactness: "every sequence has a convergent subsequence" is generalised to "every net has a convergent subnet". – Daniel Fischer Aug 26 '15 at 13:49
  • This question has come up many times before. See, e.g., here: http://math.stackexchange.com/questions/371928/what-should-be-the-intuition-when-working-with-compactness/1165378#1165378 . It is an "existential" question indeed. – Christian Blatter Aug 26 '15 at 17:39

9 Answers9

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One of my favorite textbooks is Klaus Janich's Topology, and he has a nice motivation for compactness I feel, namely why we should care about. This is in addition to my comment about compact subsets of a Hausdorff space being essentially like finite point sets. But he writes:

In compact spaces, the following generalization from "local" to "global" properties is possible: Let $X$ be a compact space and $P$ a property that open subsets of $X$ may or may not have, and such that if $U$ and $V$ have it, then so does $U\cup V$. Then if $X$ has this property locally, i.e. every point has a neighborhood with property $P$, then $X$ itself has property $P$.

This is nice, but it is slightly advanced, and he gives some examples that follow like a continuous/locally bounded map from a compact space to $\mathbb{R}$ is bounded, and some discussions of locally finite covers and manifolds (honestly, I like this book after the fact of learning topology, not to learn from).

Hope that helps somewhat.

Moya
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  • Yes, Jänich's book is great. I did not understand compactness when I first read that passage (which was indeed my first contact with the concept), but I believe almost nobody understands compactness immediately. – leftaroundabout Aug 26 '15 at 09:19
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According to Munkres, the original definition of compactness is a space which satisfies the Bolzano-Weierstrauss property holds. That is, if every infinite subset has a limit-point.

Unfortunately, it turns out, this conception of compactness,sometimes called limit point compactness, doesn't have all the useful properties that compactness has.

For example, the continuous image of a limit point compact space need not be limit point compact. Also, a limit point compact subspace of a Hausdorff space need not be closed.

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    I am looking for regular compactness, not any variants of it. – Christopher King Aug 26 '15 at 02:45
  • Yes. I understand that. I'm trying to show you some of the reasons why the current definition of compactness is preferable to the original. – Tim Raczkowski Aug 26 '15 at 02:47
  • I think history is great motivation. It explains where concepts came from in the first place. – mathematician Aug 26 '15 at 02:52
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    This definition can actually be generalized to a topological context using the notion of a converging net, which generalizes the notion of a converging sequence. A set is compact iff any net contains a converging subnet. When I was first introduced to topology, we sometimes gave the equivalent definition using nets, it might lead to some more intuition because you can reuse your intuition about limits, at the same time the notion of a net is confusing. Of course this won't help OP but it's still interesting I hope. – blue Aug 26 '15 at 16:18
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An equivalent def'n is that if $ F$ is a non-empty family of closed sets with the F.I.P. (Finite Intersection Property) then $\cap F \not = \phi $ . This generalizes the idea of limits , and you can show that many results, e.g. on bounded closed subsets of $ R^n$ , using this property, so it is seen to be a useful tool that a space is compact. Once you show some additional consequences, e.g. that a continuous image of a compact set is compact, you can show how to apply them, e.g. in analysis, showing that an extremum exists, (hence the Mean Value Theorem in calculus). So you get easier results and new ones, from the compactness.

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For boundedness: You can give an exercise so that the students need to show that a bounded metric can induce the same topology as an unbounded one (at least you can easily show that for metric spaces with $d(x,y)$ and $\frac{d(x,y)}{1+d(x,y)}$). So, boundedness is not really a topological property.

If each open covering has a finite subcovering (and using that bounded metric instead of the regular one), you can associate compact sets with sets that are not too big, having something like an idea of boundness. In fact, they behave "pointwise" ($T_2$ and compact implies $T_4$, $f(K)$ compact again, for $K$ being compact, etc.) As you are asking for a motivation, I think the metric space should be fine.

mdot
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Topological spaces are made out of open sets.

Sometimes, you have the occasion to write your topological space as a union of open sets: e.g. because whatever you are trying to study is easy to understand when restricted to just one of the sets.

If you can motivate that it is useful to do this sort of thing, then the usefulness of the usual definition of compactness is almost self-evident; a finite union is much easier to work with than an infinite union.

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The motivation for the definition of compactness is that that condition is extremely useful. Essentially every proof of every fact about the Riemann integral on the line, for example, depends on it.

Definitions capture useful properties which allow us to prove useful things — the good ones, at least.

If you want to motivate the definition, give the definition and immediate proceed to prove useful things with it. If you are looking for an a priori reason that justifies the definition, well, there is none.

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    By the time students are exposed to the pure point-set-topology definition of co pactness they should have seen many times situations where compactness is used. If not, then show them that. – Mariano Suárez-Álvarez Aug 26 '15 at 02:39
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I think the better way to motivate compactness is in real analysis, a set $F$ is said be a compact set if it bounded and closed. Is very easy to imagine something compact like this one. The general definition, in general topological spaces, is motivate by Bolzano-Weierstrass theorem.

Euler88 ...
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  • But for metric spaces, compactness is the same as sequential compactness. – Christopher King Aug 26 '15 at 02:59
  • In general metric spaces a set $F$ may be closed and bounded and isn't compact. The only thing is that in $\mathbb{R}^n$ the three notions are equivalent, so $\mathbb{R}^n$ gives you a motivation for generalize the definition. – Euler88 ... Aug 26 '15 at 03:02
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Well I like to think of compactness like this because it gives me the idea of a compact set in terms of closed sets.

A collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for every finite subcollection

$$\{C_1,\cdots C_n\}$$ of $\mathcal{C}$ , $\bigcap_{i=1}^{n} C_i$ should be nonempty.

Now if in a topological space every collection $\mathcal{C}$ of closed sets in $X$ have the finite intersection property , then the space $X$ is compact.

Well instinctively you would always think of a compact set as a closed set(though we have examples where compact sets are not closed.. One lies in your backyard!!) here is a definition involving closed sets.

Also you can use this definition to prove the tychonoff theorem

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    The way you stated the connection between the finite intersection property and compactness seems wrong. For every collection $C$ of closed sets in $X$ which have the finite intersection property, $\cap C$ should be nonempty. – Massimo Ortolano Aug 26 '15 at 14:56
  • Thanks for showing that. I overlooked that @MassimoOrtolano –  Aug 26 '15 at 15:54
  • @Massimo wasn't asking you to replace "is" with "should be" -- the point is that compactness says that every collection of closed sets that has the finite intersection property has nonempty intersection. But even otherwise, I don't see what you're trying to say with the answer -- what does "thinking of compact sets as closed" have to do with using a definition that involves the phrase "closed sets"? – Abhimanyu Pallavi Sudhir Aug 29 '19 at 05:55
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A compact set is the "next best thing to finite"- every compact set is both closed and bounded (in a metric space), just like a finite set. In a metric space, given any point, p, there are both minimum and maximum distances from p to points in the compact set.

user247327
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