So much of the properties of compact sets are motivated by finite sets, to the point that thinking of compact sets as topologically finite sets may yield some deeper understanding. But finite sets have the intuitive property that every subset of a finite set is also compact(also finite), why is it that compact sets give this up?

It is easy to state that they just do and provide the example $I=(0,1)$ and $K=[0,1]$ as an example, but the problem with that is it really only helps illuminate how compact sets work in the euclidean spaces. It isn't generally true in all spaces that open sets aren't compact, or that closed and bounded sets are compact. So there is a problem generalizing the ideas.

The heart of my question is: Given a subset $E$ of a compact set $K$ why isn't it compact?

Simple Answer: Because there is an open cover of $E$ which has no finite subcover

The deeper question: Why?

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    $\begingroup$ I would say that the question illustrates clearly why one shouldn't take the analogy between finite sets and compact sets too far. Finite sets are very very special compact sets. In particular, the phenomenon of accumulation points belonging to the set is lost entirely simply because a finite set has no accumulation points. Removing an accumulation point is enough to turn a compact set into a non-compact one, showing why one would not ever expect compact to be hereditary to all subsets. $\endgroup$ Commented Feb 12, 2019 at 18:47
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    $\begingroup$ It seems to me that compact sets are motivated by closed intervals, not by finite sets. What are the serious applications of finite sets in analysis? $\endgroup$
    – saulspatz
    Commented Feb 12, 2019 at 18:47
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    $\begingroup$ Why are subsets of compact sets not necessarily compact? Because the definition of compactness allows counterexamples to exist. $\endgroup$
    – Aweygan
    Commented Feb 12, 2019 at 18:49
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    $\begingroup$ No need for a separation axiom there. Any closed subset of a compact set is compact, full stop. $\endgroup$
    – jmerry
    Commented Feb 12, 2019 at 18:56
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    $\begingroup$ Respectfully, I would like to mention that every time during my education that I was consumed with a question such as this one, people couldn't wait to tell me that there was no "why" to be found. They never convinced me. $\endgroup$
    – Eric Auld
    Commented Feb 12, 2019 at 22:35

4 Answers 4


There's a nice symmetry/duality here stemming from the fact that "finite" has two generalizations in the topological category.

Images and subsets of finite sets are finite.

Discrete topological spaces generalize finite sets in that subspaces inherit the property.

Compact topological spaces generalize finite sets in that images inherit the property.

A topological space that is discrete and compact is finite.

  • $\begingroup$ Is there a way to define Discrete and Compact in that way somehow? $\endgroup$ Commented Feb 13, 2019 at 1:02
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    $\begingroup$ Does this answer the OPs question? Compact sets generalize the image property, but why don’t they satisfy the heredity property? $\endgroup$ Commented Feb 13, 2019 at 2:10
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    $\begingroup$ @SantanaAfton imagine you are a primitive human trying to develop a language, and your tribe is comprised of adult foragers. Because everyone is large and a human, you choose the single word "person" to describe a large human. However, you eventually need to distinguish between a baby and an adult, so you invent the notion of "size". Now an adult is a "large person" while a baby is a "small person". Then one day you are being attacked by a tiger and need a way to describe it to your tribe. There are no other tigers for a size comparison, and it is not human, so you are stumped... $\endgroup$ Commented Feb 13, 2019 at 7:21
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    $\begingroup$ ... so you invent the concept of "type", distinguishing between "humans" and "not-humans". Now you can say your tribe is comprised of "large humans", a baby is "small", and a tiger is "non-human". Though the tiger part of that analogy could be improved, this is exactly the situation at hand with finiteness. The notion of finiteness is quite natural, and so we defined it first, but eventually we came across structures that are akin enough to finiteness to share common properties but are quite distinct. With enough time, we boiled the main two properties down to compactness and discreteness. $\endgroup$ Commented Feb 13, 2019 at 7:24
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    $\begingroup$ It seems a bit inaccurate to describe “discrete” as a generalisation of “finite” to the topological category, since discrete spaces correspond to arbitrary sets, not just finite ones. $\endgroup$ Commented Feb 13, 2019 at 11:37

One way to think of it is that compactness means that sequences (more generally, nets) cannot "run away". There are two types of "running away":

  1. Running away to infinity (failure of the set to be bounded, or some analog*. Failure of the sequence/net to have any limit point at all).
  2. Running away to a limit point which isn't in the set (failure of a set to be closed. The sequence/net has a limit point(s) in an ambient space, but that point is missing from the set in question).

If I have a compact space $X$, and I remove a point, $X-\{p\}$ may suddenly permit the second type of "running away".

At first it seemed awkward to me to phrase these things in terms of sequences and nets, because sequences and nets are "discrete" objects describing a continuous thing. But one can always phrase everything in terms of nets/filters of open sets, not of points. That can make it seem a bit more natural.

In any case, the basic point of a compact set is that it does not allow you to play a certain kind of game with infinity.

*For example, a uniform space is compact iff totally bounded and Cauchy complete, which are exactly analogous to conditions 1 and 2 above.

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    $\begingroup$ "running away" is cute. +1 $\endgroup$
    – fleablood
    Commented Feb 12, 2019 at 22:16
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    $\begingroup$ Your two types of running away are actually the same. You can add points to an ambient space providing limits for those "unbounded" sequences. $\endgroup$ Commented Feb 13, 2019 at 17:56
  • $\begingroup$ @PaulSinclair They can be unified, but I find sometimes refinement can be more enlightening than unification. (But why not both) $\endgroup$
    – Eric Auld
    Commented Nov 9, 2023 at 6:43
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    $\begingroup$ I don't see it as unifying disparate ideas, but recognizing that what you thought were disparate ideas were actually the same within the confines of area of discussion (in this case, topology). In the post, you talked about removing a point $p$ from $X$ permitting the second type of running away. But from a strictly topological viewpoint, "bounded" means "contained within a compact set", and removing $p$ results neighborhoods of $p$ becoming unbounded. It is only when non-topological concerns are considered that your two forms of running away become distinct. $\endgroup$ Commented Nov 9, 2023 at 13:08

What you may be after is the idea of a "pre-compact" subset, that is, a subset whose closure is compact. Then every subset of a pre-compact set is pre-compact; as are finite unions of them. A compact set is a closed pre-compact set.

For metric spaces, there is a closely connected concept of "totally bounded" subsets; a complete totally bounded set is compact.

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    $\begingroup$ This is not quite an answer to the question. Would have been better placed as a comment. $\endgroup$ Commented Feb 12, 2019 at 18:52

Because a subset of a compact set is smaller than the compact set, the subset might have a different open cover that does not cover the compact set.

This open cover may not have a finite subcover.

Example: Let $A = [0,1]$ and $B = (0, 1] \subset A$.

Now $U = \{U_i| U_i= (\frac 1i, 1.1)\}$ is an open cover of $B$ but it is not an open cover of $A$. (And $U$ does not have a finite subcovering of $A$.)

To extend $U$ so that that it will cover $A$ we must add an open set that contains $0$. Call that $U_{\alpha}$ and $0 \in U_{\alpha}$ and now $U \cup \{U_{\alpha}\}$ is an open cover of $A$.

But $U_{\alpha}$ is open so there is an $r > 0$ so that $N_r(0) = (-r, r) \subset U_\alpha$. But we can find an $n > \frac 1r$ or in other words $0 < \frac 1n < r$.

So $(0, \frac 1n] \subset U_{\alpha}$ so $(0, \frac 1n]$ is covered but the single open set $U_{\alpha}$. Without $U_{\alpha}$ and with only $U = \{U_i = (\frac 1i, 1.1)\}$ we would have needed an infinite number of $U_i| i > n$ to cover $(0, \frac 1n]$. But with $U_{\alpha}$ we don't need ANY of them anymore.

So ... throw them away! We are left with $\{U_\alpha\} \cup \{U_i|i \le n; n > \frac 1r\}$ and that is a finite subcover of $A$. And of $B$.

But the point is. Without the requirement that there is an open set containing $0$ we wouldn't have a situation where a single open set must "do the work" of an infinite number of open sets which a non-compact set without the point $0$ could require.

... more explicitly with maybe too much detail...

So $A \setminus U_\alpha \subset (\frac 1n, 1]\subset B$. And $(\frac 1n, 1]$ is covered by the finite subclass $\{U_i| i \le n\}$ and we don't need $\{U_i| i > n\}$ any more because $U_\alpha$ covers everything in $A$ that was not covered in $\{U_i|i > n\}$.

(Namely $U_{\alpha}$ covers $\{0\} \cup (0, \frac 1n]$ whereas without $U_\alpha$ we needed ALL of $\{U_i| i > n\}$ to cover $(0, \frac 1n]$)

So $\{U_i|i \le n\} \cup \{U_\alpha\} \subset U \cup \{U_\alpha\}$ is a finite subcover of $A$. (even that $U$ had no finite subcover of $B$.

  • $\begingroup$ In my opinion, the fact that $0$ belongs to the set we are considering is the key of all. Namely, the behaviour and the effects a single neighborhood of $0$ has. I suppose it is related with the accumulation points of compact sets. Nice answer @fleablood. $\endgroup$
    – Dog_69
    Commented Feb 12, 2019 at 21:19
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    $\begingroup$ That's exactly how I see it @Dog_69. In terms of "compact" and and finite subcovers we don't have a good terminology for the points in $A\setminus B$ where $B$ is an non-compact subset... but in essence; the set $A\setminus B$ contain accumulation points of $B$ that aren't in $B$ and any open set containing $A\setminus B$ makes any infinite subcover of $B$ unnecessary. $\endgroup$
    – fleablood
    Commented Feb 12, 2019 at 22:15

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