Note to the reader: when I first started this post, I thought it would be a few easy paragraphs. The size began to swell, so I am taking this post in several bite sized parts. This is part I.

Pretty much everyone knows what a compact space is. But not everyone is up on equivalent definitions and on how to prove that the arbitrary product of compact spaces is compact.

The genesis of this blog post is David Wright’s *Proceedings of the American Mathematical Society* paper (1994) on Tychonoff’s Theorem.

Since I am also writing for my undergraduate topology class, I’ll keep things elementary where possible and perhaps put in more detail than a professional mathematician would have patience for.

I should start by saying why this topic is dear to me: my research area is knot theory; in particular I studied embeddings of the circle into the 3-dimensional sphere , which can be thought of as the “compactification” of ; basically one starts with and then adds a point and declares that the neighborhoods of this new point will be generated by sets of the following form: The reason we do this: we often study the complement of the embedded circle, and it is desirable to have a compact set as the complement.

I’ve also studied (in detail) certain classes of embeddings of the real line into non-compact manifolds; to make this study a bit more focused, I insist that such embeddings be “proper” in that the inverse image of a compact set be compact. Hence compactness comes up again, even when the objects of study are not compact.

**So, what do we mean by “compact”?**

Instead of just blurting out the definition and equivalent formulations, I’ll try to give some intuition. If we are talking about a subset of a metric space, a compact subset is one that is both closed and bounded. Now that is NOT the definition of compactness, though it is true that:

Given a set , is compact if and only if is both closed (as a topological subset) and bounded (in that it fits within a sufficiently large closed ball). In compact subsets can be thought of as selected finite unions and arbitrary intersections of closed intervals. In the higher dimensions, think of the finite union and arbitrary intersections of things like closed balls.

Now it is true that if is continuous, then if is a compact topological space, then is compact (either as a space, or in the subspace topology, if is not onto.

This leads to a big theorem of calculus: the Extreme Value Theorem: if is continuous over a compact subset then attains both a maximum and a minimum value over .

Now in calculus, we rarely use the word “compact” but instead say something about be a closed, bounded subset. In the case where we usually say that , a closed interval.

So, in terms of intuition, if one is thinking about subsets of , one can think of a compact space as a domain on which any continuous real valued function always attains both a minimum and a maximum value.

**Now for the definition**

We need some terminology: a collection of open sets is said to be an open cover of a space if and if a collection of open sets is said to be an open cover of if A finite subcover is a finite subcollection of the open sets such that .

Here is an example: is an open cover of in the subspace topology. A finite subcover (from this collection) would be

Let be a topological space. We say that is a compact topological space if any open over of has a finite subcover. If we say that is a compact subset of if any open cover of has a finite subcover.

Prior to going through examples, I think that it is wise to mention something. One logically equivalent definition is this: A space (or a subset) is compact if every cover by open basis elements has a finite subcover. Here is why: if is compact, then ANY open cover has a finite subcover, and an open cover by basis elements is an open cover. On the other hand: if we assume the “every open cover by open basis elements has a finite subcover” condition: then if is an open cover, then we can view this open cover as an open cover of the basis elements whose union is each open . This open cover of basis elements has a finite subcover of basis elements..say . Then for each basis element, choose a single for which . That is the required open subcover.

Now, when convenient, we can assume that the open cover in question (during a proof) consists of basic open sets. That will simplify things at times.

**So, what are some compact spaces and sets, and what are some basic facts?**

Let’s see some compact sets, some non compact sets and see some basic facts.

1. Let be any topological space and a finite subset. Then is a compact subset. Proof: given any open cover of choose one open set per element of which contains said element.

2. Let have the usual topology. Then the integers is not a compact subset; choose the open cover is an infinite cover with no finite subcover. In fact, ANY unbounded subset in the usual metric topology fails to be compact: for with choose ; clearly this open cover can have no finite subcover.

3. The finite union of compact subsets is compact (easy exercise).

4. If is compact and is a Hausdorff topological space () then is closed. Here is why: let and for every choose open where . Now is an open set which contains and has a finite subcover Note that each is an open set which contains and now we have only a finite number of these. Hence which is disjoint from which contains . Because was an arbitrary point in , is open which means that is closed. Note: this proof, with one minor addition, shows that a compact Hausdorff space is regular () we need only show that a closed subset of a compact Hausdorff space is compact. That is easy enough to do: let be an open cover for ; then the collection is an open cover for , which has a finite subcover. Let that be where each . Now since does not cover does.

So we have proved that a closed subset of a compact set is compact.

5. Let (or any infinite set) be given the finite complement topology (that is, the open sets are the empty set together with sets whose complements consist of a finite number of points). Then ANY subset is compact! Here is why: let be any set and let be any open cover. Choose . Since is a finite set of points, only a finite number of them can be in , say . Then for each of these, select one open set in the open cover that contains the point; that is the finite subcover.

Note: this shows that non-closed sets can be compact sets, if the underlying topology is not Hausdorff.

6. If is continuous and onto and is compact, then so is . Proof: let cover and note that covers , hence a finite number of these open sets cover: . Therefore covers . Note: this shows that being compact is a topological invariant; that is, if two spaces are homeomorphic, then either both spaces are compact or neither one is.

7. Ok, let’s finally prove something. Let have the usual topology. Then (and therefore any closed interval) is compact. This is (part) of the famous Heine-Borel Theorem. The proof uses the least upper bound axiom of the real numbers.

Let be any open cover for . If no finite subcover exists, let be the least upper bound of the subset of that CAN be covered by a finite subcollection of . Now because at least one element of contains and therefore contains for some . Assume that . Now suppose , that is is part of the subset that can be covered by a finite subcover. Then because for some then which means that , which means that isn’t an upper bound for .

Now suppose ; then because there is still some where . But since then and so . So if can be covered by then is a finite subcover of which means that was not an upper bound. It follows that which means that the unit interval is compact.

Now what about the closed ball in ? The traditional way is to show that the closed ball is a closed subset of a closed hypercube in and so if we show that the product of compact spaces is compact, we would be done. That is for later.

8. Now endow with the lower limit topology. That is, the open sets are generated by basis elements . Note that the lower limit topology is strictly finer than the usual topology. Now in this topology: is not compact. (note: none of are compact in the coarser usual topology, so there is no need to consider these). To see this, cover by and it is easy to see that this open cover has no finite subcover. In fact, with a bit of work, one can show that every compact subset is at most countable and nowhere dense; in fact, if is compact in the lower limit topology and there exists some where .