Just realised (two seeks ago, but only gotten around to finish this blog posting now) that an argument used to prove a proposition in a project I am working on is wrong. After reducing the problem to its core I found that it is something quite elementary. So today’s post would be of a different flavour from the ones of recent past.

**Question** Let be topological spaces. Let be a bounded, continuous function. Is the function continuous?

Intuitively, one may be tempted to say “yes”. Indeed, there are plenty of examples where the answer is in the positive. The simplest one is when we can replace the infimum with the minimum:

**Example** Let the space be a finite set with the discrete topology. Then is continuous.

*Proof* left as exercise.

But in fact, the answer to the question is “No”. Here’s a counterexample:

**Example** Let with the standard topology. Define

which is clearly continuous. But the infimum function is roughly the Heaviside function: if , and if .

So what is it about the first example that makes the argument work? What is the different between the minimum and the infimum? A naive guess maybe that in the finite case, we are taking a minimum, and therefore the infimum is attained. This guess is not unreasonable: there are a lot of arguments in analysis where when the infimum can be assumed to be attained, the problem becomes a lot easier (when we are then allowed to deal with a minimizer instead of a minimizing sequence). But sadly that is not (entirely) the case here: for every , we can certainly find a such that . So attaining the infimum point-wise is not enough.

What we need, here, is **compactness**. In fact, we have the following

**Theorem** If are topological spaces and is compact. Then for any continuous , the function is well-defined and continuous.

*Proof* usually proceeds in three parts. That follows from the fact that for any fixed , is a continuous function defined on a compact space, and hence is bounded (in fact the infimum is attained). Then using that the sets and form a subbase for the topology of , it suffices to check that and are open.

Let be the canonical projection , which we recall is continuous and open. It is easy to see that . So continuity of implies that this set is open. (Note that this part does not depend on compactness of . In fact, a minor modification of this proof shows that for any family of upper semicontinuous functions , the pointwise infimum is also upper semicontinuous, a fact that is very useful in convex analysis. And indeed, the counterexample function given above is upper semicontinuous.)

It is in this last part, showing that is open, that compactness is crucially used. Observe that . In other words an open set. This in particular implies that there exists a “box” neighborhood contained in . Now using compactness of , a finite subset of all these boxes cover . And in particular we have

and hence is open. Q.E.D.

One question we may ask is how *sharp* is the requirement that is compact. As with most things in topology, counterexamples abound.

**Example** Let be any uncountably infinite set equipped with the co-countable topology. That is, the collection of open subsets are precisely the empty set and all subsets whose complement is countable. The two interesting properties of this topology are (a) is not compact and (b) is hyperconnected. (a) is easy to see: let be some countably infinite subset of . For each let . This forms an open cover with not finite sub-cover. Hyperconnected spaces are, roughly speaking, spaces in which all open nonempty sets are “large”, in the sense that they mutually overlap a lot. In particular, a continuous map from a hyperconnected space to a Hausdorff space must be constant. In our case we can see this directly: suppose is a continuous map. Fix . Let be open neighborhoods of . Since is continuous, is open and non-empty (by the co-countable assumption). Therefore for any pairs of neighborhoods. Since is Hausdorff, this forces to be the constant map. This implies that for any topological space , a continuous function is constant along , and hence for any , we have is continuous.

One can try to introduce various regularity/separation assumptions on the spaces to see at what level compactness becomes a crucial requirement. As an analyst, however, I really only care about topological manifolds. In which case the second counterexample up top can be readily used. We can slightly weaken the assumptions and still prove the following partial converse in essentially the same way.

**Theorem** Let be Tychonoff, connected, and first countable, such that contains a non-trivial open subset whose closure is not the entire space; and let be paracompact, Lindelof. Then if is noncompact, there exists a continuous function such that is not continuous.

**Remark** Connected (nontrivial) topological manifolds automatically satisfy the conditions on and except for non-compactness. The conditions given are not necessary for the theorem to hold; but they more or less capture the topological properties used in the construction of the second counterexample above.

**Remark** If is such that every open set’s closure is the entire space, we must have that it is hyperconnected (let be a closed set. Suppose is another closed set such that . Then and vice versa, but is open, so . Hence cannot be written as the union of two proper closed subsets). And if it is Tychonoff, then is either the empty-set or the one-point set.

**Lemma** For a paracompact Lindelof space that is noncompact, there exists a countably infinite open cover and a sequence of points such that if .

*Proof*: By noncompactness, there exists an open cover that is infinite. By Lindelof, this open cover can be assumed to be countable, which we enumerate by and assume WLOG that . Define and inductively by: and choose .

*Proof of theorem*: We first construct a sequence of continuous functions on . Let be a non-empty open set such that its closure-complement is a non-empty open set ( exists by assumption). By connectedness , so we can pick in the intersection. Let be a sequence of points converging to , which exists by first countability. Using Tychonoff, we can get a sequence of continuous functions on such that and .

On , choose an open cover and points per the previous Lemma. By paracompactness we have a partition of unity subordinate to , and by the conclusion of the Lemma we have that . Now we define the function

which is continuous, and such that . But by construction , which combined with the fact that shows the desired result. q.e.d.