### Continuity of the infimum

#### by Willie Wong

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.

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I had been working on a problem where I needed to know why your first theorem was true. Thanks for this article!

Would you mind explaining the following steps in the proof of your first theorem? “And in particular we have {x} \times Y…” and everything that follows — i.e., I don’t understand the last two lines of your proof.

I was thinking that the second-to-last line of your proof (the line I quoted part of) should be true regardless of the compactness of Y. Can you explain why we need compactness here? One reason for my confusion is the fact that the open sets V_(x,y) are no longer mentioned. I also am confused about why, in the second-to-last line in the proof you consider the intersection of all the U_(x,y_i).

Perhaps most importantly, I totally fail to see how the last line (proof conclusion) follow from the second-to-last line. Can you help me understand?

Thank you very much for posting this on your blog.

Compactness is only used to extract a finite covering when given an open cover. This is so that the intersection is open. If infinitely many are needed, the infinite intersection of open sets need not be open. I encourage you to follow through the proof of the theorem while thinking about the previous example, this way you will see why compactness is important.

Now why do we take the intersection? The point is that you start with and want to conclude that some neighbourhood of is also satisfies the same condition. We know that for each fixed there is a small neighbourhood as desired because is continuous. But the width of this neighbourhood may be non-uniform. At some points this neighbourhood can be allowed to be quite large and at some points this neighbourhood can be small. By taking the intersection we are looking for the largest neighbourhood of such that holds for any point in this neighbourhood and also any . Again, draw a picture of the example of the non-compact case, and it should be pretty clear. (Alternatively, you can try proving the same statement in metric spaces or just on and and maybe you will see the mechanism clearer.)

For the final implication: as mentioned before, implies that . The displayed formula shows that there exists some open (due to compactness of !) set about this such that . This implies that contains an open neighbourhood of . So that it is open as needed.

Thank you very much for the explanation. If I’m not mistaken, it sounds like you must use the compactness of Y in the last paragraph of your explanation as well in order to prove that N_x \times Y \subset f^{-1}((b,\infty)) implies that N_x is a subset of g^{-1}((b,\infty)). (Sorry if my LaTeX doesn’t work…I’m not sure how to use it on this website.)

Anyway, I think I understand now. Thank you so much for taking the time to help.

Hi Matthew, sorry for the late reply. Great that you understood. To use LaTeX on WordPress.com blogs, you have to use the usual $…$ incantation, but with the word “latex” included just inside the first dollar sign (no spaces).

[…] assumption; continuity would suffice. The statement below is taken from Willie Wong’s blog where a proof is […]