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
![\displaystyle f(x,y) = \begin{cases} 1 & x > 0 \\ 0 & x < -e^{y} \\ 1 + x e^{-y} & x\in [-e^{y},0] \end{cases}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%28x%2Cy%29+%3D+%5Cbegin%7Bcases%7D+1+%26+x+%3E+0+%5C%5C+0+%26+x+%3C+-e%5E%7By%7D+%5C%5C+1+%2B+x+e%5E%7B-y%7D+%26+x%5Cin+%5B-e%5E%7By%7D%2C0%5D++%5Cend%7Bcases%7D&bg=fff&fg=444444&s=0&c=20201002)
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.
Bubbles Bad; Ripples Good 