3
$\begingroup$

Let $X$ be a Polish space. $C_B(X)$ is the space of bounded continuous functions $X\to\mathbb{R}$ equipped with the strict topology, which is the finest locally convex topology that agrees with the compact-open topology on all norm-bounded subsets of $C_B(X)$.

I've already looked through a few papers on it, most notably "Bounded Continuous Functions on a Completely Regular Space" by Sentilles, which had a lot of handy results. Note that by Theorem 5.8, the assumption that $X$ is Polish implies that the strict, substrict, and superstrict topologies analyzed in that paper are all identical.

The particular conjecture I'm trying to show is that, for this topology, compactness is equivalent to sequential compactness. The obvious way to do this is to show that $C_B(X)$ is first-countable and Lindelof, as those two conditions are fulfilled by many spaces, and suffice to prove the usual "sequential compactness iff compactness" theorem. However, I'm not having much luck with establishing either of those, or finding papers which talk about whether the strict topology fulfills them.

Two intermediate results which might be of use are that $C_B(X)$ with the strict topology is separable, and that it can be non-metrizable.

EDIT: The Eberlein-Smulian theorem seems like it's probably applicable here, as it's a functional analysis theorem about how compactness equals sequential compactness for the weak topology even when first-countability can't be assumed. Sadly, it relies on the original space being a Banach space, which is something I don't think we can assume here. I'm starting to think that either first-countability or Lindelofness will actually fail here and that any solution to this will somehow be a variant of the Eberlein-Smulian theorem, but I got a bit lost in the details of the proof when trying to adapt it to this case.

$\endgroup$
1
  • $\begingroup$ This is a version of Eberlein's theorem stated in Grothendieck's book Topological Vector Spaces (page 209): Let $E$ be a quasi-complete LCTVS. A subset of $E$ is weakly relatively compact if and only if every subsequence admits a weak clusterpoint. $\endgroup$ Mar 19 at 12:07

1 Answer 1

0
$\begingroup$

Alright, I think I've got a proof of compactness being equivalent to sequential compactness (all countable sequences have a convergent subsequence with a limit point in the set). It relies on a series of reductions to easier and easier proof targets, until we get to something we can actually prove.

First, definitions.

A set $A$ is countably relatively compact if, given any sequence in it, that sequence has a cluster point (which may or may not be in $A$)

A set $A$ is sequentially compact if, given any sequence in it, there's a subsequence that converges to a point that lies in $A$

A set $A$ is compact if, given any open cover of it, there's a finite subcover.

A space is angelic if $A$ being countably relatively compact implies that $\overline{A}$ (topological closure) is compact, and that all points in $\overline{A}$ have a sequence from $A$ that limits to them.

Reduction step 1: From Weakly Compact Sets by Klaus Floret, on page 40, there's a theorem that in angelic spaces, sequential compactness = compactness. That's exactly what we're going for, so our proof goal shifts to showing that $C_B(X)$ is angelic.

Reduction Step 2: From Weakly Compact Sets by Klaus Floret, on page 40, there's a theorem that, if $\phi:Y\to Z$ is a continuous injection, and $Y$ is a regular space, and $Z$ is angelic, then $Y$ is angelic. In particular, we'll be letting $Y$ be $C_B(X)$ with the strict topology, and $Z$ being $C_B(X)$ with the weak topology, and $\phi$ being the identity function. It's clearly an injection, and continuous because the weak topology has fewer opens than the strict topology. Thus, our proof goal shifts to showing that $C_B(X)$ with the weak topology is an angelic space, and that $C_B(X)$ with the strict topology is a regular space.

Reduction Step 3: From Corollary 1.11.iii in "On Compactness in Locally Convex Spaces" by B. Cascales and J. Orihuela, $C_B(X,\mathbb{R})$ with the weak topology induced by the strict topology is an angelic space, as long as $X$ has a dense K-analytic subset, and $\mathbb{R}$ is in their specified class of nicely-behaved locally convex vector spaces. The latter holds because they were bragging about how large their class of vector spaces was, and $\mathbb{R}$ is, in a sense, the nicest-behaved vector space there is, so if their stunningly general class of vector spaces doesn't include $\mathbb{R}$, something has gone horribly wrong. So, anyways, we've reduced the problem to showing that $C_B(X)$ with the strict topology is a regular space, and that $X$ has a dense K-analytic subset.

Reduction Step 4: From "Compact Coverings for Baire Locally Convex Spaces" by J Kakol and M Lopez Pellicer, in the introduction, $X$ being analytic implies it's K-analytic, and a space $X$ is analytic if there's a continuous surjection $\mathbb{N}^{\mathbb{N}}\to X$. Also, by the Wikipedia article for Baire Space (set theory), every Polish space $X$ fulfills that property. Our space is Polish, so by this argument, the entire space $X$ counts as a dense K-analytic subset of $X$. The problem has now been reduced to just showing that $C_B(X)$ with the strict topology is a regular space.

Reduction Step 5: We'll prove that if $C_B(X)$ with the strict topology has the following property, it's regular. The relevant property is...

There's a local basis of the constant-0 function consisting of open sets, $\mathbb{B}$, where, for every $U\in\mathbb{B}$, there's a open set $U'$ s.t. $0\in U'\subseteq U$, and for all $f\not\in U$, there's an open neighborhood of $f$ that's disjoint from $U'$. (strict topology used throughout)

Assume this property holds. Then, given any point $f$ and closed set $C$ with $f\not\in C$, we can generate disjoint open neighborhoods of $f$ and $C$ (establishing that $C_B(X)$ is normal), as follows. $C^c$ is an open that contains $f$. Thus, $C^c-f$ (shifting our open set) is an open neighborhood of 0. Then, we can pick a smaller open from the local basis of 0, call it $U$. Then, we generate the indicated set $U'$ from our assumption. And now, $U'+f$ and $(\overline{U'}+f)^{c}$ will act as our disjoint open neighborhoods for the point $f$ and closed set $C$, respectively.

Verifying openness and disjointness is pretty easy, as is verifying that $f\in U'+f$, because $U'$ was an open neighborhood of 0. Establishing that $C\subseteq(\overline{U'}+f)^{c}$ is more difficult. We'll do this by showing that $\overline{U'}\subseteq U$, because if we can show that, then it'd mean that $\overline{U'}+f\subseteq U+f\subseteq(C^c-f)+f=C^c$ (remember, $U$ was selected to be an open subset of $C^c-f$). And then, from that, it means that $\overline{U'}+f\cap C^c=\emptyset$, and from there, that $C\subseteq(\overline{U'}+f)^c$, which is what we wanted to show.

That just leaves showing that $\overline{U'}\subseteq U$, which can be done, because by our starting assumption, all the $f\not\in U$ have an open neighborhood $U_f$ that doesn't intersect $U'$, which witnesses that $f$ is not included in the intersection of closed neighborhoods of $U'$. Thus, the closure of $U'$ can't include anything from outside of $U$, and everything works, we've proven that $C_B(X)$ with the strict topology is normal.

And so, all that remains is just proving that assumption, that there's a local basis of the constant-0 function consisting of open sets, $\mathbb{B}$, where, for every $U\in\mathbb{B}$, there's a open set $U'$ s.t. $0\in U'\subseteq U$, and for all $f\not\in U$, there's an open neighborhood of $f$ that's disjoint from $U'$.

Reduction Step 6: Actually proving that damn thing. We'll be using Theorem 2.4a from Bounded Continuous Functions on a Completely Regular Space, by F Dennis Sentilles, that explicitly gives a local basis. It is as follows. It's parametrized by a family $\{K_n,a_n\}_{n\in\mathbb{N}}$, where the $K_n$ are compact subsets of the space $X$, and $a_n$ are a sequence of numbers s.t. $\min_{n}a_n>0$ and $\liminf_{n\to\infty}a_n=\infty$. The open set corresponding to a family like that is $$\{f|\forall n:f(K_n)\subseteq(-a_n,a_n)\}$$ These fulfill the property to be a strict-open, that, when restricted to any norm-bounded subset of functions, it looks like an open set from the compact-open topology.

Where were we? Ah right, taking one of them, some arbitrary $U$, and generating a new strict-open neighborhood $U'$ where, for every $f\not\in U$, there's a strict-open neighborhood of $f$ that doesn't intersect $U'$.

So, pick an arbitrary $U$, alternately writeable as $$U=\{f|\forall n:f(K_n)\subseteq(-a_n,a_n)\}$$ For some sequence of compact sets and numbers. Now, we can pick a $U'$ from the same local basis of 0 as follows. $$U':=\{f|\forall n:f(K_n)\subseteq\left(-\frac{a_n}{2},\frac{a_n}{2}\right)\}$$ And, given some $f\not\in U$, here's how we generate its associated open neighborhood $U_f$. Given $f$, there must be some $n_f$ where $f(K_{n_f})\not\subseteq(-a_{n_f},a_{n_f})$. Let $x_f$ be some point from $K_{n_f}$ that lands outside that open. Then, let $U_f$ be defined as $$\{g|g(x_f)\in\left(f(x_f)-\frac{a_{n_f}}{4},f(x_f)+\frac{a_{n_f}}{4}\right)\}$$

Alright, let's get going. Obviously, $0\in U'\subseteq U$. $U'$ and $U$ were both picked from the local basis of open neighborhoods of 0, so they're strict-open. All the $U_f$ are open in the compact-open topology, which is coarser than the strict topology so they're strict-open. Obviously, for any $f$, it's in $U_f$. That just leaves showing that if $f\not\in U$, that $U_f\cap U'=\emptyset$.

Let's say that there was a point in both sets, some function $g$. Then, $g(x_f)\in\left(f(x_f)-\frac{a_{n_f}}{4},f(x_f)+\frac{a_{n_f}}{4}\right)$ (since it's in $U_f$). And also, $x_f$ was picked to be a point where $f(x_f)\not\in(-a_{n_f},a_{n_f})$. If you work it out, $f(x_f)$ went either high or low, and in either case, $g(x_f)$ is distant from 0. In particular, we have $g(x_f)\not\in\left(-\frac{3a_{n_f}}{4},\frac{3a_{n_f}}{4}\right)$. However, we have $g(x_f)\in\left(-\frac{a_{n_f}}{2},\frac{a_{n_f}}{2}\right)$ since $g\in U'$. Contradiction.

So, no point $g$ can be in both sets, $U_f$ and $U'$, they're disjoint. Since $f$ was arbitrary not in $U$, we have that every $f\not\in U$ has an open neighborhood $U_f$ that's disjoint from $U'$. And this argument goes through regardless of which $U$ we picked from the base, and we finally hit our proof target and everything's done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.