This post is about two partial converse results to the main theorem from my last post. Both of these results were established by John rather than me.

At present I don’t know if the full converse is true. I have more converse results than are present in this post but we never did prove the full form.

Theorem: Let X be a topological space. The following three conditions are equivalent.

1. X is normal
2. X is \(OCA(V)\) for any finite dimensional V
3. X is \(OCA(\mathbb{R})\)

Proof:
The implication \(1 \implies 2\) follows from our previous theorem. The implication \(2 \implies 3\) is obvious. The implication \(3 \implies 1\) is thus the only one we need to prove.

Let \(F, G \subseteq X\) be disjoint closed sets. Define \(f : X \to \mathbb{R}\) as \(f|_F = -1\), (f|_G = 1\) and \(f(x) = 0\) elsewhere.

Claim: \(\rho(f) \leq \frac{1}{2}\). Proof: \(f(G^c) \subseteq \overline{B}(-\frac{1}{2}, \frac{1}{2})\), \(f(F^c) \subseteq \overline{B}(\frac{1}{2}, \frac{1}{2})\) and because the two sets are disjoint and closed this forms an open cover of the space.

Therefore if X is \(OCA(\mathbb{R})\) we can find a continuous function \(g\) with \(||f – g|| \leq \frac{3}{4}\). Then the sets \(\{ x : g(x) < 0\}\) and \(\{x : g(x) > 0 \}\) are disjoint open sets containing F and G respectively.
QED

We can also detect countable paracompactness fairly straightforwardly.

Theorem: Let X be a topological space. Let \(c_0\) be the space of real valued sequences which converge to 0 given the uniform metric. The following three conditions are equivalent.

1. X is countably paracompact
2. X is \(OCA(V)\) for any separable V
3. X is \(OCA(c_0)\)

Proof:
Again we need only prove the implication \(3 \implies 1\).

Let \(U_n\) be a countable open cover of \(X\). Define \(f(x) = e_m\) where \(e_n\) is the sequence that is 1 at position n and 0 elsewhere and \(m = \min \{ k : x \in U_k \}\).

Claim: \(\rho(f) \leq \frac{1}{2}\). Proof: \(f(U_n) = \overline{B}(\frac{1}{2}(e_0 + \ldots + e_n), \frac{1}{2})\).

Now let \(||g – f|| < \frac{3}{4}\) and let \(V_n = U_n \cap \{x : g(x)_n > \frac{1}{4}\}\).

This is an open cover: If \(m = \min \{ k : x \in U_k \}\) then we have \(f(x)_m = 1\), so \(g(x)_m \geq \frac{1}{4}\) and thus \(x \in V_m\). It is obviously a refinement of \(U_n\)

Further, \(V_n\) is locally finite. Let \(x \in X\) and pick \(W \ni U\) open such that \(||f(x) – f(y) < \frac{1}{8}\) for \(y \in W\). Now pick \(N\) such that \(|f(x)| < \frac{1}{8}\) for \(n \geq N\). Then for \(y \in W\) and \(n \geq N\) we have \(|g(y)_n| < \frac{1}{4}\) so \(y \not\in V_n\). Thus \(\{n : V_n \cap W \neq \emptyset\} \) is finite. Thus we have found a locally finite refinement of our open cover and \(X\) is countably paracompact. QED There's a generalization of the proof for \(c_0\) I came up with that I'm a little dubious about the value of but which does allow one to establish various other converse results that fall short of full paracompactness.

In a previous blog post I looked at the question “How well can you approximate a set of bounded real-valued functions by a single continuous one?”.

This post looks at the generalization that’s at the heart of what I’m writing up: How does this work when we replace \(\mathbb{R}\) with an arbitrary normed space?

I won’t be looking at the question of whether a center exists (it turns out that the question is better answered independently of this theory). Instead we will generalize the result that $$\frac{1}{2} \sup\limits_x (\mathcal{F}^*(x) – \mathcal{F}_*(x)) \leq r(\mathcal{F}, C(X))$$ and that this inequality was an equality when \(X\) was normal.

The functions \(\mathcal{F}^*(x)\) and \(\mathcal{F}_*(x)\) don’t really make sense for arbitrary vector spaces (they depend inherently on the ordered structure of \(\mathbb{R}\)), but their difference does. The following lemma will help point out how to generalize them:

Theorem: Define the oscillation of a family \(\mathcal{F}\) at a point \(x\) as $$\omega_x(\mathcal{F}) = \inf\{ \mathrm{diam}(\mathcal{F}(U)) : U \textrm{ open }, U \ni x \} $$

where \(\mathcal{F}(A) = \bigcup\limits_{f \in \mathcal{F}} f(A) \). (This is a natural generalization of the usual definition).

Then $$ \mathcal{F}^*(x) – \mathcal{F}_*(x) = \omega_x(\mathcal{F})$$

Proof:

Let \(\epsilon > 0\).

Let \(U \ni x\) such hat \(\mathrm{diam}(\mathcal{F}(U)) \leq \omega_x(\mathcal{F}) + \epsilon\). Then for \(u, v \in U\) and \(f, g \in \mathcal{F}\) we have \(|g(x) – f(u)| \leq \omega_x(\mathcal{F}) + \epsilon\). So \(f(u) \leq g(v) + \omega_x(\mathcal{F}) + \epsilon\). Taking the supremum and infimum we have
$$\mathcal{F}^*(x) \leq \sup \mathcal{F}(u) \leq \inf \mathcal{F}(v) + \omega_x(\mathcal{F}) + \epsilon \leq \mathcal{F}_*(x) + \omega_x(\mathcal{F})(x) + \epsilon$$

Hence $$\mathcal{F}^*(x) – \mathcal{F}_*(x) \leq \omega_x(\mathcal{F})(x) + \epsilon$$

Letting \(\epsilon \to 0\) we get $$\mathcal{F}^*(x) – \mathcal{F}_*(x) \leq \omega_x(\mathcal{F})(x)$$

The other direction of the inequality follows similarly (because in \(\mathbb{R}\), \(\mathrm{diam}(A) = \sup A – \inf A \).
QED

In particular $$r(\mathcal{F}, C(X)) \geq \sup\limits_x \omega_x(\mathcal{F})$$, with equality if \(X\) is normal.

This theorem suggests an obvious generalization: When is it the case that \(d(\mathcal{F}, C(X, V)) = \sup\limits_x \omega_x(\mathcal{F})\)?

It turns out the answer is “rarely”, but mostly because this is the wrong generalization. The fact that the oscillation so well defines the approximability for the real case is because we can find good centers – in \(\mathbb{R}\) we have \(\mathrm{diam}(A) = 2 r(A)\). In general this relation does not hold. This leads us to define the following modification:

Define the radial oscillation of a family \(\mathcal{F}\) at a point \(x\) as $$\rho_x(\mathcal{F}) = \inf\{ r(\mathcal{F}(U)) : U \textrm{ open }, U \ni x \} $$

Write \(\rho(\mathcal{F}) = \sup\limits_x \rho_x(\mathcal{F})\).

The following theorem is fairly immediate from the definition:

Theorem: $$\rho(\mathcal{F}) \leq r(\mathcal{F}, C(X, V))$$

Proof:

Let \(\mathcal{F} \subseteq B(g, R)\)

Let \(x \in X\). Pick \(U \ni x\) such that for \(u \in U\), \(||g(u) – g(x)|| < \epsilon\). Then for \(u \in U\) and \(f \in \mathcal{F}\) we have \(||f(u) - g(x)|| \leq R + \epsilon\). Therefore letting \(\epsilon \to 0\) we have \(\rho_x(\mathcal{F}) \leq R\). Taking the infimum over possible values of R we have the desired result. QED So the question is this: For what pairs X, V is this inequality an equality? We saw that if \(V = \mathbb{R}\) then X being normal was sufficient. Is normal generally sufficient? We will say that X is OCA(V) ("has optimal continuous approximations into V") if this is an equality for every bounded family \(\mathcal{F}\). Advance warning: Unfortunately we were never able to resolve the full answer to this. We have some quite nice sufficient conditions, and a bunch of partial necessary conditions, but we were never able to get a general characterization that satisfied us. Still, I think some of the partial results we got are quite interesting. That depressing note aside, here's one of our main sufficiency conditions: Theorem: Let \(\kappa\) be a cardinal (including a finite one), \(X\) a topological space with the property that every open cover of fewer than \(\kappa\) open sets has a subordinate partition of unity and V a normed space such that every bounded subset of V has an \(\epsilon\)-net of size \(< \kappa\) for any \(\epsilon > 0\). Then X is OCA(V).

This has a number of immediate corollaries.

Corollary: If \(X\) is paracompact then \(X\) is OCA(V) for any V.
Corollary: If \(X\) is countably paracompact then \(X\) is OCA(V) for any separable V.
Corollary: If \(X\) is normal then \(X\) is OCA(V) for any finite dimensional V.

In order to prove the result we will need the following lemma:

Lemma: Let \(U_\alpha\) be an open cover of X such that \(\mathcal{F}(U_\alpha) \subseteq \overline{B}(v_\alpha, R)\). Let \(p_\alpha\) be a partition of unity subordinate to \(U_\alpha\). Let $$g(x) = \sum\limits_\alpha p_\alpha(x) v_\alpha$$

Then \(\mathcal{F} \subseteq \overline{B}(g, R)\).

Proof:

Let \(x \in X\). Then \(g(x)\) is a convex combination of \(v_\alpha\) such that \(f(x) \in \overline{B}(v_\alpha, R)\). Thus \(v_\alpha \in \overline{B}(x, R)\). Balls are convex, hence \(g(x) \in \overline{B}(x, R))\). QED

We will now prove the theorem:

Let \(\epsilon > 0\) and let \(\{u_\alpha : \alpha \in A \}\) be an \(\epsilon\)-net with \(|A| < \kappa\). By the definition of \(rho\) for every \(x \in X\) we may find an open set \(V_x\) such that \(\mathcal{F}(V_x) \subseteq \overline{B}(c, \rho(\mathcal{F}) + \epsilon)\). But then we have \(||c - v_\alpha|| \leq \epsilon\) for some \(v_\alpha\), so in fact \(\mathcal{F}(V_x) \subseteq \overline{B}(v_\alpha, \rho(\mathcal{F}) + 2\epsilon)\) Now let $$U_\alpha = \bigcup \{ V_x : \mathcal{F}(V_x) \subseteq \overline{B}(v_\alpha, \rho(\mathcal{F}) + 2\epsilon) \}$$ We have \(\mathcal{F}(U_\alpha) \subseteq \overline{B}(v_\alpha, \rho(\mathcal{F}) + 2\epsilon)\). Further because \(|A| < \kappa\), by hypothesis we can find a subordinate partition of unity \(p_\alpha\). We now apply our lemma and get a continuous function g such that \(\mathcal{F} \subseteq B(g, \rho(\mathcal{F}) + 2 \epsilon)\). Hence \(r(\mathcal{F}, C(X, V)) \leq \rho(\mathcal{F}) + 2\epsilon\). Letting \(\epsilon \to 0\) the result is proved. QED

Notes

• While the exact form of this theorem is probably novel, none of the details are. Unfortunately I can no longer find my notes on the relevant literature, as we arrived at this ourselves and then found the existing work later.
• The original form of this particular proof used a continuous selection theorem from A unified theorem on continuous selections by Michael and Pixley. We ended up unpicking the proof and proving it directly in order to get the tight cardinal bounds.
• This is in some sense our “most general” result about OCA. There will be a few other theorems which give some examples not covered by this, but this covers far more cases than others
• One natural question that emerges is to what extent is the converse result true. In particular if X is OCA(V) for every V, is X paracompact? We have some partial results of this form (which I’ll cover in a later post), but we never managed to get a full converse. As far as I know it’s an open problem

Here’s a result I’m going to need later for the random collection of theorems I’m going to be presenting from my old paper. It’s extremely standard, this is just a presentation of a proof.

When I’ve previously proved it I’ve used Fodor’s lemma, but I’ve just realised that there’s actually a simpler and slightly nicer proof by directly using some of the machinery you’d normally use to prove Fodor’s lemma. It’s not really substantially different, I just think it’s a little tidier.

The key result of this post is that every continuous function from an uncountable cardinal to a metric space is eventually constant. We’ll need to start with some set theoretic machinery about cardinals which we’ll use to prove this and then we’ll move on to some more purely topological arguments.

Let \(\kappa\) be an uncountable cardinal. The cofinality of \(\kappa\) is the smallest cardinality of an unbounded subset of \(\kappa\).

A club set in \(\kappa\) is an unbounded subset of \(\kappa\) that is closed under taking suprema of subsets.

Theorem: Let \(\kappa\) be an uncountable cardinal with \(cf(\kappa) > \aleph_0\). Let \(A, B \subseteq \kappa\) be club. Then \(A \cap B\) is club.

Proof:

It’s clear that the intersection is closed under suprema, so it remains only to show that that it is unbounded.

Let \(x \in \kappa\). Pick \(a_0 \in A\) with \(a \geq x\). Now recursively pick \(b_n \geq a_n\) and \(a_{n+1} \geq b_n\) with \(b_n \in B\) and \(a_n \in A\). We can do this because both sets are unbounded.

Because \(cf(\kappa) > \aleph_0\) any sequence must be bounded above (by definition: Otherwise it would be a countable unbounded set). Now let \(y = \sup a_n = \sup b_n\) (by construction the two sequences must have the same supremum). Then because \(A\) and \(B\) are closed under suprema we must have \(y \in A \cap B\). Therefore for any \(x \in \kappa\) there exists \(y \in A \cap B\) with \(y \geq x\). Therefore \(A \cap B\) is unbounded. QED

We can now extend this straightforwardly:

Theorem: Let \(\kappa\) be an uncountable cardinal with \(cf(\kappa) > \aleph_0\). Let \(A_n \subseteq \kappa\) be club. Then \(\bigcap A_n\) is club.

Proof:

By induction on the previous theorem finite intersections of club sets are club. Therefore \(A_1 \cap \ldots \cap A_n\) is club. Therefore for \(x \in \kappa\) we may pick \(a_n \in A_1 \cap \ldots \cap A_n\) with \(a_0 \geq x \) and \(a_{n+1} \geq a_n\).

Then \(a = \sup a_n \in \bigcap A_n \) and \(a \geq x\). Therefore \(\bigcap A_n\) is club. QED

This will give us a surprising (at least the first time you see it, but it’s fairly widely known) result about continuous functions.

Theorem: Let \(\kappa\) be a cardinal with uncountable cofinality. Let \(X\) be a metric space and let \(f : \kappa \to X\) be continuous when \(\kappa\) is given the order topology. Then there exists \(\alpha \in \kappa\) such that \(f|_{[\alpha, \kappa)}\) is constant.

Proof:

First note that any closed subset of \(\kappa\) is closed under suprema, so any unbounded subset closed under the order topology is club.

Secondly note that \(\kappa\) is sequentially compact (as any countable subset is contained in a compact subset). Therefore its image under \(f\) is sequentially compact and, since \(X\) is a metric space, thus compact. In particular it is separable. We may assume \(f\) is surjective and hence that \(X\) is separable.

Let \(x_1, \ldots, x_n, \ldots\) be a dense subset of \(x\). For \(\epsilon > 0\) the sets \(A(n, \epsilon) = \{ \alpha : f(\alpha) \in \overline{B}(x_n, \epsilon) \}\) form a countable cover of \(\kappa\) so by the pigeonhole principle at least one of them must be uncountable. Let \(C_\epsilon\) pick some \(A(n, \epsilon)\) such that it is uncountable. Then \(C_\epsilon\) is a club-set with \(\mathrm{diam}(f(C_\epsilon)) \leq 2 \epsilon\).

Let $$C = \bigcap C_{\frac{1}{n}}$$

By the preceding theorem, \(C\) is a club set. Necessarily \(\mathrm{diam}(f(C)) = 0\), so \(f(C) = \{x\}\) for some \(x\).

Now, consider the set \(M_\epsilon = \{ \alpha : d(f(\alpha), x) \geq \epsilon \}\). This set must necessarily be bounded, as otherwise it would intersect \(C\), which would contradict the definition. But \( \{ \alpha : f(\alpha) \neq x \} = \bigcup M_{\frac{1}{n}}\), which is a countable union of countable sets and thus bounded. Hence the set of points where f is not equal to x is bounded, and thus f is eventually constant as desired.

QED

Let \(V\) be a metric space and \(A, W \subseteq V\). Define

\[ r(A, W) = \inf \{ R : \exists w \in W, A \subseteq \overline{B}(w, R) \} \]

Where \(\overline{B}(w, R)\) is the closed ball of radius \(R\) centered on \(w\).

i.e. we’re looking at the radius of \(A\) when the center is restricted to lie in \(W\).

When \(W = V\) we will simply write \(r(A)\)

A \(W\)-center for \(A\) is a point \(w \in W\) such that \(A \subseteq \overline{B}(w, r(A, W))\). Note that it is not guaranteed that such a point exists (example: Let \(V = W = [-1, 1] \setminus \{0\}\). \(r(V) = 1\) but V has no center).

The case we’re interested in for this post is where we have some topological space \(X\) and set \(V = l^\infty(X)\), the space of bounded functions \(X \to \mathbb{R}\), and \(W = C(X)\), the subspace consisting of all continuous functions.

Why are we interested in this? Good question. I have absolutely no practical reason for studying this question and am not aware of any applications. There were three reasons I found it interesting:

1. It can be regarded as the question “Given a bunch of possibly discontinuous functions, how well can you simultaneously approximate them by a continuous function” (this is the problem I started with which lead to the whole paper, although with only one discontinuous function)
2. Studying radius is a fairly natural geometric problem and \(C(X)\) is a natural example of a Banach space
3. There’s a lot of surprisingly neat theory involved in the question

In this post we’ll establish three results:

1. A very nice characterization of \(r(\mathcal{F}, C(X))\)
2. If \(X\) is normal then every bounded \(\mathcal{F} \subseteq l^\infty(X)\) has a \(C(X)\)-center
3. A theorem about how the C(X) radius of \(\mathcal{F} \subseteq l^\infty(X)\) can be determined from “sufficiently large” subsets (see later)

A characterization of \(r(\mathcal{F}, C(X))\) and the existence of centers

Let \(\mathcal{F} \subseteq l^\infty(X)\) be bounded. We define the following functions:
$$\mathcal{F}^*(x) = \inf\limits_{U \ni x} \sup \bigcup\limits_{f \in \mathcal{F}} f(U) $$ $$\mathcal{F}_*(x) = \sup\limits_{U \ni x} \inf \bigcup\limits_{f \in \mathcal{F}} f(U) $$

where U is restricted to ranging over open sets.

Clearly for \(f \in \mathcal{F}\) we have \(\mathcal{F}_*(x) \leq f(x) \leq \mathcal{F}^*(x) \). The idea is that these functions provide better behaved tight bounds on \(\mathcal{F}\). Specifically:

Theorem: \(\mathcal{F}^*(x)\) is upper semicontinous. \(\mathcal{F}_*(x)\) is lower semicontinuous.
Proof:

We’ll just show that \(\mathcal{F}^*(x)\) is upper semicontinous. The other case will follow similarly.

So we must show that the set \(A = \{ x : \mathcal{F}^*(x) < t \}\) is open. Let \(x \in A\). Then there exists \(U \ni x\) such that \(\forall f \in F, y \in U, f(y) < t\) (by definition of \(\mathcal{F}^*\)). But then for any \(y \in U\) we must have \(\mathcal{F}^*(y) < t\). Therefore \(A\) contains an open neighbourhood of x. x was arbitrary, hence A must be open. QED Believe it or not, this is almost all we need to construct centers: Theorem: \(\frac{1}{2} \sup\limits_x (\mathcal{F}^*(x) – \mathcal{F}_*(x)) \leq r(\mathcal{F}, C(X))\)

If \(X\) is normal then this inequality is an equality and \(\mathcal{F}\) has a C(X) center.

Proof:

It may seem that this theorem is two unconnected results, but in fact our proof of the latter will simply fall out of the proof of the former.

Let \(R = r(\mathcal{F}, C(X))\) and \(S = \frac{1}{2} (\sup\limits_x \mathcal{F}^*(x) – \mathcal{F}_*(x))\).

We will first show \(S \leq R\).

Let \( \epsilon > 0 \). Let \(g \in C(X)\) be such that \(\mathcal{F} \subseteq \overline{B}(g, R + \epsilon)\).

Fix \(x \in X\) and find \(U \ni x\) open such that for \(y \in U, |f(x) – f(y)| < \epsilon\). Then for \(y \in U, f \in \mathcal{F}\) we have \(|f(y) - g(x)| \leq |f(y) - g(y)| + |g(y) - g(x)| \leq R + 2\epsilon\). This we must have \(\mathcal{F}^*(x) \leq g(x) + R + 2 \epsilon\) and \(\mathcal{F}_*(x) \geq g(x) - R - 2 \epsilon\). Hence we have \(\mathcal{F}^*(x) - \mathcal{F}_*(x) \leq 2R + 4\epsilon\). But \(\epsilon\) was arbitrary, so we must have $$\mathcal{F}^*(x) - \mathcal{F}_*(x) \leq 2R$$ and hence \(S \leq R\). Now assume \(X\) is normal. From the definition, we have $$\mathcal{F}^*(x) - S \leq \mathcal{F}_*(x) + S$$ The left hand side is upper semicontinuous, the right hand side is lower semi continuous, therefore by the Katětov–Tong insertion theorem there exists a continuous function \(g\) with $$\mathcal{F}^*(x) – S \leq g \leq \mathcal{F}_*(x) + S$$

Now let \(f \in \mathcal{F}\)

We have $$f(x) – S \leq \mathcal{F}^*(x) – S \leq g \leq \mathcal{F}_*(x) + S \leq f(x) + R$$

Or, rearranging, $$g(x) – S \leq f(x) \leq g(x) + S$$

Hence \(||f – g|| \leq R\), and \(\mathcal{F} \subseteq \overline{B}(g, S)\)

Thus \(R \leq S\) and hence \(R = S\)

But additionally we have shown that actually \(\mathcal{F} \subseteq \overline{B}(g, R)\), and thus \(g\) is a C(X)-center for \(\mathcal{F}\).

QED

Finding radii from radii of subsets

The diameter of a set is very nicely behaved in terms of the diameter of its subsets: It is the supremum of the diameter of all two-element subsets simply by matter of its basic definition. The radius is not so well behaved.

Consider the Banach space \(c_0\), the space of all sequences converging to 0. Consider the set \(E = \{ e_n \}\) where \((e_n)_k\) is 1 if \(k \leq n\) and 0 elsewhere. Every finite subset of this has radius half (it is contained in the ball of radius half around \(\frac{1}{2} e_N\) for some sufficiently large N), but the overall set has radius \(1\) because any center for it with smaller radius than that would not converge to 0.

It’s fairly easy to turn this into an example of a \(C(X)\) with similar behaviour.

Let \(X = \{ \pm \frac{1}{n} : n \in \mathbb{N} \} \cup \{ 0 \} \}\) and let \(\mathcal{F} = \{f_n\}\) where \(f_n(x) = \mathrm{sign}(x)\) if $|x| \geq \frac{1}{n}\), else \(f_n(x) = 0\).

Then once again any finite subset is contained within \(\overline{B}(\frac{1}{2} f_N, \frac{1}{2})\) for some sufficiently large N, but \(\mathcal{F}^*(0) = 1\) and \(\mathcal{F}_*(0) = -1\) so \(r(\mathcal{F}) = 1\).

(The idea of this example is that this space looks like two copies of the space of convergent sequences glued together by requiring that they converge to the same thing. We then force that value to be 0 by considering sequences where the desired center converges to opposite things).

So radius with centers in \(C(X)\) isn’t determined by finite subsets in general (incidentally, it is for centers in \(l^\infty(X)\)). How large do the subsets need to be in order to determine it?

There turns out to be a nice theorem in this regard.

First we’ll need a definition:

Let \(X\) be a topological space. The character of a point \(x \in X\) written \(\chi(x)\) is the smallest cardinality of a neighbourhood base of \(x\). The character of the space X, written \(\chi(X)\), is \( \sup\limits_{x \in X} \chi(x)\).

Lemma: Let \(x\) be a point with \(\chi(x) \leq \kappa\) and let \(\mathcal{F} \subseteq l^\infty(X)\). There exists \(\mathcal{G} \subseteq \mathcal{F}\) with \(|\mathcal{G}| \leq \kappa\), \(\mathcal{G}^*(x) = \mathcal{F}^*(x)\) and \(\mathcal{G}_*(x) = \mathcal{F}_*(x)\).

Proof:

We’ll construct \(\mathcal{G}_n\) of size \(\leq \kappa\) with \(\mathcal{G}^*(x) \geq \mathcal{F}^*(x) – \frac{1}{n}\) and \(\mathcal{G}_*(x) \leq \mathcal{F}_*(x) + \frac{1}{n}\). Then \(\mathcal{G} = \bigcup\limits_n \mathcal{G}_n\) will produce the desired set.

In fact it will suffice to construct a set \(\mathcal{H}\) with \(\mathcal{H}^*(x) \geq \mathcal{F}^*(x) – \frac{1}{n}\). We can then repeat an identical construction to get the other condition and take the union of the two sets.

So, let \(\{U_\alpha : \alpha < \kappa \}\) be a neighbourhood base for \(x\). By definition of \(\mathcal{F}^*(x)\) for each \(\alpha\) we can find \(f_\alpha \in \mathcal{F}\) with \(\sup f_\alpha(U_\alpha) \geq \mathcal{F}^*(x) - \frac{1}{n}\). Let \(\mathcal{H} = \{ f_\alpha \}\). Then if \(U \ni x\) is open it contains some \(U_\alpha\) and thus \(\sup f_\alpha(U) \geq \mathcal{F}^*(x) - \frac{1}{n}\). Hence \(\mathcal{H}^*(x) \geq \mathcal{F}^*(x) - \frac{1}{n}\) as desired. QED This lemma contains the bulk of the work we'll need to prove the following theorem: Theorem: Let \(X\) be a topological space with \(\chi(X) \leq \kappa\). Let \(\mathcal{F} \subseteq l^\infty(X)\). There exists \(\mathcal{G} \subseteq \mathcal{F}\) with \(|\mathcal{G}| \leq \kappa\) and \(r(\mathcal{G}, C(X)) = r(\mathcal{F}, C(X))\).

Proof:
Let \(R = r(\mathcal{G}, C(X))\).

We’ll use our previous characterization of the radius. So we can find \(x_n\) with \(\mathcal{F}^*(x_n) – \mathcal{F}_*(x_n) \geq R – \frac{1}{n}\). As per the lemma we can find a subset \(\mathcal{G}_n\) with \(|\mathcal{G}_n| \leq \kappa\), \(\mathcal{G}_n^*(x_n) = \mathcal{F}^*(x_n)\) and \(\mathcal{G}_{n*}(x_n) = \mathcal{F}_*(x_n)\). Then \(r(\mathcal{G}_n) \geq R – \frac{1}{n}\) hence if \(\mathcal{G} = \bigcup \mathcal{G}_n\) then \(r(\mathcal{G}) = r(\mathcal{F})\) as desired.
QED.

In particular if \(X\) is first countable then the radius is determined by countable subsets.

It’s worth noting that this is not an equivalence. Easy counter examples happen because there are first countable spaces with non first countable compactifications.

It is however tight in that for a regular uncountable cardinal \(\kappa\) if we take the order topology on two copies \(\{ a_\alpha : \alpha \leq \kappa \}\) and \(\{ b_\alpha : \alpha \leq \kappa \}\) with the quotient that \(a_\kappa = b_\kappa\). We can then construct an example exactly like our modification of the \(c_0\) example (this relies on the fact that every function in \(C(\kappa)\) is eventually constant).

References

• The basic result about centers existing isn’t at all new, but I can’t track down the exact reference right now (various forms of it were proved at various times)
• The inspiration for the theorem about characters comes from a mathoverflow answer by Sergei Ivanov. I have no idea if it’s new or not. I expect not though. It’d be surprising if no one else had proven it or something like it before.