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.

- X is normal
- X is \(OCA(V)\) for any finite dimensional V
- 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.

- X is countably paracompact
- X is \(OCA(V)\) for any separable V
- 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.

rogerI am glad there are people in the world such as yourself that excel at this kind of thing. Autarch of Algorithms indeed.

R