David R. MacIver

I’ve imported my old maths blog, A Mathematician’s Scratchpad. It’s available under the category mathematics. Unfortunately I can’t get the LaTeX to compile. The old LaTeX-render plugin I used no longer works with the latest version of wordpress, and I’m experiencing a host of problems with the new one (not least among them “It makes my site shit-slow”), so I’ve disabled it. Hopefully it should be reasonably readable anyway.

Most likely none of this stuff is particularly interesting to anyone who reads this blog currently, but I’m enjoying the nostalgia trip. :-)

Woo hoo. Blog is back. :-)

Here’s a new silly proof I spotted recently.

Theorem: Let [tex]f : \omega_1 \to \mathbb{R}[/tex] be continuous. Then [tex]f[/tex] is eventually constant.

Proof:

This proof assumes the that [tex]2^{\aleph_0} > \aleph_1[/tex]. The result doesn’t actually need this though, which is one of the main reasons this proof is silly.

So, [tex]f[/tex] is continuous. [tex]\omega_1[/tex] is countable compact, thus so is [tex]f(\omega_1)[/tex]. But [tex]\mathbb{R}[/tex] is a metric space, so countably compact subsets are compact. But every compact subspace of [tex]\mathbb{R}[/tex] has cardinality [tex]\aleph_0[/tex] or [tex]2^{\aleph_0}[/tex]. We know that [tex]2^{\aleph_0} > \aleph_1[/tex], so it can’t be [tex]2^{\aleph_0}[/tex]. Hence it has cardinality [tex]\aleph_0[/tex]. By the pigeon hole principle we must have [tex]f[/tex] being constant on some uncountable set. But for any [tex]t[/tex] the set [tex]\{ x : f(x) = t \}[/tex] is closed, with these being disjoint for distinct values of [tex]t[/tex]. You can’t have disjoint uncountable closed sets in [tex]\omega_1[/tex], so all but one of these sets must be countable. Thus take an upper bound for all the countable subsets, say [tex]y[/tex]. [tex]f[/tex] is constant on [tex] \[y, \omega_1) [/tex].

I can’t actually take much credit for this. I made the initial conjecture that sparked this theorem, but my friend John Bytheway was the first person to codify the actual theorem and prove it. My proof of it is pretty independent of his, but I probably wouldn’t have come up with it unless I already knew the theorem was true. (The neccesary part of the theorem is stolen from him, but the sufficiency part is entirely mine).

Note: In this post I take the slightly unconventional approach that a Normal space needn’t be [tex]T_1[/tex]. i.e. a Normal topological space is one in which any two disjoint closed sets have disjoint neighbourhoods, but points needn’t be closed. I could invent a new term for this, like quasinormal, but I really can’t be bothered. Feel free to replace ‘normal’ with ‘quasinormal’ wherever you see it if this makes you more comfortable. It may be readily verified that the usual proof of Urysohn’s lemma in no way depends on the closedness of points, so Urysohn’s lemma holds for this new definition of normal.

First we need some preliminary definitions.

Definition 1:

Let [tex]X[/tex] be a topological space and [tex]f : X to mathbb{R}[/tex] a function, not neccesarily continuous. For [tex]x in X[/tex] we define the oscillation of [tex]f[/tex] at [tex]x[/tex] to be

[tex]omega_f(x) = inflimits_{U ni x} diam f(U)[/tex]

where [tex]U[/tex] ranges over open sets.

We define the total oscillation of [tex]f[/tex] to be

[tex]delta(f) = suplimits_{x in X} omega_f (x) [/tex]

We now prove some preliminary results about [tex]delta[/tex]

Proposition 2:

Let [tex]X[/tex] be an arbitrary topological space and consider [tex]B(X)[/tex] the Banach space of bounded functions on [tex]X[/tex] and [tex]C(X)[/tex] the closed subspace consisting of the continuous bounded functions. Consider [tex]delta : B(X) to [0, infty)[/tex]

1. [tex]delta(f) = 0[/tex] iff [tex]f[/tex] is continuous.
2. [tex]delta(f + g) leq delta(f) + delta(g)[/tex]
3. [tex]delta(tf) = |t| delta(f)[/tex]
4. [tex]delta(f) leq 2 ||f||[/tex]
5. [tex]delta(f) leq 2 d(f, C(X))[/tex]

These are all perfectly trivial to prove, so I’m not going to bother.

What I noticed is that in almost every case I considered the final inequality was in fact an equality. I could prove a weaker result for Compact hausdorff spaces (I showed that [tex]||f|| leq delta(f)[/tex], but I couldn’t do any better, so I passed this over to John to see what he could come up with. He came up with the following theorem, which is the main theorem of this post:

[tex]delta(f) = 2 d(f, C(X))[/tex] for every [tex]f in B(X)[/tex] iff [tex]X[/tex] is normal.

I’ll prove this in two parts. For the right to left implication I’ll in fact prove something stronger:

Theorem 4:

Let [tex]X[/tex] be a normal topological space and let [tex]f in B(X)[/tex]. Then there is a (not usually unique) continuous function [tex]h[/tex] with [tex]||f – h|| = 2 delta(f)[/tex].

As is my wont, the proof of this will precede by a couple clever definitions and then drop out as practically a corollary of a Big Theorem.

The big theorem in question is the following, which is due to Tong. I conjectured it, tinkered around with proving it for a bit, went to look up something about semicontinuous functions in Engelking’s general topology book and saw the theorem staring out at me from one of the exercises.

Theorem 5:

Let [tex]X[/tex] be a normal topological space. Let [tex]f, g : X to mathbb{R}[/tex] be upper semicontinuous and lower semicontinuous respectively with [tex]f leq g[/tex]. There is a continuous function [tex]h[/tex] with [tex]f leq h leq g[/tex].

Note the direction of the inequality and which are upper and lower semicontinuous respectively. If you reverse this then the theorem becomes false.

I’m not actually going to prove this here – I’ve not yet totally sorted out the details of the proof in my mind. It’s basically a modified version of the standard proof of Urysohn’s lemma, but with additional constraints on the sets constructed. (Update: See here for a proof.)

Anyway, we now make some more definitions:

Definition 6:

Let [tex]f in B(X)[/tex]. Define

[tex]f^*(x) = inflimits_{U ni x} sup f(U)[/tex]

[tex]f_*(x) = suplimits_{U ni x} inf f(U)[/tex]

Proposition 7:

These satisfying the following properties:

1. [tex]f_* leq f leq f^*[/tex]
2. [tex]f_*[/tex] is lower semicontinuous. [tex]f^*[/tex] is upper semicontinuous.
3. If [tex]f[/tex] is upper semicontinuous then [tex]f^* = f[/tex]. Similary if [tex]f[/tex] is lower semicontinuous then [tex]f_* = f[/tex].
4. The maps [tex]f to f^*[/tex] and [tex]f to f_*[/tex] are monotone with respect to the pointwise ordering.
5. If [tex]g, h[/tex] are lower, upper semicontinuous respectively and [tex]g leq f leq h[/tex] then [tex]g leq f_* leq f leq f^* leq h[/tex]
6. [tex]omega_f(x) = f^*(x) – f_*(x)[/tex]

Again, these are all really very easy to prove (assuming you prove them in order), so I’m not going to do it. I’ll actually not use most of these, but those that I don’t use are of independent interest. i.e. they’re cool. :-)
Now, we have:

Proof of theorem 4:

Note that [tex]delta(f) = sup (f^*(x) – f_*(x)[/tex]. Consequently we have that

[tex] f^* – frac{1}{2}delta(f) leq f_* + frac{1}{2} delta(f) [/tex]

Now, the left hand side is upper semicontinuous and the right hand side is lower semicontinuous. Thus we have an interpolating continuous function [tex]h[/tex].

So

[tex] f^* – frac{1}{2}delta(f) leq h leq f_* + frac{1}{2} delta(f) [/tex]

But we have that [tex]f leq f^*[/tex] and [tex]f_* leq f[/tex].

So

[tex] f – frac{1}{2}delta(f) leq h leq f + frac{1}{2} delta(f) [/tex]

i.e. [tex]||f – h|| leq frac{1}{2} delta[/tex]
But we know that [tex]||f – h|| geq frac{1}{2} delta[/tex] by our very first proposition. Hence we have equality.

QED

Lemma 8:

If for every [tex]f in B(X)[/tex] we have [tex]d(f, C(X)) = frac{1}{2} delta(f)[/tex] then [tex]X[/tex] is normal.

Proof:

This proof is entirely John’s.

Let [tex]A, B[/tex] be disjoint closed sets. Define [tex]f : X to mathbb{R}[/tex] by [tex]f|_A = -1[/tex], [tex]f|_B = 1[/tex] and [tex]f(x) = 0[/tex] otherwise.

By considering appropriate neighbourhoods we may see that [tex]omega_f(x) leq 1[/tex] for every [tex]x[/tex]. i.e. [tex]delta(f) leq 1[/tex]. Consequently we have [tex]d(f, C(X)) leq frac{1}{2}[/tex] and may find a continuous function [tex]h[/tex] with [tex]||f – h|| leq frac{3}{4}[/tex].

Thus we have that [tex]h|_A leq – frac{1}{4}[/tex] and [tex]h|_B geq frac{1}{4}[/tex]. Composing with some appropriate function from [tex][-1, 1] to [0, 1][/tex] we get a continuous function [tex]g[/tex] which separates the two closed sets. Then [tex]g^{-1}([0, frac{1}{2}))[/tex] and [tex]g^{-1}((frac{1}{2}, 1])[/tex] are appropriate disjoint neighbourhoods of [tex]A[/tex] and [tex]B[/tex].

QED

Here begins a new series, akin to the “Silly Proofs” series. Obscure results which are cool, but which you probably haven’t heard of.

Suppose we’ve got a pair of measurable spaces (sets with a sigma algebra on them) [tex]X, Y[/tex]. We make [tex]X times Y[/tex] by taking the sigma algebra generated by sets of the form [tex]A times B[/tex]. In the case where we have topologies on [tex]X[/tex] and [tex]Y[/tex] and are giving them their Borel algebras, we might suppose this agrees with the Borel algebra of the product. Alas, ’tis not so! It does in the second countable case, but in general not:

Theorem (Nedoma’s Pathology): Let [tex]X[/tex] be a measurable space with [tex]|X| > 2^{aleph_0}[/tex]. Then the product algebra on [tex]X^2[/tex] does not contain the diagonal.

In particular, if [tex]X[/tex] is Hausdorff then the diagonal is a closed set in the product topology which is not contained in the product algebra.

The proof proceeds by way of two lemmas:

Lemma: Let [tex]X[/tex] be a set, [tex]mathcal{A} subseteq P(X)[/tex] and [tex]U in sigma(mathcal{A})[/tex]. There exist [tex]A_1, ldots, A_n, ldots[/tex] with [tex]U in sigma{ A_1, ldots, A_n, ldots }[/tex]

Proof: The set of [tex]U[/tex] satisfying the conclusion of the theorem is a [tex]sigma[/tex] algebra containing [tex]mathcal{A}[/tex].

Lemma: Let [tex]U subseteq X^2[/tex] be measurable. Then [tex]U[/tex] is the union of at most [tex]2^{aleph_0}[/tex] sets of the form [tex]A times B[/tex].

Proof:

By the preceding lemma we can find [tex]A_n[/tex] with [tex]U in sigma { A_m times A_n }[/tex]

For [tex]x in {0, 1}^{mathbb{N}}[/tex], define [tex]B_x = bigcap C_n[/tex] where [tex]C_n = A_n[/tex] if [tex]x_n = 1[/tex] and [tex]A_n^c[/tex] otherwise.

Sets of the form [tex]B_x times B_y[/tex] then form a partition of of [tex]X^2[/tex]. Thus the sets which can be written as a union of sets of the form [tex]B_x times B_y[/tex] form a [tex]sigma[/tex] algebra. This contains each of the [tex]A_m times A_n[/tex], and so contains [tex]U[/tex]. Thus [tex]U = bigcup { B_x times B_y : B_x times B_y subseteq U }[/tex]. There are at most [tex]2^{aleph_0}[/tex] sets in this union. Hence the desired result.

Finally we have the proof of the theorem:

Let [tex]D[/tex] be the diagonal. Suppose [tex]D[/tex] is measurable. Then [tex]D[/tex] is the union of at most [tex]2^{aleph_0}[/tex] sets of the form [tex]A times B[/tex]. Because [tex]|D| > 2^{aleph_0}[/tex] at least one of these sets must contain two points. Say [tex](u, u)[/tex] and [tex](v, v)[/tex]. But then it also contains [tex](u, v) notin D[/tex]. This is a contradiction.

QED

To be honest, this theorem doesn’t seem that useful to me. But knowing about it lets you avoid a potential pitfall – when you’re dealing with measures on large spaces (e.g. on [tex]{0, 1}^{kappa}[/tex], which it’s really important to be able to do when you’re playing with certain forcing constructions) things are significantly less well behaved than you might hope.

I went to an interesting maths conference this week (Set theory and its neighbours meets the Cameleon), and I’m probably going to write a report on it at some point. This isn’t that report.

This is just a quick note to say that the stuff about Boolean algebras and operator algebras is at least known, if not well known, and has been for a good few decades. Oh well. The noncommutative stuff probably hasn’t – I’m going to email the guy whose talk was on a related subject and ask him about it.

David