A Mathematician’s Scratchpad

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 . 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 be a topological space and a function, not neccesarily continuous. For we define the oscillation of at to be

where ranges over open sets.

We define the total oscillation of to be

We now prove some preliminary results about

Proposition 2:

Let be an arbitrary topological space and consider the Banach space of bounded functions on and the closed subspace consisting of the continuous bounded functions. Consider

1. iff is continuous.
2. 
3. 
4. 
5. 

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 , 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:

for every iff 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 be a normal topological space and let . Then there is a (not usually unique) continuous function with .

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 be a normal topological space. Let be upper semicontinuous and lower semicontinuous respectively with . There is a continuous function with .

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 . Define

Proposition 7:

These satisfying the following properties:

1. 
2. is lower semicontinuous. is upper semicontinuous.
3. If is upper semicontinuous then . Similary if is lower semicontinuous then .
4. The maps and are monotone with respect to the pointwise ordering.
5. If are lower, upper semicontinuous respectively and then
6. 

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 . Consequently we have that

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

So

But we have that and .

So

i.e.
But we know that by our very first proposition. Hence we have equality.

QED

Lemma 8:

If for every we have then is normal.

Proof:

This proof is entirely John’s.

Let be disjoint closed sets. Define by , and otherwise.

By considering appropriate neighbourhoods we may see that for every . i.e. . Consequently we have and may find a continuous function with .

Thus we have that and . Composing with some appropriate function from we get a continuous function which separates the two closed sets. Then and are appropriate disjoint neighbourhoods of and .

QED

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