Workin' for the Something

Hey, I started real work again, which is sweet, because as much as I like doing nothing all day, I'd rather do it at home where it is less obvious that I am doing nothing and I don't have to change pants or anything. But, whatever, work has started out well and that is good.


This is completely unrelated, but I thought it was sweet. That was my lunch a couple days back, which consisted of miso soup, a salad I made out of vegetables that I'm not sure the translation of, kimchi flavored cucumber pickles and melon creme soda, which I bought out of a vending machine. It was in a 1.5 liter bottle. I couldn't figure out a way to combine those last two sentences without making it sound like either I or the vending machine was inside a bottle.

I was messing with continuity and compactness and whatnot, and I came upon a little claim that said that any continuous function mapping from a compact metric space to any old metric space is necessarily a uniformly continuous function. This result is called Heine-Cantor or something, I can't remember. As always, definitions because I know lots of people are reading this and just don't want to go all the way to wikipedia.

Metric space:
A topological space in which the topology is defined (or induced, or whatever you prefer) by a metric. Brilliant, so what is a metric, and what is a topology? A topology is just the set of subsets on the space that we call open sets. Open sets (and thus, topologies) need to meet certain criteria, but that doesn't matter right here. Just accept that metrics do a good job of setting them up, but aren't the only way to do it.

A metric is really just a distance function. That is, it takes in two points in the space and returns a non-negative real number. It has to meet these conditions, which if you think about for a second are really just what we mean when we say distance:

1. d(x, y) is non-negative, and d(x, y) = 0 iff x = y; This just says that distance is positive, unless it is the distance between a point and itself, which should be zero.

2. d(x, y) = d(y, x); it is nice if the distance between two points is the same regardless of direction.

3. d(x, y) + d(y, z) > or = d(x, z); the shortest distance between two points is going from one to the other. This is called the triangle inequality.

To make it a bit more concrete, just think of a metric as being absolute value of the difference, or really just how you would intuitively find distance given a picture.

Continuous: Remember your calculus here; a function f is continuous at x if for any real number r > 0, there exists a real number s > 0 such that if d(x, y) < s, d(f(x), f(y)) < t. If f is continuous at all x, then f is said to be continuous.

That's one that gets calculus students all the time. Since I don't have Greek character keys, I had to substitute for epsilon and delta, but if you've ever taken one of these courses, you should have recognized it. What does it really mean? It means that we can get arbitrarily close to the function value at x [that is, f(x)] and still be able to find a neighborhood around x that maps entirely closer to f(x) than whatever value we chose.

I'm sort of glossing over the fact that you can have different metrics on the domain and range spaces, but I figured you can figure that out and I don't have subscript buttons. Furthermore, we don't even need a metric space to define continuity, but it will do for right now.

Uniform continuous: Also usually throws people at first because it looks like continuity. So what is the difference? Delta, or s, as above is independent of x. What does that mean? It means that if we choose some arbitrarily small epsilon, or r, then we can find s as before, but that this s (some s that we can find will have this property anyway) is sufficient for any x. So we can move our s-neighborhood to some other point, say y, and it will map to within the r-neighborhood of f(y). Clear as mud?

Compact: A metric space is compact if any open cover of the space has a finite subcover. This just means that if we choose some of our open sets (possibly an infinite number of them), and that by "unioning" them, we cover the metric space, we are guaranteed to be able to keep only some finite number of these open sets and still be covering the space. I know, it's a little out there.

So, on to the problem. I am thinking this way might work. First, let's choose our arbitrarily small r. Now, assuming f is continuous, that means that at each x, we can find our s value that meets that nice condition. Of course, we don't know that f is uniformly continuous (as this is what we'd like to prove!), so we must assume that s depends on x, so let's write s = s(x).

Here I've got a little issue that I haven't bothered to work out all the details of. Since s is a real number, and number less than s is going to work too, so how do we properly define this function? Experience tells me we generally have a largest s that works for a given x, which makes sense if you think about real numbers, so I suggest choosing that number. Of course, it is possible that any s may work, but I think then we can just choose whatever number we want and it won't really matter.

Anyway, now let's consider a set we just created implicitly. That is, let's look at each of the s-neighborhoods of x (x can take any value in our original domain space, which we might as well call X). It's fairly obvious that x is an element of it's own s-neighborhood, and it also turns out that an s-neighborhood is necessarily open [this results from how we define the topology via the metric], so the set of s-neighborhoods of x forms an open cover for X. If we assume X is compact, as we are doing, then we know we can keep only a finite number of these s-neighborhoods and still cover X. Since the set of s's that we are considering is now finite, we know there's a smallest member, which I'm tempted to call s'.

It seems now to me that this s' should be sufficient for any x in X, but I haven't worked out the details. Feel free to work it out for yourself if you want. And congrats if you made it through this post.