Solving the Evaluation Map Problem for Metric Spaces X and Y

[radou]

Homework Statement

This problem is a bit of a digression (at least it seems so) from the problems about imbeddings I'm dealing with currently (and I yet have a few more to complete).

Let X and Y be spaces. Define e : X x C(X, Y) --> Y with e(x, f) = f(x). e is called the evaluation map, and C(X, Y) is the set of all continuous functions from X to Y. If d is a metric for Y and if C(X, Y) has the corresponding uniform topology, then e is continuous.

The Attempt at a Solution

Now, if I defined a metric on X x C(X, Y), and if I considered the topology induced by this metric, the solution is quite easy, at least it seems so. But the problem is, I can't do that, right? And even if I'd do so (if the problem would be formulated that way), I don't see where the uniform topology on C(X, Y) jumps in.

So, any suggestions on how to attack this?

 

[micromass]

No that's right, you can't just put a metric on X\times C(X,Y). X carries a natural topology, C(X,Y) carries the uniform topology. so X\times C(X,Y) carries the product topology. You don't even know if this space is metrizable!

I guess there is no other option than to apply the definiiton of continuity: take a open neighbourhood U of f(x), and show that there exists open neighbourhood V of x and W of f such that e(V\times W)\subseteq U. Of course, since C(X,Y) is metrizable, we can always choose W a ball around f with a certain radius...

 

[micromass]

And Y is also metrizable, so U can also be taken as a ball...

 

[radou]

Btw, what does "natural topology" mean in this context? Is it like "some" topoogy, but we don't need/can't specify it?

 

[micromass]

Yes, X just carries a topology. But we don't specify which one...

 

[micromass]

Let me give an informal proof for this. I'll let you fill in the details:

Take (x,f) in X\times C(X,Y) and take U an open neighbourhood of f(x). By continuity of f, there exists an open neighbourhood V of X such that f(V)\subseteq U. If, a function g lies "close enough" to the function f, then g(V)\subseteq U. These functions which lie "close enough" give you an open set W of C(X,Y). This gives us e(V\times W)\subseteq U...

 

[radou]

OK, this may be a bit fast, but it seems to me that for an ε-ball in Y, be can take an ε-ball in C(X, Y), and U can be any open set in X?

Since if we take B'(y, ε), and B(f, ε) = {g in C(X, Y) : sup{d(f(x), g(x)) < ε, x is in X, d is the bounded metric}}. B is a collection of functions such that for any x in X, the upper inequality holds. Hence e(U x B) is contained in B'(y, ε)!

 

[micromass]

So I take (z,g) in your U\times B(x,\epsilon). You claim that e(z,g)\in B^\prime(f(x),\epsilon). But why is d^\prime (g(z),f(x))&lt;\epsilon?

 

[radou]

Well, e(z, g) is in B'(f(x), ε) since for all x in X, sup {d(f(x), g(x))} < ε. Ahh, I just realized that it won't really work, since d is the bounded metric! (I should have called it d1 or something)

 

[micromass]

Yes, you have got d(f(x),g(x))&lt;\epsilon and d(f(z),g(z))&lt;\epsilon. But you want to show that d(f(x),g(z))&lt;\epsilon. Here you got two variables!

Of course, you can apply the triangle inequality to obtain d(f(x),g(z))&lt;d(f(x),f(z))+d(f(z),g(z)). The second term can be bounded by \epsilon. But you'll need to bound the first term to!

 

[radou]

OK, I'll start over, I got a bit lost in the notation, which wasn't consistent from the beginning.

Let z = (x, f) be a point in X x C(X, Y) and e(z) = f(x) its image in Y. Let B'(f(x), ε) be given.

First, consider the ε-ball in C(X, Y) around f, B(f, ε) = {g in C(X, Y) : sup {d1(g(x), f(x)) : x in X} < ε}.

Now, let U be an open neighborhood of x in X. Let z' = (x', g) be a point of U x B. Then e(z') = g(x'). Now, g is a function such that sup {d1(g(x), f(x)) : x in X} < ε. Here's a logical argument I'm not sure about: We took any ε > 0. If g is a function such that, for some x in X, d(f(x), g(x)) > 1, then sup {d1(g(x), f(x)) : x in X} = 1. Hence, for any ε > 0 we have 1 < ε, which is a contradiction! Hence, d(f(x), g(x)) < 1 for all x in X, if it is in B(f, ε). So, the metric d1 in our case coincides with the metric d, and hence d(f(x), g(x')) < ε.

Now, this may be terribly wrong, and it probably is. But I'm not ashamed to write it down and get my reasoning straightened.

 

[micromass]

Seems good, but I've got two problems.

Hence, for any ε > 0 we have 1 < ε, which is a contradiction!

Contradiction with what? You didn't assume \epsilon\leq 1 anywhere... However, since only the small \epsilon count, you can easily choose \epsilon\leq 1

hence d(f(x), g(x')) < ε.

Why is this? Even if you choose g=f, then you're not even sure that d(f(x),f(x^\prime))&lt;\epsilon... You'll need to adjust your U for that...

 

[radou]

Seems good, but I've got two problems.



Contradiction with what? You didn't assume \epsilon\leq 1 anywhere... However, since only the small \epsilon count, you can easily choose \epsilon\leq 1

Well, it seemed like a contradiction with an arbitrary choice of ε!

Why is this? Even if you choose g=f, then you're not even sure that d(f(x),f(x^\prime))&lt;\epsilon... You'll need to adjust your U for that...

Yes, but isn't g a function such that for all (by definition of the uniform metric) x in X, the upper inequality holds, so the choice of U doesn't matter?

 

[micromass]

Well, it seemed like a contradiction with an arbitrary choice of ε!

Yes, ε is arbitrary, but it is fixed to. Nothing prevents us from choosing ε=2 in the beginning. The point is that the proof must work for all possible ε, even ε=2...

Yes, but isn't g a function such that for all (by definition of the uniform metric) x in X, the upper inequality holds, so the choice of U doesn't matter?

No, g is a function such that d(g(x),f(x))<ε for all x. So you know that d(g(x),f(x))<ε and d(g(x'),f(x'))<ε. But you do NOT know that d(g(x'),f(x))<ε, since you are using two different variables here: x and x'...

An example, take g=f, this g certainly satisfies the condition d(f(x),g(x))<ε for all x. But it does not in general satisfy d(f(x),g(x'))<ε for all x and x'!. You'll need some assumption on x and x' for that...


Hope this was clear...

 

[radou]

Ah, of course! That ε-argument was really stupid, I see it now!

And I see that I was confusing variables again, too.

Basically, we have want to show that d(f(x), g(x')) < ε. Now, the triangle inequality can remedy this, as you mentioned in an earlier post. And we can remedy the term d(f(x), f(x')), too, since f is continuous. We can choose a neighborhood W of x which will guarantee this.

If all this is true, there remains only one thing to discuss.

We know that d1(f(x'), g(x')) < ε. But d1 is the bounded metric. Any I want to work out the triangle inequality in the metric d!

i.e. I need:

d(f(x), g(x')) <= d(f(x), f(x')) + d(f(x'), g(x') < ε, for all x in W. W can be adjusted however I please. But I need to relate d1 to d somehow.

 

[micromass]

Ah, yes. The way to fix that is a standard trick:

You have chosen ε arbitrary, and you need to find V and W such that e(V\times W)\subseteq B(f(x),\epsilon).

- If ε<=1, then there is no problem, since the metric d1 will coincide with the metric d.
- If ε>1, then you just search V and W such that e(V\times W)\subseteq B(f(x),1). Finding such V and W should be no problem, we did that in the previous point. If we found those V and W, then

e(V\times W)\subseteq B(f(x),1)\subseteq B(f(x),\epsilon)

So this V and W are good for our ε to, even if ε>1...

 

[radou]

Ah, OK!

So, I'll write it all down now, for convenience.

Let ε > 1 be given. First, choose B(f, 1/2) around f. Then, since f is continuous, for 1/2 choose a neighborhood W of x such that d(f(x), f(x')) < 1/2.

Now we have:

d(f(x), g(x')) < = d(f(x), f(x')) + d(f(x'), g(x')) < 1/2 + 1/2 < 1. Now clearly e(W x B) is contained in B'(f(x), ε).

If ε <= 1, we simply take B(f, ε/2) around f and W such that d(f(x), f(x')) < ε/2.

 

[micromass]

Yes, that is correct! Nice!

 

[radou]

Yes, that is correct! Nice!

OK, finally!

Thanks for your patience!