# Evaluation map problem

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## 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?

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...

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

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

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

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 $$e(V\times W)\subseteq U$$...

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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)!

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 $$d^\prime (g(z),f(x))<\epsilon$$?

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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)

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

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

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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. 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 $$d(f(x),f(x^\prime))<\epsilon$$... You'll need to adjust your U for that...

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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 $$d(f(x),f(x^\prime))<\epsilon$$... 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?

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...

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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.

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...

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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.

Yes, that is correct! Nice!

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Yes, that is correct! Nice!

OK, finally!

Thanks for your patience! 