# The space of continuous functions.

• MathematicalPhysicist
In summary: Actually, I think that's the best way to do it.In summary, we are given a compact space X, a compact metric space (Y,p), and a closed subset F of C(X,Y) (the continuous functions space). We must prove that F is compact. To do this, we can use Tychonoff's theorem, which states that the space of all functions X->Y is compact under the product topology. Since F is a closed subset of this space, it must also be compact.
MathematicalPhysicist
Gold Member
Let X be a compact space, (Y,p) a compact metric space, let F be a closed subset of C(X,Y) (the continuous functions space) (i guess it obviously means in the open-compact topology, although it's not mentioned there) which satisifes:
for every e>0 and every x in X there exists a neighbourhood U of x s.t for every f in F:
f(U) is a subset of B_e(f(x)) which is an open ball with radius e.
prove that F is compact.

well i want to prove that C(X,Y) is compact and then it will follow that F is compact.

now let U(W(C,U))=C(X,Y) which is an arbitrary covering of C(X,Y), i.e for every C compact in X and every U open in Y s.t f(C) is contained in U.
now i need to find a finite covering, now because Y is compact and metric then it is covered by a finite union of open balls, because f in F is continuous then f^-1(B_e(f(x))
is open in X, so finite union of them cover the entire set X, so because every compact set in X is actually covered by a finite union of f^-1(B_e(f(x))) and also every every open set in Y is covered by a finite union of open balls, then C(X,Y) is covered by a finite union of W(C_i,B_e_j(f(x))) s.t C_i is compact in some f^-1(B_e_j(f(x)) i.e f(C_i) is contianed in B_e_j(f(x))), so C(X,Y) is compact and then also F is compact as beiing a closed subset of C(X,Y).

am i right or way off?

loop quantum gravity said:
well i want to prove that C(X,Y) is compact and then it will follow that F is compact.

How does that follow?

well there's a theorem that if X is a compact topological space and A a closed subset of it then A is compact in X.

What is the continuous functions space? Which one? Define an open set or subbasis element of it?

I've heard of compact-open topology but not open-compact topology. Maybe it's the reverse, but we would have to prove it's a topology first, which I don't think it is.

If X is a topological (or metric) space and M is a metric space, then the set of all bounded continuous functions from X to M forms a metric space if we define the metric as above: d(f, g) = supx in X d(f(x), g(x)) for any bounded continuous functions f and g. If M is complete, then this space is complete as well.

http://en.wikipedia.org/wiki/Metric_space

loop quantum gravity said:
well i want to prove that C(X,Y) is compact and then it will follow that F is compact.

C(X,Y) isn't compact.

instead of telling me I'm wrong, you could have also pointed me to a possible solution, could you not?

anyway, jason, open-compact is the same as compact-open, i just changed the order.

okay, i see it now, thanks jason.

wait a minute gel, if C(X,Y) is metric, and because Y is complete it's also complete, then by the condition given on F, C(X,Y) is absolutely bounded, then by a thoeorem we learned in class, C(X,Y) must be compact, cause a metric space is compact iff it's complete and absolutely bounded (absolutely bounded, or totally, is that there's the metric space has a finite e>0 net.

So what i said is wrong?
just got the wrong terminology on absolutely bounded (should be totally bounded), it's a problem that you learn the terms in hebrew and after you look at the definitions in munkres with the theorems. (the lecturer is awfull, an entire lecture today was only with figures, with no formalism, and a lot of handwaving of the worst kind, we learned on manifolds, a topic which he didn't prove anything in it rigorously).

loop quantum gravity said:
So what i said is wrong?
It is wrong. You can't conclude that C(X,Y) is totally bounded based on what you're given. What you could try to do is prove that F is totally bounded, although I don't think you will find this very fruitful. Instead, maybe you could try the Arzela-Ascoli theorem.

loop quantum gravity said:
instead of telling me I'm wrong, you could have also pointed me to a possible solution, could you not?
Yes, but I was in a rush, and I think it's useful to know when you're going about something the wrong way.

loop quantum gravity said:
wait a minute gel, if C(X,Y) is metric, and because Y is complete it's also complete, then by the condition given on F, C(X,Y) is absolutely bounded, then by a thoeorem we learned in class, C(X,Y) must be compact, cause a metric space is compact iff it's complete and absolutely bounded (absolutely bounded, or totally, is that there's the metric space has a finite e>0 net.

yes C(X,Y) is complete, so F is too. C(X,Y) isn't absolutely bounded though. Maybe you meant to say that F is absolutely bounded, which I think it is (if I understand your definition correctly - what is an e>0 net? is it just a cover by open balls of radius e?) Then compactness of F follows.

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Maybe I'm missing something, but the condition on F doesn't really look like it implies total boundedness easily. It's saying that F is equicontinuous, hence my inclination to try Arzela-Ascoli.

morphism said:
Maybe I'm missing something, but the condition on F doesn't really look like it implies total boundedness easily. It's saying that F is equicontinuous, hence my inclination to try Arzela-Ascoli.

Well looking at wikipedia statement of http://en.wikipedia.org/wiki/Arzelà-Ascoli_theorem" , it does imply that F is compact

Let X be a compact metric space, Y a metric space. Then a subset F of C(X,Y) is compact if and only if it is equicontinuous, pointwise relatively compact and closed.

In the original question, it said that F is closed and equicontinuous. Also, Y is compact => F is pointwise compact. So the original question was almost a restatement of the Arzela-Ascoli theorem anyway.

I can also see that F will be compact from Tychonoff's theorem. The space of all functions X->Y (not just continuous ones) is compact under the product topology, and F is closed with this topology.

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gel said:
I can also see that F will be compact from Tychonoff's theorem. The space of all functions X->Y (not just continuous ones) is compact under the product topology, and F is closed with this topology.
That's an interesting observation, but is it enough here? The topology C(X,Y) gets from the compact-open subbase here is the topology of uniform convergence, i.e. the topology generated by the uniform metric, whereas the topology it gets as a subspace of the product space Y^X is the topology of pointwise convergence, which in general is strictly weaker than the former. So certainly F is going to be closed in the product topology if it's to be closed in the compact-open topology, but for the converse we need something stronger. Or did you have something else in mind?

Edit:
Evidently the equicontinuity of F can be our "something stronger", because it reduces the compact-open topology on F to the product topology. <Proof removed.> So I guess it is enough to prove that F is closed in the product topology.

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eventually i realized before handing in the assignment that I should show that F is complete and totally bounded, i hope my proof is rigorous enough, well the fact that F is complete follows from the fact that Y is complete then C(X,Y) is complete then F is complete, and the condition that F satisfies assures us it's totally bounded and it's metric then it's compact.

## 1. What is the space of continuous functions?

The space of continuous functions refers to a set of all functions that are continuous on a given interval or domain. These functions have no breaks, jumps, or sudden changes in their values. They are smooth and can be represented by a continuous line on a graph.

## 2. What is the importance of the space of continuous functions?

The space of continuous functions is essential in many areas of mathematics and science, including calculus, analysis, and physics. It allows us to model and study real-world phenomena and make accurate predictions. It is also a fundamental concept in functional analysis, which is the study of spaces of functions.

## 3. How is the space of continuous functions different from the space of differentiable functions?

The main difference between the space of continuous functions and the space of differentiable functions is that the former contains all continuous functions, while the latter only includes functions that have a well-defined derivative at every point. This means that all differentiable functions are continuous, but not all continuous functions are differentiable.

## 4. Can all functions be represented in the space of continuous functions?

No, not all functions can be represented in the space of continuous functions. Functions that have discontinuities, such as step functions or functions with vertical asymptotes, cannot be represented in this space. However, they can be approximated by a sequence of continuous functions, making the space of continuous functions a dense subset of the set of all functions.

## 5. How is the space of continuous functions related to the concept of compactness?

The space of continuous functions is a compact space, meaning that it is closed and bounded. This property is important because it guarantees that certain properties, such as continuity and uniform convergence, are preserved when working with functions in this space. It also allows us to use powerful tools, such as the Arzelà–Ascoli theorem, to study the behavior of functions in this space.

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