# Metric space of continuous & bounded functions is complete?

## Homework Statement

The book I'm using provided a proof, however I'd like to try my hand on it and I came up with a different argument. I feel that something might be wrong.

Proposition: Let ##<X,d>## be a metric space, ##<Y,D>## a complete metric space. Then ##<C(X,Y), \sup D>## is a complete metric space. For emphasis, the metric of ##C(X,Y)## is ##\sup D[f(x),g(x)]## such that ##x \in X##. Also, ##C(X,Y)## is the space of bounded and continuous functions from ##X## to ##Y##.

N\A

## The Attempt at a Solution

Let ##(f_k(x))## be a Cauchy sequence in the complete metric space ##Y##. Hence, ##(f_k(x))\rightarrow g(x)## for some ##g(x) \in Y##. Keep in mind that this means, given ##\epsilon \gt 0##, for ##M\in\Bbb{N}##, ##\forall k\geq M##, we have ##D[g(x),f_k(x)] \lt \epsilon##, ##\forall x\in X##. Since ##f,g\in C(X,Y)## is bounded, then ##D[g(z),f_k(z)]=\sup\{D[g(x), f_k(x)] \vert x \in X\}## for some ##z \in X## and ##\forall k \in \Bbb{N}##. But note that ##\forall k\geq M##, ##\sup\{D[g(x), f_k(x)] \vert x \in X\} \lt \epsilon##. So ##C(X,Y)## must be complete.

Last edited:

member 587159

That is, let ##(f_k)_k## be a Cauchy sequence in ##C(X,Y)##.

You then have to show that this sequence of functions converges in ##C(X,Y)##. So, you have to check a couple of things:

(1) It must have a limit.
(2) The limit must be bounded
(3) The limit must be a continuous function.

I think you have showed (1) and (2) (although I'm not too sure), but you certainly didn't show (3).

Terrell
(1) It must have a limit.
(2) The limit must be bounded
(3) The limit must be a continuous function.
Now this I don't understand. I thought we only need to show (1)? Where did this criterion come from? The book did use the outline you provided, but I don't get why.

member 587159
Now this I don't understand. I thought we only need to use the definition of convergence to show that ##(f_k)_k## converges in ##C(X,Y)##? Where did this criterion come from?

Simply the definition of convergence of a sequence applied on this case:

Let ##(Z,d)## be a metric space. Let ##(z_n)## be a sequence in ##Z##. Let ##z \in Z##. Then ##z_n \to z## iff $$\forall \epsilon > 0: \exists N: \forall n \geq N: d(z_n,z) < \epsilon$$

The ##z \in Z## part is crucial here.

As an example, the sequence ##(1/n)## does not converge in ##\mathbb{R} - \{0\}##, but it does converge (to 0) in ##\mathbb{R}## (all sets here equipped with the usual metric).

Terrell
Simply the definition of convergence of a sequence applied on this case:
As an example, the sequence ##(1/n)## does not converge in ##\mathbb{R} - \{0\}##, but it does converge (to 0) in ##\mathbb{R}## (all sets here equipped with the usual metric).
So convergence due to boundedness and completeness due to continuity?

member 587159
So boundedness due to convergence and continuity due to completeness?

Not quite. You have to show that the space of continuous and bounded functions is complete.

Thus, any Cauchy sequence must have a limit in this space. That the limit must be an element of the space means that the limit must be continuous and bounded.

Thus, any Cauchy sequence must have a limit in this space. That the limit must be an element of the space means that the limit must be continuous and bounded.
Got it! Thanks! Somehow I keep thinking that if the limit exists, then it's continuous when I know that it's not!

member 587159
Got it! Thanks! Somehow I keep thinking that if the limit exists, then it's continuous when I know that it's not!

No problem! Here's a starter for a correct attempt:

Let ##(f_n)_n## be a Cauchy sequence in ##C(X,Y)##. Define

##f: X \to Y: x \mapsto (\lim_{n \to \infty} f_n(x))##.

Show that:

(1) ##f## is well-defined.
(2) ##f## is continuous
(3) ##f## is bounded
(4) ##f_n \to f## in sup-norm.

___________________________

As an interesting side remark: if ##X## is compact, the boundedness assumption can be left out the problem statement.

Terrell
Show that:

(1) ##f## is well-defined.
(2) ##f## is continuous
(3) ##f## is bounded
(4) ##f_n## \to ##f## in sup-norm.
Are you referring to ##f## as the limit of ##(f_n)_n##? Isn't every function in ##C(X,Y)## already continuous and bounded?

member 587159
Are you referring to ##f## as the limit of ##(f_n)_n##? Isn't every function in ##C(X,Y)## already continuous and bounded?

I have defined ##f## in my post. It's the function such that ##f(x) = \lim_n f_n(x)##. I claim that this is the limit of the given Cauchy sequence (so yes, it will turn out to be the limit you are looking for), but that's for you to verify.

Terrell