Metric space of continuous & bounded functions is complete?

In summary: 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...figure out.
  • #1
Terrell
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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##.

Homework Equations


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.
 
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  • #2
I don't follow your reasoning.

You have to start with a Cauchy sequence in ##C(X,Y)##.

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).
 
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  • #3
Math_QED said:
(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.
 
  • #4
Terrell said:
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).
 
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  • #5
Math_QED said:
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?
 
  • #6
Terrell said:
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.
 
  • #7
Math_QED said:
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!
 
  • #8
Terrell said:
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.
 
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  • #9
Math_QED said:
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?
 
  • #10
Terrell said:
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.
 
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1. What is a metric space?

A metric space is a mathematical concept that describes a set of points and a distance function between those points. The distance function, also known as a metric, measures the distance between any two points in the set.

2. What is a continuous function?

A continuous function is a function that does not have any "jumps" or "breaks" in its graph. In other words, the graph of a continuous function can be drawn without lifting the pen or pencil from the paper.

3. What does it mean for a function to be bounded?

A function is bounded if its values do not exceed a certain limit. In other words, there is a finite number that serves as an upper and lower bound for the values of the function.

4. How is completeness defined in a metric space?

A metric space is complete if every Cauchy sequence in that space converges to a point within that space. In other words, the space contains all of its limit points.

5. How does the completeness of the metric space of continuous and bounded functions impact mathematical analysis?

The completeness of this metric space is crucial in mathematical analysis, as it ensures the existence of solutions to certain problems. Additionally, it allows for the use of the Cauchy convergence criterion, which simplifies the process of proving convergence of sequences and series in this space.

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