# Thread: Proof that a function space is complete

1. ## Proof that a function space is complete

Another one that I'm kind of uncomfortable with my solution:

B(X) vector space of all bounded functions in X. Show that it's going to be defined under the sup norm:

d(f,g) = sup (mod(f(x)-g(x))

What I did was to show that every Cauchy sequence in B(X) converges to an f, (so mod(f-fn) gets arbitrarily small with n --> infinity);

and the terms fn and fm get arbitrarily close to each other with a high n. So for a high enough n, we have

mod(f) <= mod(fn) + mod(f - fn) < E, for an arbitrary E, so f , the limit of the cauchy sequence, is a bounded function, belonging to B(X). Hence, the space is complete.

I feel that I'm missing something here, since i didn't use the sup norm directly.

2. What are the domain and codomain of the functions in your space?

3. Domain is X, codomain is not defined in the question.

Domain is X, codomain is not defined in the question.
It has to be any metric space $\displaystyle X$ and the codomain needs to be $\displaystyle \mathbb{R}$.

Now, you've taken analysis before, right?

5. Originally Posted by Drexel28
It has to be any metric space $\displaystyle X$ and the codomain needs to be $\displaystyle \mathbb{R}$.

Now, you've taken analysis before, right?

I'm taking my first analysis course right now. Sorry if my questions are too basic.

I'm taking my first analysis course right now. Sorry if my questions are too basic.
I guess that's my fault for not asking the question clearly. Your question is actually a rephrasing (with slight modification) of a theorem you should have covered when you did uniformly convergent sequences of functions.

7. The codomain could be any Banach space couldn't it?

edit: does X have to be a metric space?

The codomain could be any Banach space couldn't it?
Yes.

edit: does X have to be a metric space?
It could be any topological space.

My reasoning for my response was the OPs previous posts. But, noticing that he said "mod" maybe not a Banach space and $\displaystyle \mathbb{C}$?

9. Does it have to be a topological space?

It's possible I'm being old-fashioned, but I think it's in the spirit of the question. Usually one talks at most generally about $\displaystyle \mathcal{C}\left[X,\mathcal{B}\right]$ with $\displaystyle X\in\text{TOP}$ (:P) since all of the fun things you can do with it are in that context. If you want to try to do it without, go ahead.