Banach space C(X)

1. Sep 26, 2008

jostpuur

I already know how to prove that if $$M$$ is a compact metric space, then

$$C(M) = \{f\in \mathbb{C}^M\;|\; f\;\textrm{continuous}\}$$

with the sup-norm, is a Banach space, but now I encountered a claim, that actually metric is not necessary, and $$C(X)$$ is Banach space also when $$X$$ is a compact Hausdorff space.

My proof concerning $$C(M)$$ follows the following steps:
(1) Show that $$C(M)$$ is a norm space.
Let $$f_1,f_2,f_3,\cdots\in C(M)$$ be some Cauchy-sequence.
(2) Show that a pointwise limit $$f=\lim_{n\to\infty}f_n$$ exists in $$\mathbb{C}^M$$.
(3) Show that $$f$$ is continuous.
(4) Show that $$\|f-f_n\|\to 0$$ when $$n\to\infty$$.

All other steps work without metric, except the third one. This is how I did it:

Let $$x\in M$$ and $$\epsilon >0$$ be arbitrary. One has to find $$\delta >0$$ so that $$f(B(x,\delta))\subset B(f(x),\epsilon)$$. First fix $$N\in\mathbb{N}$$ so that $$\|f_i-f_j\| < \epsilon / 3$$ for all $$i,j\geq N$$. Then fix $$\delta > 0$$ so that $$f_N(B(x,\delta))\subset B(f_N(x),\epsilon /3)$$. Then for all $$y\in B(x,\delta)$$

$$|f(x) - f(y)| \leq |f(x) - f_N(x)| \;+\; |f_N(x) - f_N(y)| \;+\; |f_N(y) - f(y)| < \epsilon$$

But how do you the same without the metric, with the Hausdorff property only?

2. Sep 26, 2008

jostpuur

Whoops, I just realized that I have not encountered the claim, that C(X) would be a Banach space. I made a mistake when reading one thing......

Well, I'll be listening if somebody has something to say about the issue anyway

Last edited: Sep 26, 2008
3. Sep 26, 2008

morphism

Just use open sets instead of balls; it all works out the same.

4. Sep 26, 2008

jostpuur

But because open sets don't have any natural radius, the radius cannot be divided by three either.

edit: Oooh... but the balls are in $$\mathbb{C}$$!

5. Sep 26, 2008

morphism

ding ding ding

Oh, as an aside, do you know how the Hausdorff property is being used?

6. Sep 26, 2008

jostpuur

No. Is it used at all?

7. Sep 26, 2008

Nope!