Proving C(X) is a Banach Space for Compact Hausdorff X

[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?

 

[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

 

[morphism]

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

 

[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 $$\mathbb{C}$$!

 

[morphism]

ding ding ding

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

 

[jostpuur]

ding ding ding

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

No. Is it used at all?

 

[morphism]

Nope!