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

<br /> C(M) = \{f\in \mathbb{C}^M\;|\; f\;\textrm{continuous}\}<br />

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

<br /> |f(x) - f(y)| \leq |f(x) - f_N(x)| \;+\; |f_N(x) - f_N(y)| \;+\; |f_N(y) - f(y)| &lt; \epsilon<br />

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!