I already know how to prove that if [tex]M[/tex] is a compact metric space, then(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

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

[/tex]

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

My proof concerning [tex]C(M)[/tex] follows the following steps:

(1) Show that [tex]C(M)[/tex] is a norm space.

Let [tex]f_1,f_2,f_3,\cdots\in C(M)[/tex] be some Cauchy-sequence.

(2) Show that a pointwise limit [tex]f=\lim_{n\to\infty}f_n[/tex] exists in [tex]\mathbb{C}^M[/tex].

(3) Show that [tex]f[/tex] is continuous.

(4) Show that [tex]\|f-f_n\|\to 0[/tex] when [tex]n\to\infty[/tex].

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

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

[tex]

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

[/tex]

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

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# Banach space C(X)

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