- #1
jostpuur
- 2,112
- 19
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 >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)
<br /> |f(x) - f(y)| \leq |f(x) - f_N(x)| \;+\; |f_N(x) - f_N(y)| \;+\; |f_N(y) - f(y)| < \epsilon<br />
But how do you the same without the metric, with the Hausdorff property only?
<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 >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)
<br /> |f(x) - f(y)| \leq |f(x) - f_N(x)| \;+\; |f_N(x) - f_N(y)| \;+\; |f_N(y) - f(y)| < \epsilon<br />
But how do you the same without the metric, with the Hausdorff property only?
