- #1
jostpuur
- 2,116
- 19
I already know how to prove that if [tex]M[/tex] is a compact metric space, then
[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?
[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?