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Banach space C(X)

  1. Sep 26, 2008 #1
    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?
     
  2. jcsd
  3. Sep 26, 2008 #2
    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 :smile:
     
    Last edited: Sep 26, 2008
  4. Sep 26, 2008 #3

    morphism

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    Just use open sets instead of balls; it all works out the same.
     
  5. Sep 26, 2008 #4
    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 [tex]\mathbb{C}[/tex]!
     
  6. Sep 26, 2008 #5

    morphism

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    ding ding ding :wink:

    Oh, as an aside, do you know how the Hausdorff property is being used?
     
  7. Sep 26, 2008 #6
    No. Is it used at all?
     
  8. Sep 26, 2008 #7

    morphism

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