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Metric space of all bounded real functions is complete

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Let X be a non-empty set and let C be the set of all bounded real functions defined on X, with the metric induced by the supremum norm: d(f,g) = ||f - g|| = sup |f(x)-g(x)| , x in X.
    Show that the metric space (C,d) is complete.
    Hint: if [tex] \{f_{n}\} [/tex] is a cauchy sequence, then [tex] \{f_{n}(x)\} [/tex] is a cauchy sequence for all x in X.

    2. Relevant equations



    3. The attempt at a solution

    let [tex] \{f_{n}\} [/tex] be a cauchy sequence in (C,d). From the definition, we have for all x in X: [tex] |f_{n}(x) - f_{m}(x)| \leq ||f_{n} - f_{m}|| [/tex] so [tex] \{f_{n}(x)\} [/tex] is cauchy for all x in X, hence it converges (because range of f is complete). Let f denote the function whose value at x is the limit of [tex] f_{n}(x) [/tex] as n goes to infinity. Thus for every e > 0 and x in X, there exists N(x,e) > 0 such that [tex] |f_{n}(x) - f(x)| < e [/tex] whenever n > N(x,e).

    Now correct me if I'm wrong, but I cannot deduce from the last statement what the problem is asking for, unless f is uniformly convergent over X, that is, N(x,e) is independent of x. But that can't be true in general. Is there something I'm missing?

    This problem is taken from Simmons' Introduction to Topology and Modern Analysis. However in Rudin's Principles of Mathematical Analysis, a book that I went through a few months before, I read this:

    C(X) denotes the set of all complex-valued, continuous, bounded functions with domain X. [...]
    A sequence [tex] {f_{n}} [/tex] converges to f with respect to the metric of C(X) (induced by the supremum norm as above) if and only if fn -> f uniformly on X.

    Is there something I'm missing or is the result the original problem is asking for is wrong?
     
    Last edited: Dec 1, 2008
  2. jcsd
  3. Dec 1, 2008 #2

    Office_Shredder

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    Keep in mind also that {fn} Cauchy means that fn gets uniformly close to each other (which is where you can get the uniform convergence through clever use of the triangle inequality)
     
  4. Dec 1, 2008 #3
    ok, thank you. I figured it out :)
     
  5. Dec 3, 2008 #4
    So once you have that fn --> f uniformly on X, how do you get that the space is complete?
     
  6. Dec 3, 2008 #5

    Office_Shredder

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    symbol, that's it.... by definition it's complete if every Cauchy sequence converges.
     
  7. Dec 3, 2008 #6
    By definition, a metric space M is complete if every Cauchy sequence in M converges in M.
    So in this case we would have to show that every Cauchy sequence in (C,d) converges in (C,d).
    So if fn --> f under the sup norm, I guess we have to show that f is in (C,d).
     
  8. Dec 3, 2008 #7

    Office_Shredder

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    You have to show it's bounded, but that's pretty easy since each fn is bounded, and fn converges uniformly to f... so we know for some n, |fn(x)-f(x)|<1 for all x, and if |fn(x)|<M for some M for all x, |f(x)|<M+1 necessarily
     
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