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Math Help - compactness

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    compactness

    The Hilbert cube Q is the subset of l given by {a : NR││a(i)│2-i}.
    Show that Q is a compact subspace of l∞ (with the norm a=supnN│a(n)│).


    can anyone please help me to do this question ?? ,,
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  2. #2
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    Quote Originally Posted by jin_nzzang View Post
    The Hilbert cube Q is the subset of l given by {a : NR││a(i)│2-i}.
    Show that Q is a compact subspace of l∞ (with the norm a=sup_{nN}│a(n)│).
    Let \mathcal B denote the unit ball of l^\infty. Then \mathcal B consists of all sequences (x_n) of real numbers such that x_n\in[-1,1] for all n. Give \mathcal B the topology that it has as the product of an infinite number of copies of [1,1]. In this topology, a sequence ((x^{(k)})) converges to ((x^{(0)})) if it converges coordinatewise. That is to say, if x^{(k)}_n\to x^{(0)}_n for each coordinate n.

    The map x_n)\mapsto(2^{-n}x_n)" alt="fx_n)\mapsto(2^{-n}x_n)" /> takes \mathcal B onto Q. The idea of the proof is to show that f is continuous from \mathcal B (with the above product topology) to Q (with the topology from the l^\infty norm). By Tychonoff's theorem, any product of compact spaces is compact. So \mathcal B is compact, and hence Q, as a continuous image of a compact space, is also compact.

    To see that f is continuous, suppose that x^{(k)}\to x^{(0)} in \mathcal B, and let \varepsilon>0. Choose N so that 2^{-N}<\varepsilon/2. Then |(f(x^{(k)}) - f(x^{(0)}))_n| = 2^{-n}|x^{(k)}_n - x^{(0)}_n| < \varepsilon for all n>N. But (by the definition of the product topology in \mathcal B) there exists K such that |x^{(k)}_n - x^{(0)}_n| < \varepsilon for all k>K and all n with 1≤n≤N. Therefore \|f(x^{(k)}) - f(x^{(0)})\|_\infty \leqslant \varepsilon, so that f(x^{(k)})\to f(x^{(0)}) in Q.

    [That is essentially the argument given here to show that the Hilbert cube is compact for the topology that comes from the l^2-norm, except that the Wikipedia page completely ducks the central part of the proof, which is to show that the map f is continuous.]
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