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Math Help - Compactness in a metric space

  1. #1
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    Compactness in a metric space

    Problem:

    Let X = the set of real number sequence  \{ X_{n} \} ^{ \infty }_{n=1} such that  \sum ^{ \infty } _{n=1} |X_{n}| < \infty

    Define  d( \{ X_{n} \} , \{ Y_{n} \} ) = \sum ^{ \infty } _{n=1} |X_{n}-Y_{n}|.

    Let A= \{ \{ X_{n} \} \in X : \sum ^{ \infty } _{n=1} |X_{n}| \leq 1 \}

    Let N \in N, let T= \{ (x^{k} \in X : k \geq N \Rightarrow x^{k} = 0 \}

    Define \hat {T} = T \cap A , is  \hat {T} compact?

    My proof:

    First, we claim that there exist a subsequence of every sequence in  \hat {T} that converges in X.

    Suppose that  \{ X_{n} \} is a sequence in  \hat {T} , note that  \sum ^{N} _{k=1} |X_{n}^{k}| \leq 1

    Now,  \{ X_{n} \} = \{ X_{n} ^{1}, X_{n}^{2} , X_{n}^{2} , X_{n}^{3} , ... , X_{n}^{k-1}, 0 , 0 , 0 , . . . , 0 \}

    Note that  \sum ^{N} _{n=1} |X_{n}^{1}| \leq 1 , so  \{ X_{n}^{1} \} is bounded. By a theorem, we know that there exist a subsequence  \{ X_{n_{i}}^{1} \} of  \{ X_{n}^{1} \} such that  X_{n_{i}}^{1} converges to a point in R, denote by  X_{0}^{1} .

    Now, consider the component  \{ X_{n_{i}}^{2} \} in the subsequence  \{ X_{n_{i}} \} , which is also bounded. Thus there exist a subsequence  \{ X_{n_{i_{p}}}^{2} \} of  \{ X_{n_{i}}^{2} \} such that  X_{n_{i_{p}}}^{2}  converges to a point in R, this time denote by  X_{0}^{2} .

    Note that the component  \{ X^{1}_{n_{i_{p}}} \} is a subsequence of  \{ X^{1}_{n_{p}} \} , thus it also converges to the point  \{ X_{0}^{1} \} .

    Continue in the manner, we find a subsequence of  \{ X_{n} \} , denote by  \{ X_{n_{q}} \} for simplification purpose, such that  X^{k}_{n_{q}} \longrightarrow X^{k}_{0} \\\\\ \forall k \in N.

    Now, consider  \lim _{q \rightarrow \infty } d( X_{n_{q}}, X_{0} ) = \lim _{q \rightarrow \infty } \sum ^{N} _{k=1} |X_{n_{q}}^{k} - X_{0}^{k} | = 0 . Thus proves that  \{ X_{n_{q}} \} \longrightarrow \{ X_{0} \} \in X .

    Now, we claim that  \{ X_{0} \} is a point in  \hat {T} .

    Assume to the contrary that  \{ X_{0} \} is not a point in  \hat {T} . We have  \sum ^{N}_{k=1} |X_{0}^{k}| > 1

    Note that  \sum ^{N}_{k=1} |X_{n_{i}}^{k}| \leq 1 since it is a sequence in  \hat {T} .

    But,  \sum ^{N}_{k=1} |X^{k}_{n_{i}} - X_{0}^{k} | > 0 , in which contradicts our assumption. Therefore we show that  \{ X_{0} \} \in \hat {T} , thus proves  \hat {T} is compact.

    Q.E.D.

    Please check my proof, thank you!
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  2. #2
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    Yes, that argument looks correct.

    Another approach would be to look at the map f: T\to \mathbb{R}^N defined by f(x) = (x_1,x_2,\ldots,x_N), where x = (x_1,x_2,\ldots,x_N,0,0,\ldots). This map is a homeomorphism from T (with the metric d) to \mathbb{R}^N with the euclidean metric. The set f(T∩A) is a closed, bounded subset of \mathbb{R}^N, hence compact, from which it follows that T∩A is also compact.
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