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Thread: 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 $\displaystyle \{ X_{n} \} ^{ \infty }_{n=1} $ such that $\displaystyle \sum ^{ \infty } _{n=1} |X_{n}| < \infty $

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

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

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

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

    My proof:

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

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

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

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

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

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

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

    Now, consider $\displaystyle \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 $\displaystyle \{ X_{n_{q}} \} \longrightarrow \{ X_{0} \} \in X $.

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

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

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

    But, $\displaystyle \sum ^{N}_{k=1} |X^{k}_{n_{i}} - X_{0}^{k} | > 0 $, in which contradicts our assumption. Therefore we show that $\displaystyle \{ X_{0} \} \in \hat {T} $, thus proves $\displaystyle \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 $\displaystyle f: T\to \mathbb{R}^N$ defined by $\displaystyle f(x) = (x_1,x_2,\ldots,x_N)$, where $\displaystyle x = (x_1,x_2,\ldots,x_N,0,0,\ldots)$. This map is a homeomorphism from T (with the metric d) to $\displaystyle \mathbb{R}^N$ with the euclidean metric. The set f(T∩A) is a closed, bounded subset of $\displaystyle \mathbb{R}^N$, hence compact, from which it follows that T∩A is also compact.
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