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Complete metric space proof
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[QUOTE="Incand, post: 5652704, member: 525632"] [h2]Homework Statement [/h2] Let ##E## be a metric subspace to ##M##. Show that ##E## is closed in ##M## if ##E## is complete. Show the converse if ##M## is complete. [h2]Homework Equations[/h2] A set ##E## is closed if every limit point is part of ##E##. We denote the set of all limit points ##E'##. A point in ##p\in M## is a limit point to ##E\subseteq M## if ##\forall \epsilon > 0## ##\exists q \in E \cap B(p,\epsilon)## [h2]The Attempt at a Solution[/h2] We want to show that ##E' \subseteq E##. Take ##p \in E'## then clearly we can choose ##p_n \in B(p,1/n)## so that ##p_n \in E##. But then for all ##\epsilon > 0## ##d(p_n,p_m)\le d(p_n,p)+ d(p_m,p)<\epsilon## for ##n,m \ge N_\epsilon = 2/\epsilon## i.e. ##(p_n)## is a cauchy sequence. But then it must converge to some ##\lambda \in E##. However ##\lambda = p## since ##d(p,\lambda) = d(p_n,p)+d(p_n,\lambda)< \epsilon## Is this a correct proof? Should I use a similar approach to the second part? [/QUOTE]
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Complete metric space proof
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