Let $X$ be a countably infinite compact subset of real line . Let $C(X)$ be the space of all real valued continuous functions on $X$ equipped with the sup norm . Let $V$ be the subspace of $l^\infty$ defined by $V:=\{(x_n)\in l^\infty |(x_n)$ is convergent $\}$ . Then is it true that $C(X)$ is isometrically isomorphic with $V$ ?
The space of real valued continuous functions on a countably infinite compact set of real line , isomorphic with the space of convergent sequences?
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functional-analysis
metric-spaces
normed-spaces
isometry
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0Is it possible to find a bijective map between them? – 2017-01-30
1 Answers
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It is trivial to see that $V$ is isometrically isomorphic to $C(X_0)$, where $$X_0=\left\{\frac1n:\ n\in\mathbb N\right\}\cup\{0\}.$$ Now, by the Banach-Stone Theorem, $C(X)$ is isometrically isomorphic to $C(Y)$ if and only if $X$ and $Y$ are homeomorphic.
The answer to your question is no, then, as one can easily construct countable compact subsets of $\mathbb R$ that are not homeomorphic to $X_0$. For instance, $X_0$ has a single accumulation point, so any $X_1$ with countably many accumulation points will do.
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0how is $V$ is isometrically isomorphic with $C(X_0)$ ? – 2017-01-31
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0$\{x_n\}\longmapsto \left(\frac1n\mapsto x_n\right)$. – 2017-01-31