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This question already has an answer here:

In a theorem I am reading about closed subspace the author states that an infinite dimensional subspace need not be closed.

What is an example of infinite dimensional subspace that is not closed?

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marked as duplicate by Carsten S, E. Joseph, Dominik, Alex M., user228113 Dec 23 '16 at 11:11

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  • $\begingroup$ Perhaps better stated as "In a a normed linear space $X$ an infinite dimensional subspace need not be closed in $X.$ $\endgroup$ – zhw. Dec 23 '16 at 0:34
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Take $C([0,1],\|\|_{\infty})$ and the subset of polynomials. Every continuous function is a limit of polynomials by Stone Weirstrass. Thus the subset of polynomial functions of $C([0,1])$ is dense, thus it is not closed.

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    $\begingroup$ nit picky: "dense and a proper subset, thus not closed". The whole space is closed and dense $\endgroup$ – user2520938 Dec 23 '16 at 9:42
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Let $\ell^2$ be the space of all square-summable real (or complex) sequences $x = (x_1,x_2, \ldots)$ with norm $\|x\| = \displaystyle ( \sum |x_i|^2)^{1/2}$. Let $V \subset \ell ^2$ be the subspace of all sequences with all but finitely many entries equal to zero. Then $V$ is infinite-dimensional but not closed. It is not closed because its closure contains the limit point $(1,1/2, 1/3, \ldots)$

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