# How to show that linear span in $C[0,1]$ need not be closed [duplicate]

Possible Duplicate:
Non-closed subspace of a Banach space

Let $X$ be an infinite dimensional normed space over $\mathbb{R}$. I want to find a set of vectors $(x_k)$ such that the linear span of $(x_k)$ of vectors is not closed.

I feel like the set $P$, which conists of polynomials in $X=C[0,1]$ (with the sup-norm) would be a good candidate since the Weierstrass approximation theorem yields that the span of $P$ is dense in $X$. How can I show that this span is not closed?

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## marked as duplicate by Norbert, Davide Giraudo, Jason DeVito, tomasz, Ｊ. Ｍ.Oct 12 '12 at 12:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Do you know what "closed" and "dense" mean? –  Chris Eagle Oct 10 '12 at 12:23

## 1 Answer

You can take your favourite convergent sequence of polynomials (e.g. partial sums of $\exp x = \displaystyle \sum_{n = 0}^\infty \frac{x^n}{n!}$) and then prove that the limit is not in the span.

This proves the span is not sequentially closed. Since $X$ is normed, it follows that the span isn't closed.

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