Closure of the set of all polynomial with variable $x\in [0,1]$

Let ${P}$ denote the set of all polynomial with variable $x\in [0,1]$, I need to know what is the closure of ${P}$ in $C[0,1]$?
Well, Stone-Weierstrass theorem says: If $f\in C[0,1]$ then there exists a sequence of polynomials $p_n(x)$ which converges uniformly to $f$. So can I say that $closure{P}=C[0,1]$?

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If you mean closure with respect to the supremum norm, that's exactly the statement of Stone-Weierstrass theorem. –  Davide Giraudo Jul 19 '12 at 13:21
yes with supnorm –  Une Femme Douce Jul 19 '12 at 13:22

Yes. The Stone-Weierstrass approximation theorem tells you that $P$ is dense in $C[0,1]$ with respect to $\|\cdot\|_\infty$ and a set $D$ is dense in a set $S$ if the closure of $D$ equals $S$. (by definition)

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Talking about the definition of being dense. Can't we say that $[0,1)$ is dense in $[0,1)$? –  Ilya Jul 19 '12 at 13:23
@Ilya Yes of course, the entire space is always dense in itself. –  Rudy the Reindeer Jul 19 '12 at 13:26
@HenningMakholm Well in the given case the whole space is complete with respect to $\|\cdot\|_\infty$. But the definitions I've seen of "dense set" don't put any requirement on the space, in particular, they don't require the space to be complete. –  Rudy the Reindeer Jul 19 '12 at 13:32
@MattN: Sorry -- had gotten myself very confused there. –  Henning Makholm Jul 19 '12 at 14:27
@HenningMakholm Np : ) I'm glad I'm not the only one to whom this happens. –  Rudy the Reindeer Jul 19 '12 at 14:33

I thought Stone-Weierstrass theorem says $C[0,1]$ is the uniform closure of $P$? A sequence in $P$ may converge to a discontinuous function on $[0,1]$.

With respect to the norm defined the in answer above we can say it is the closure of $P$.

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