Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $closure{P}=C[0,1]$?

share|cite|improve this question
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 – Un Chien Andalou Jul 19 '12 at 13:22
up vote 3 down vote accepted

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)

share|cite|improve this answer
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 $P$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.