$P_n[0,1]$ be the set of all polynomial of degree atmost $n$ with supnorm is it a closed in $C[0,1]$? $P_n[0,1]$ be the set of all polynomial of degree atmost $n$ with supnorm is it a closed in $C[0,1]$? and $P[0,1]$ is set of all polynomials in $C[0,1]$  I know which is dense in $C[0,1]$ as we know by Weirstrass Polynomial approximation theorem any continous function can be uniformly approximated by sequence of polynomials. 
Thank you for  help. 
 A: $P_n[0,1]$ is an $n+1$-dimensional subspace of $C[0,1]$. In any topological vector space, a finite-dimensional subspace is closed. So, yes, $P_n[0,1]$ is closed in $C[0,1]$ with respect to the uniform norm (sup norm). In fact, the same is also true for any other norm that makes $C[0,1]$ a topological vector space.
The fact that finite-dimensional subspaces are closed is fairly well known, I think. There's a proof in these notes. Please see theorem 3G on page 8. The proof seems very long and roundabout, to me. I'll look for something shorter/simpler.
A nicer proof (in my opinion) can be found here, paraphrased from Bourbaki, which is where I must have learned all this, I suppose, many decades ago.
Just discovered that the question is a duplicate of this one, which has good answers.
A: All finite dimensional normed spaces of the same fixed dimension are isomorphic to each other. In your case, $P_n[0,1]$ is an $n+1$-dimensional normed space and so is isomorphic to $\ell_{\infty}^{n+1}$, which is clearly closed.
You are correct when you say that any continuous function can be uniformly approximated by a sequence of polynomials, but this does not mean that $P_n[0,1]$ is not closed. If $f \in C[0,1]$, we do not necessarily have that $f \in P_n[0,1]$ or $\overline{P_n[0,1]}$, for some fixed $n$. Instead we have that
$$
f \in \overline{ \bigcup_n P_n[0,1]}.
$$
If $p_k$ is a sequence of polynomials converging to $f$ in the uniform norm, then we should expect that the degree of $p_k$ goes to $\infty$ as $k\to \infty$ (unless $f$ happens to be a polynomial).
