Why are polynomials in $[0,1]$ not isomorphic to $C[0,1]$. I'm learning about Stone-Weisstrass Theorem and I learn that the closure of the polynomials is $C[0,1]$, I don't know where I read that they the polynomials in $[0,1]$ where isomorphic to $C[0,1]$. This only could happend if the polynomials in $[0,1]$ are closed but if I'm not mistaked they are not.
 A: No, they do not form a closed set. If they did, then the closure of the set $P$ of the polynomial functions from $[0,1]$ into $\mathbb R$ would be $P$ itself, not $C[0,1]$, and the Stone-Weierstrass theorme would be false.
If you want a concrete example, consider the polynomials $P_n(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}$. They converge to the exponential function in $C[0,1]$, and that function is not polynomial.
A: The space of polynomials has countable infinite dimension (a basis is given by the $x^n$), while the space of continuous functions has uncountable infinite dimension.
The ring of polynomials is an integral domain, while the space of continuous functions is not (see for example Ring of all continuous functions from reals into reals is not integral domain - you can easily adopt this to $[0,1]$).
A: Not sure why the isomorphism claim, or even what type of isomorphism is meant here: purely algebraic, an isometry in normed spaces (and if so, in what norm), etc..
In the sup norm, the polynomials do form a dense subset of ${\cal C}[0,1]$, by Stone-Weierstrass.  However, as you remarked, this set is not closed, and I think it even has no interior points.  
To prove the absence of interior points, pick a polynomial $p(x)$ and an $\epsilon > 0$.  
Now define $f_{\epsilon}(x) = 1 + {\epsilon} \;$(triangle wave).
And, finally, let
$$
g(x) = f_{\epsilon}(x) p(x).
$$
Notice that $g$ is continuous on $[0, 1]$, but is not a polynomial because of the cusps.  Intuitively, $g$ was constructed by taking $p(x)$ and "perturbing it with small-height cusps".
We see that
$$
|g(x) - p(x)| = {\epsilon} \; |\mbox{(triangle wave)}(x)| \leq {\epsilon}
$$
It follows that $p(x)$ has no open neighborhood consisting entirely of polynomials.
