Why the set of polynomials is not closed in $C[0,1]$ This question starts from the following theorem: "Every finite dimensional subspace $Y$ of
a normed space $X$ is closed in $X$" (Kreyzig 2.4-3)
So the given counterexample for an infinite dimension subspace, is that if we take $X=C[0,1]$ and $Y = span\{x_0,x_1,x_2...\}$, where $x_j = t^j$, then $Y$ is not closed.
I guess this is the case because for example a sine function can be expressed as the limit point of a sequence of polynomials that converge uniformly (Taylor series). But why is the sine function not considered as a polynomial generated by an infinite basis? After all, we are explicitly saying $Y$ is infinite dimensional. For an infinite dimensional basis case, I would expect to have a mapping between every sequence to a polynomial in $C[0,1]$. If this is not the case, what exacly does it mean for $span\{x_0,x_1,x_2...\}$ to be considered infinite dimensional? If I input an infinite basis, it seems like it is not longer a polynomial.
I am confused.
 A: Even when a vector space is infinite-dimensional, each vector is written as a finite linear combination of basis vectors. Similarly, the definition of $\textrm{span}(S)$ is "the set of all finite linear combinations of elements from $S$".
Thus, even though the sine function can be written as an infinite series, we still define the set of all polynomials to be those functions that can be written as a finite sum of monomials.
The reason for not allowing infinite linear combinations is that this requires some notion of convergence and limits. While in specific cases this is possible, for a general abstract vector space there is no notion of convergence. Even in the case of the polynomials converging to the sine function, this convergence is only uniform on a compact set, not uniform over $\mathbb{R}$, so there is still some choice to be made for how to define convergence.
A: It's not just sine. The Weierstrass approximation theorem makes clear that every function in $C[0, 1]$ can be uniformly approximated by a sequence of polynomials, even something like $f(x) = |x - 0.5|$, a blancmange curve, Cantor's devil's staircase, or the Minkowski question mark function.
If we want to treat the sine function as a polynomial because it can be written as the uniform limit of a sequence of polynomials, all of those examples would have to be considered "polynomials" also, because they are also uniform limits of some sequence of polynomials. This broadens the definition of polynomial beyond even what I suspect OP would be willing to consider.
Functions like sine, that can be written as an "infinite polynomial" (i.e. are equal to their Taylor series at every point in the interval), are a special subset of $C[0, 1]$ called analytic functions. $A[0, 1]$, the set of analytic functions on $[0, 1]$, is much bigger than the set of polynomial functions $\Bbb{R}[x] \cap C[0, 1]$. For instance, while the set of all polynomials has a countable basis as an $\Bbb{R}$-vector space--the standard one--$A[0, 1]$ does not have such a countable basis, since $f_c(x) = 1/(x + c)$, $c > 0$, is an uncountable linearly independent subset of $A[0, 1]$.
A: To talk about convergence or closure, we first need to equip $C[0,1]$ with a topology. Which may be induced by a norm, though it's not mandatory. Here, since you are mentioning uniform convergence, define the norm on $C[0,1]$ by $||f||_\infty=\sup_{[0,1]}|f|$, so that $C[0,1]$ is now a Banach space. And convergence is defined as uniform convergence, i.e. $f_n\to f$ in $C[0,1]$ iff $\sup_{[0,1]} |f_n-f|\to 0$.
The subset $P$ of polynomials is closed in $C[0,1]$, iff for any sequence of polynomials $(p_n)_n$ that converges in $C[0,1]$, i.e. $p_n\to p$, then $p\in P$.
However, the Stone-Weierstrass theorem tells us that for any continuous function $f$ in $C[0,1]$, there is a sequence of polynomials $p_n$ such that $p_n\to f$, i.e. $||p_n-f||_\infty\to 0$. Take any $f$ in $C[0,1]$ that is not a polynomial, and you have a counterexample of the condition above, which proves that $P$ can't be closed in $C[0,1]$.
In fact, Stone-Weierstrass theorem means that with this definition of convergence (i.e. for this topology), $\overline P=C[0,1]$, that is, polynomials are dense in $C[0,1]$.
By the way we also have that $\mathring P=\emptyset$, since every polynomial $p$ is trivially the (uniform) limit of a sequence in $C[0,1]\backslash P$, for instance $f_n(x)=p(x)+\frac1n\exp(x)$. And $\mathring P\ne P$ proves that $P$ is not open.
