Proving that the space $\mathbb{P}$ of all polynomials is not finite dimensional Just want to know if my approach to this problem is correct:
Suppose that the space P of all polynomials is finite dimensional. Then we can call it $\mathbb{P_{n}}$ with $n<\infty$.
Therefore the standard basis of $\mathbb{P_{n}}$ has $n+1$ elements such that {$\overrightarrow v_{1},...,\overrightarrow v_{n+1}$}={$1,...,x^n$}.
But since $\mathbb{P_{n}}$ is the space of all polynomials, then it contains a vector $\overrightarrow v=x^{n+1}$.
However, $\not\exists r_{1},...,r_{n+1}$ such that $$1*r_{1}+...+x^{n}*r_{n+1}=x^{n+1}$$
Then {$1,...,x^n$} is not a basis for $\mathbb{P_{n}}$, we have contradiction.
$\therefore \mathbb{P}$ is not finite dimensional.
 A: Hint/Solution: $\{1,x,x^2,x^3,\dots\}$ is a linearly independent subset of $\mathbb P$.
Alternate Solution: Note that  for each $n \in \bf N$,the vector space of polynomials of degree less than or equal to $n$ sits inside $\mathbb P$ Hence the dimension of $\mathbb P$ is greater than any natural number. Hence $\mathbb P$ is not finite dimensional
A: Let $p_1,p_2,\dots,p_n$ be polynomials. It is not restrictive to assume they are nonzero, so let $d$ be the maximum degree of those polynomials.
Prove that $x^{d+1}$ does not belong to the span of $\{p_1,\dots,p_n\}$. Thus no finite set can span the space of all polynomials.

About your proof. I would grade it as incorrect; however, your idea is similar to what I sketched above. Your error is in assuming that if $\mathbb{P}$ is finite dimensional then $\{1,x,\dots,x^n\}$ must be a basis. Your argument shows it isn't, but it doesn't exclude that another basis exists.
You could salvage it by saying that if $n=\dim\mathbb{P}$, then, since $\{1,x,\dots,x^{n-1}\}$ must be a basis, because it is linearly independent and has as many elements as the dimension (which is possibly what you were attempting to say).
In any case, this would be using much more than needed: what you just need to show is that no finite subset of $\mathbb{P}$ spans it.
A: All polynomial $p(x)=a_nx^n+.....+a_1x+a_0$ is a linear combination of the monomials $x^k;\space k=0,1,.....,n$.
Besides $p=0\iff p(x)=0$ for all $x\iff a_k=0;\space k=0,1,.....,n$ because $p(x)=0$ has just $n$ roots. This shows that the monomials $x^k$ are linearly independant (over the field of scalars).
Thus, since the exponent $n$ is not bounded, $\mathbb{P}$ is of infinite dimension.
