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.
Best Answer
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