Define the set $P$ by \begin{equation*} P=\setparts{p}{p(x)\text{ is a polynomial in }x} \end{equation*}
Our operations will be the same as those defined for $P_n$ (Example VSP).
With no restrictions on the possible degrees of our polynomials, any finite set that is a candidate for spanning $P$ will come up short. We will give a proof by contradiction (Proof Technique CD). To this end, suppose that the dimension of $P$ is finite, say $\dimension{P}=n$.
The set $T=\set{1,\,x,\,x^2,\,\ldots,\,x^n}$ is a linearly independent set (check this!) containing $n+1$ polynomials from $P$. However, a basis of $P$ will be a spanning set of $P$ containing $n$ vectors. This situation is a contradiction of Theorem SSLD, so our assumption that $P$ has finite dimension is false. Thus, we say $\dimension{P}=\infty$.