Show that the set of polynomial functions is not finitely spanned The set of the polynomial functions $P(\Bbb R)$ is a subset of functions in the form of $$f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n.$$
I wish to show that the set of polynomial functions is not finitely spanned. 
The following is what I have done: 
Assume, by contradiction, that the set of polynomial functions is finitely spanned. Then, take the basis of the polynomial functions, i.e., the set $\{1,x,x^2,\ldots,x^n\}$. This is so because: 


*

*Each $x^i$ is linear independent of $x^j$ if $i\neq j$, that is we cannot find a scalar $c\in\Bbb R$ such that $x^i=c\cdot x^j$; 

*Each polynomial function $f(x)$ can be written in the form of $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$, a linear combination of the vectors $1,x,x^2,\ldots,x^n$. 


Next, consider the function $g(x)=x^{n+1}$, which is clearly a polynomial function, and we assume it can written as a linear combination of the vectors $1,x,x^2,\ldots,x^n$, i.e., $$g(x)=x^{n+1}=\lambda_1+\lambda_2x+\cdots+\lambda_nx^n.$$
I wish to show this is not achievable, but I got stuck on how to show this explicitly since it is such a obvious result.
 A: First, your attempt will only show that the finite sets $\{ 1, x, \dotsc, x^n \}$ cannot span $P(\mathbb R)$; you need to show that no finite set of polynomials $\{f_1(x),\dotsc,f_n(x)\}$ can span $P(\mathbb R)$.
Your idea of considering $x^{n+1}$ can be adapted to this situation.
Use the fact that any non-zero polynomial has only finitely many roots to show that distinct polynomials cannot induce the same function.
A: You're on the right track! Once you assume it's finitely spanned by some set $S$, you know that there is a polynomial $p\in S$ of maximum degree, say $\deg p = n$. (You implicitly assume that $\operatorname{span}(S) = \operatorname{span}\{1,x,\ldots, x^n\}$, but this might not be the case. You might have $\operatorname{span}(S)\subsetneq\operatorname{span}\{1,x,\ldots, x^n\}$! It does suffice to consider the case where the two spans are equal, because this is the "worst" that would happen, but in either case, the key idea is the same.) Then the polynomial $x^{n+1}$ cannot be in $\operatorname{span}(S)$, because $\deg(p + q)\leq\max\{\deg p, \deg q\}$. (This last equation seems to be exactly the piece you're looking for.)
A: If $x^{n+1}$ could be written as a linear combination of a polynomial of order $n$, then the sum of the two, i.e.
$$\lambda_0+\lambda_1x+\cdots+\lambda_nx^n-x^{n+1}$$
would need to equal zero for non-zero $\lambda_i$. This is clearly impossible.
A: This solution uses the variable $n$ in several different places, for several different purposes.  Also, there is no need to introduce the standard basis; you are given a (supposed) basis already, you merely need to find some polynomial that is not expressible with that basis.  See @Stahl's suggestion for how to find one.
A: Let $\{p_1, \ldots, p_n\}\subset P(\mathbb R)$ be a finite subset.
Let $d=\max\{\deg(p_1),\ldots, \deg(p_n)\}$ be the macimal degree of its elements.
Then for any $f\in\langle p_1,\ldots p_n\rangle$ we observe that $x\mapsto \left|\frac{f(x)}{x^d}\right|$ is bounded on $(1,\infty)$ (because the expression simplifies to a combination of $1, x^{-1},x^{-2},\ldots$ ), whereas $x\mapsto \left|\frac{x^{d+1}}{x^d}\right|$ is not bounded on $(1,\infty)$. We conclude that $\langle p_1,\ldots p_n\rangle\ne P(\mathbb R)$.

Note that we make iuse of specific properties of $\mathbb R$. If we consider the set of polynomial (or arbitrary) functions $F\to F$ for a finite field $F$, then this space is clearly $\#F$-dimensional. In this setup, one really must distinguish between the infinte-dimensional space of polynomials and the finite-dimensional space of polynomial functions.
