Prove if $W \subset V$ is a subspace such that $T(W)\subset W$, then $V/W$ is finite-dimensional. Let $V$ denote the vector space of polynomials in one variable with coefficients in $\mathbb{R}$, and let $$T(f(x)) = xf(x).$$
Prove that if $W \subset V$ is a subspace such that $T(W)\subset W$, then $V/W$ is finite-dimensional.
First I let $f(x) \in W,$ such that $a_0+a_1x+a_2x^2+...+a_nx^n$, and $T(f(x)) = a_0x+a_1x^2+a_2x^3+...+a_nx^{n+1} \in W$. So I think $W$ is the set contains polynomials such that $a+ax+ax^2+ ... +ax^n +...ax^\infty$. But I am not sure about it, and how it is related to finite-dimension. 
Please help me to solve this problem, thank you!!

You need to show the following claim:

For all $f(x) \in V$, for all $g(x) \in W$, necessarily $f(x)g(x) \in W$.

To show this, write $f(x)= a_n x^n + \dots +a_1x + a_0$. Then
$$a_0g(x) \in W$$
$$a_1g(x) \in W \Rightarrow a_1xg(x) \in W$$
$$\vdots$$
$$a_ng(x) \in W \Rightarrow a_nxg(x) \in W \Rightarrow \dots \Rightarrow a_nx^ng(x) \in W$$
hence the sum
$$a_0 g(x) + a_1xg(x) + \dots + a_nx^ng(x) = (a_0+a_1x+ \dots +a_n x^n )g(x) = f(x)g(x) \in W$$
and the claim is proved.
From now on, we need the assumption $W \neq 0$.
Now, if you know some of ring theory, you should recognize that this means that $W$ is an ideal of $V= \Bbb{R}[x]$, hence $W=\langle m(x) \rangle$ for some generating polynomial $m(x)$, and $V/W$ has dimension $\deg m(x)$.
Otherwise, one can directly show that

$V/W$ is generated by $$1+W , x+W , \dots , x^{d-1}+W$$
  hence $\dim (V/W) \le d$.

In fact, for all $f(x) + W \in V/W$, divide
$$f(x) = q(x)m(x) + r(x)$$
where $q(x)$ is the quotient, and $r(x)$ is the remainder of the division, with $\deg r(x) < d$ or $r(x)=0$.
Then $$f(x)+W = q(x)m(x)+r(x)+W = r(x)+W$$
and you conclude.
