$\{(x_n)_n\in\mathbb R^{\mathbb N}|\text{ only finitely many components $x_i\neq 0$}\}$ isomorphic with...?

Question

Consider the vector space $(\mathbb R,\mathbb R^{\mathbb N},+)$ of all sequences in $\mathbb R$. Consider the following subset: $$ \{(x_n)_n\in\mathbb R^{\mathbb N}|\text{ only finitely many components $x_i\neq 0$}\}. $$ I can verify that this is a subspace of $\mathbb R^{\mathbb N}$. However, my book says that this subspace is isomorphic with a "well known vector space". Which one would that be?

I can imagine that the dimensions of this vectorspace is countably infinite. I know the following vectorspaces: $\mathbb R^n,\mathbb R^{n\times m},\mathbb R[X]_{\leq n},C[X]$, and so on. But I don't know which one they're referring to.

EDIT

Maybe $\mathbb R[X]$? All the real polynomials? We could identify the i-th element in our sequence with $\alpha_i X^i$. Yea... I think that's it.

Answer 1

Yes, you are correct. And one can also define a polynomial with real coefficients as a sequence $(a_n)_{n \geq 0}$ for which only a finite quantity of the $a_n$'s is non zero. Call $X = (0,1,0,\ldots)$ and you're set to go. Anyway, I digress, the isomorphism is given by $$\Phi((x_n)_n) = \sum_{n\geq 0}x_nX^n,$$and this sum is actually finite because of the convenient domain.

Answer 2

Yes, the real polynomials. Assume that the first index of the sequences is $0$ and consider the map $$(x_n)_n\mapsto\sum_{n=0}^\infty x_nX^n$$