If $X=\{1,x,x^2,\dots,x^n\}$, $sp(X)=$? Also, is it finitely generated? 
If $X=\{1,x,x^2,\dots,x^n\}$, $\operatorname{span}(X)=$? Also, is it finitely generated

Well, simply this is the set of polynomials of degree $\leq n$. So 
$$\operatorname{span}(X)=\{a_0+a_1x+a_2x^2+\dots+a_nx^n|a_i\in F, i=0,1,2,\dots,n\}=P_n(F)$$
There are an infinite number of polynomials that we can create using the linear combinations of $X$, and so the span of $X$ is infinite. But then why does my answer say that the set is finitely generated? This is confusing me.
If we can create an infinite amount of functions, why is it "finitely generated' in this case?
 A: In algebra, we say a vector space $V$ is generated by a set $S$ if it is the smallest vector space containing $S$. We also say $S$ generates $V$. We have the same sort of definition when $V$ is a group or an ideal or some other algebraic object.
For instance, the subspace of $\mathbf{R}^3$ generated by $(1,0,0)$ is $V = \{(a,0,0) : a \in \mathbf{R}\}$ because


*

*$V$ is a vector space containing $(1,0,0)$

*if $W$ is a subspace of $\mathbf{R}^3$ containing $(1,0,0)$ then it contains all scalar multiples of $(1,0,0)$ because subspaces are closed under taking scalar multiples


More generally, if $S$ is any set of vectors in a vector space $V$ then the subspace generated by $S$ is $\operatorname{span} S$ because


*

*$\operatorname{span} S$ is a vector space containing $S$

*if $W$ is a vector space containing $S$ then $W$ contains all linear combinations of elements of $S$, namely $W$ contains $\operatorname{span} S$


Given a vector space $V$, we say that $V$ is finitely generated if there is a finite set $S$ that generates $V$. For instance $\mathbf{R}^3$ is generated by $\mathbf{R}^3$ (which is infinite) and also by $S = \{(1,0,0), (0,1,0), (0,0,1)\}$ (which is finite). Since there exists a finite set that generates $\mathbf{R}^3$, namely $S$, we have that $\mathbf{R}^3$ is finitely generated.
An equivalent characterization of finitely generated vector spaces is finite dimension. If $\mathcal B$ is a finite basis for $V$ then $V$ is generated by $\mathcal B$ so $V$ is finitely generated. Conversely, if $V$ is finitely generated, say by $S$, then we can select a basis for $V$ by taking a maximally linearly independent subset of $S$.
Therefore the vector space of all polynomials is not finitely generated because it has an infinite linearly independent set, $\{1,x,x^2,\dots\}$ and therefore cannot have a finite basis. On the other hand, the vector space of all polynomials of degree $\le n$ is finitely generated because it is generated by the finite set $\{1,x,x^2,\dots,x^n\}$.
A: As a vector space over $F$, $V = \text{span} \, X$ is finitely generated by the elements $1, x, x^2, \dots, x^n$. 
For a vector space $V$ to be finitely generated means that there is a finite set $S$ for which $V = \text{span} \, S$. But $S = X$ is such a set, because $X$ has finite size $n$. 
This is different from saying whether $V$ is finite or infinite as a set. In this case, that will depend on the size of the field $F$. 
