Dimension of vector subspace $X := \{f \in C([0,1];\mathbb{R}) : f(0) = 0 \}$ I know that the space $C([0,1];\mathbb{R})$ is of infinite dimension.  I would like to know if the vector subspace $$X := \{f \in C([0,1];\mathbb{R}) : f(0) = 0 \}$$ is also of infinite dimension.
Can I use the set of vectors $\{x, x^2, x^3, \dots \}$ as the space's basis?  I know they are linearly independent, but how can I actually show that any continuous function on $[0,1]$ can be written as a linear combination of this basis?
I'm new to topology, and honestly, I prefer Calculus to this.  Thank you.
 A: Finite-dimensionality is a purely algebraic notion. There is no need for any topology to be attached to the vector space, or even the underlying field, in order to talk about finite/infinite-dimensionality. The role of topology in your proof begins and ends with you observing that the monomials are continuous functions (and hence in the space). After that, the fact that the set is infinite and linearly independent suffices to show the space is infinite-dimensional.
As an alternative, you could observe that $X$ sums directly with a one-dimensional subspace to given $C([0, 1]; \Bbb{R})$. If we let $k(x) = 1$ for all $x$ (a constant function), then any $f \in C([0, 1]; \Bbb{R})$ can be expressed as $f = (f - f(1)k) + f(1)k \in X + \operatorname{span}\{k\}$. The sum is direct because the intersection is trivial; if we had $f \in X \cap \operatorname{span}\{k\}$, then $f = \lambda k$, but $f(0) = 0$. This means that $0 = f(0) = \lambda k(0) = \lambda$, so $f = 0k = 0$.
If it were the case that $X$ were finite-dimensional, then $X \oplus \operatorname{span}\{k\} = C([0, 1]; \Bbb{R})$ would also be finite-dimensional (any basis for $X$, with $k$ added, would be a basis for their direct sum). But, this is a contradiction, so $X$ must be infinite-dimensional.
