Let me first express how I understand an infinite dimensional vector space.
A vector space V is infinite dimensional if $\forall n\in \mathbb{N} (\{e_1,\dots , e_n\} \text{ linearly independent }\Rightarrow \{e_1 , \dots , e_{n+1}\}\text{ linearly independent })$.
Having the above definition in mind, I try to prove that the vector space $C([a,b])$ of all functions from $[a,b]\to \mathbb{R}$ is infinite dimensional, with basis the set of polynomials.
Attempt at a proof: Let $n\in \mathbb{N}$ be arbitrary, and assume that $\{x , x^2 , \dots , x^n\}$ is linearly independent. My goal is to show that $\{x,x^2,\dots , x^{n+1}\}$ is linearly independent. To do that, assume that $\sum_{k=1}^{n+1}b_kx^k=0$, and now I need to show that for all $k$, $b_k=0$. My feeling is that this should proceed by contradiction, namely I should assume that there exists $k$ such that $b_k\neq 0$…
I am missing the step from here on, how do I proceed. Also is the definition of infinite dimensionality correct? Is this how one proves that a vector space is infinite dimensional?
Best Answer
Your supposed definition of infinite dimensional vector space doesn't make sense. A vector space $V$ is infinite dimensional if no finite subset of $V$ generates $V$. This is equivalent to the assertion that $V$ has an infinite subset which is linearly independent.
The space $\mathcal{C}\bigl([a,b]\bigr)$ is infinite dimensional because the set $\{1,x,x^2,x^3,\ldots\}$ is linearly independent.