In our linear algebra lectures, we proved that for any vector space $V$ over some field $\Bbb F$, there is a basis that contains no vector from a base of its subspace $M$.
I was thinking about a vector space $\mathcal C(\Bbb R)$of all real continuous functions. A function can be: even $\underline{\vee}$ odd $\underline{\vee}$ neither even nor odd.
Every even function can be written as a sum of an even and an odd function.
Every odd function can be written as a difference of an even and an odd function.
When constructing a base for $\mathcal C(\mathbb R)$, I wanted to use even functions and some that are neither even nor odd.
Let $E$ be a space of all even functions in $\mathcal C(\mathbb R)$, O space of all odd functions in $\mathcal C(\mathbb R)$ an $W$ space of the rest of the functions in $\mathcal C(\mathbb R)$.
$$\dim{\mathcal C(\mathbb R)}=\aleph_0 ?$$
Let $f$ be some arbitrary even function, $f_e \in E, f_o \in O, f_w \in W.$
Then: $$f_i=f_{e_i}+f_{o_i}$$
My greatest difficulty is that $\mathbb R$ is uncountably infinite, so my index might be nonsense.
My base for $\mathcal C(\Bbb R)$ would look like this:
$$\{(x_i,x_i',\ldots,y_i,y_i',\ldots):x_i=f_{e_i}+f_{o_i},y_i=f_{w_i},i\in I, |I|=|\Bbb R|\}$$
Does this have any sense?
Best Answer
The problem with what you're trying to achieve is that the existence of a basis for $\mathcal{C}(\mathbb{R})$ relies on the axiom of choice, which isn't constructive. While such bases do exist, it is extremely hard to provide one explicitly.
As for the cardinality of such a basis, it is $2^{\aleph_0}$.