Orthonormal basis: Countable $\infty$ vs. Uncountable $\infty$ My doubt is the following, when you create an orthonormal basis for a space, the number of coefficients in each vector, and the number of vectors is equal to the dimension of the space (at least in finite dimensional spaces).
For example the standard basis for $\mathbb{R}^3$ is $(1,0,0),(0,1,0),(0,0,1)$.
What happens when we consider infinite dimensional Hilbert spaces?
For instance, let $\mathcal{L}_2(-\pi ,\pi) $ be the collections of functions $\chi =\{x(t):-\pi \leq t \leq \pi\}$ for which $\int_{-\pi}^{\pi}|x^2(t)|dt<\infty$. Define the vector addition and scalar multiplication coordinatewise such that we end up with a Hilbert space.
My text book says that there's a set of vectors $\{z_n:n=0,\pm1,...\}$ that is a complete orthonormal sequence in $\mathcal{L}_2(-\pi ,\pi)$.
My question is, I know that every single $x(t)$ has an infinite dimension, since we are considering a continuous function, but not countable infinite, while  the number of vectors $z_n$ is clearly countable infinite. 
That makes me think, isn't necessary that the number of components in a vector of the "basis" and the number of vectors that form the latter coincide?
Hopefully you will shed some light on the problem.

You can indeed specify a member of ${\mathcal L}_2(-\pi,\pi)$ with countably many real numbers.  That doesn't say you have to: you can also choose to specify
the values $f(t)$ for all $t \in (-\pi,\pi)$ (modulo the non-uniqueness due to the fact that these are really 
equivalence classes of functions rather than functions: you can change $f$ on a set of measure $0$ and it's the same
member of $\mathcal L_2$).  But those uncountably many choices are not independent.
