Originally Posted by

**Ruun** Thanks for even showing the dimension of the space.

I'm having some trouble with your notation, I'm a completely noob on real analysis. (My motivation of studying this is to study the Hilbert Spaces of Quantum Mechanics, wich are not quite good covered at school and I think that we are missing a huge point here).

Absolutely no worries! Notation can sometimes make math unreadable, and I in particular have a penchant for using "fancy looking notation" just to look "fancy". But anways

First, let's see if I understood your post:

I think that the idea is that the theorem states (for our particular case) $\displaystyle \text{card }\mathcal{F}\leqslant 2^{\aleph_0} \mbox{ and} \text{ card }\mathcal{F}\geqslant 2^{\aleph_0} \mbox{ then } \text{card }\mathcal{F}=2^{\aleph_0}$ right? And as we know the cardinal of the basis we know the dimension.

Correct. I am appealing to something called the Schroeder-Bernstein Theorem which in essence says that the relation $\displaystyle \leqslant$ on cardinal numbers (where $\displaystyle X\leqslant Y$ means "There exists an injection $\displaystyle f:X\to Y$" is antisymmetric). And then, right again. I'm appealing to the Dimension Theorem (which, by the way, is not trivial to prove. It requires something called the ultrafilter lemma) to conclude that since the cardinality of any basis must have cardinality $\displaystyle 2^{\aleph_0}$ then $\displaystyle \dim_{\mathbb{C}}C[\mathbb{R},\mathbb{C}]=2^{\aleph_0}$

Now the notation: I don't know if $\displaystyle \mathbb{C}^{\mathbb{Q}}$ stands for the cartesian product of $\displaystyle \mathbb{C}$ , $\displaystyle \mathbb{Q}$ times ("the number of rationals times").

In essence, yes. But, with the definition of Cartesian product extended to the arbitrary case; namely that if $\displaystyle \left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ is an indexed class of sets then

$\displaystyle \displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=\left\{f:\mat hcal{A}\to\bigcup_{\alpha\in\mathcal{A}}X_{\alpha} :f(\alpha)\in X_{\alpha}\text{ for all }\alpha\in\mathcal{A}\right\}$

So, without all the jargon really $\displaystyle \mathbb{C}^{\mathbb{Q}}$ is the set of all functions from $\displaystyle \mathbb{Q}$ into $\displaystyle \mathbb{C}$.

Also I don't know wich set is $\displaystyle f_{\mid \mathbb{Q}}$.

This is just the restriction of $\displaystyle f$ to $\displaystyle \mathbb{Q}$. Namely, for a function $\displaystyle f:\mathbb{R}\to\mathbb{C}$ we can create a new function, denoted $\displaystyle f\mid_{\mathbb{Q}}$ which maps $\displaystyle \mathbb{Q}\to\mathbb{C}$ and such that $\displaystyle f_{\mid\mathbb{Q}}(x)=f(x)$ for all $\displaystyle x\in\mathbb{Q}$. In other words, we just make the domain "smaller" and consider the function on that "smaller set". It turns out, like I said, that two continuous functions on $\displaystyle \mathbb{R}$ which agree on $\displaystyle \mathbb{Q}$ must be the same function (ask if you don't see why).