Dimension of a space I'm reading a book about Hilbert spaces, and in chapter 1 (which is supposed to be a revision of linear algebra), there's a problem I can't solve. I read the solution, which is in the book, and I don't understand it either.
Problem: Prove that the space of continuos functions in the interval (0,1): $C[0,1]$, has dimension $c=\dim(\mathbb{R})$.
Solution: The solution of the book goes by proving that the size of a minimal base of the space $B$ is first $|B|\leqslant c$ and $|B|\geqslant c$, and so $|B|=c$. the proof of it being greater or equal is simple and I understand it, the problem is the other thing. The author does this:

A continuos function is defined by the values it takes at rational numbers, so $|B|\leqslant c^{\aleph_0}=c$

I don't get that.
 A: Since the rationals are dense in the reals, a continuous function $f:\mathbb R \to  \mathbb R$ is completely determined by its values on the rationals. Thus, the the function from the set of all continuous functions $f:\mathbb R \to \mathbb R$ to the set of all functions $g:\mathbb Q \to \mathbb R$ (continuous or not), given by restricting a function $f:\mathbb R \to \mathbb R$ to the rationals, is an injective function. Thus the cardinality of the former is less than or equal to the cardinality of the latter, which is $|\mathbb R|^{|\mathbb Q|}=c^{\aleph_0}=c$.
The question is concerned with the domain $[0,1]$, but there is no essential difference.
A: The point is that if $f$ and $g$ are two functions in $C[0,1]$ and the restrictions to $[0,1]\cap\mathbb Q$ of $f$ and $g$ are equal then $f$ and $g$ are actually equal.
This is a simple consequence of the density of $[0,1]\cap\mathbb Q$ in $[0,1]$ using the continuity of $f$ and $g$.
A: Note that if $f,g$ are continuous and for every rational number $q$ it holds that $f(q)=g(q)$ then $f=g$ everywhere.
This means that $|B|\leq|\mathbb{R^Q}|=|\mathbb{R^N}|=|\mathbb R|$.
Also, $|\mathbb R|$ is not necessarily $\aleph_1$. This assumption is known as the continuum hypothesis.
