Q:

How dimensions for vector space of functions are computed

Reading Introduction to Linear Algebra (Chap. 2) I encountered the following example for vector space:

The formal definition allows other things to be “vectors”-provided that addition and scalar multiplication are all right. We give three examples:
...

The space of functions $f(x)$. Here we admit all functions $f$ that are defined on a fixed interval, say $0 ≤ x ≤ 1$. The space includes $f(x) = x^2$, $g(x) = sin(x)$, their sum $(f +g)(x) = x^2 +sin(x)$, and all multiples like $3x^2$ and $−sin(x)$. The vectors are functions, and the dimension is somehow a larger infinity than for $R^∞$.

It is not clear to me why the "dimension of the vector space is somehow a larger infinity than for $R^∞$". How is this computed?

A:

If $V$ is your set of functions considered as a vector space over the field $\mathbb{F}$ ( we can suppose $\mathbb{F}=\mathbb{R}$) then , as a first step, we can say that its dimension is not finite because  for any $n \in \mathbb{N}$  the functions of the set $\{1,x,x^2,x^3,\cdots, x^n\},$ are linearly independent.
Now we can use the fact that for an infinite dimension vector space the dimension is such that : $|V|=\mbox{max}\{|\mathbb{R}|,\mbox{dim}(V)\}$, (where $|X|$ is the cardinality of $X$. See here).
In this case, since $|V|=|\mathbb{R}|^{|\mathbb{R}|} =\mathfrak{c}^{\mathfrak{c}}=2^{\mathfrak{c}} >|\mathbb{R}|$ (see here), we have
$\dim(V)=|V|=\mathfrak{c}^{\mathfrak{c}}
$

