What is the dimensionality of the space of continuous functions I got introduced to the idea of fractals, and the idea that fractals can have dimensions that are non-integers. 
This got me thinking, the space of real-valued functions has a dimensionality as well, and it seems likely to me that the space of continuous functions has a dimensionality strictly lower than that of all functions. And perhaps the space of differentiable or smooth functions has a dimensionality even lower?
Note, this question is purely out of interest, and I don't know much about this. 
 A: Continuous functions are specified by their values on $\mathbb{Q}$, so there are only $2^{\aleph_0}=c$ of them. The space of them has dimension $c$ since the $e^{kx}$ are linearly independent.
A: The set of real valued functions has cardinality $\mathfrak {c^c}$ because you can specify the value at each real independently.  The set of continuous functions has cardinality no greater than ${\mathfrak c}^{\aleph_0}=\mathfrak c$ because once you specify its value on each rational you can extend the function to the reals and has cardinality at least $\mathfrak c$ because there are that many constant functions.  
To talk of the dimension of the sets of function you need to define a topology on them.  Unlike the reals, I don't think there is a standard one.
A: I'm assuming your are talking about algebraic dimension here.
Any real vector space $V$ with $\dim V \ge \mathfrak c$ has its cardinality equal to its dimension.
Indeed, let $B$ be a basis for $V$. Every element of $V$ can be uniquely written as a finite linear combination of elements of $B$, so it is uniquely determined by a finite sequence of pairs $(\lambda_1, b_1), \ldots, (\lambda_n, b_n)$ where $\lambda_i \in \mathbb{R}$ and $b_i \in B$. Conversely, every such finite sequence is again in $V$ so we have:
$$|V| = \left|\bigcup_{n\in\mathbb{N}}(\mathbb{R} \times B)^n\right| = \aleph_0 \cdot |\mathbb{R} \times B| = \aleph_0 \cdot \mathfrak{c} \cdot |B| = |B|$$
In particular, this result holds for the spaces $\mathbb{R}^\mathbb{R}$ and $C(\mathbb{R})$ in question. Both have dimension greater than $\mathfrak c$ since $\{e^{\lambda x} : \lambda \in \mathbb{R}\}$ is a linearly independent set, as noted in the answer by @J.G.
From @Ross Millikan's answer, we see that the cardinalities of our spaces are $|C(\mathbb{R})| = c$ and $\left|\mathbb{R}^\mathbb{R}\right| = 2^\mathfrak{c}$.
We can conclude:
$$\dim C(\mathbb{R}) = \mathfrak{c}$$
$$\dim \mathbb{R}^\mathbb{R} = 2^\mathfrak{c}$$
So indeed $\dim C(\mathbb{R}) < \dim \mathbb{R}^\mathbb{R}$.
Regarding differentiable functions, and in general $C^k(\mathbb{R})$, notice that again $\left|C^k(\mathbb{R})\right| = \mathfrak{c}$. We certainly have $C^k(\mathbb{R}) \subseteq C(\mathbb{R})$ so $\left|C^k(\mathbb{R})\right| \le \mathfrak{c}$, and again, the set $\{e^{\lambda x} : \lambda \in \mathbb{R}\}$ is contained in $C^k(\mathbb{R})$ so $\left|C^k(\mathbb{R})\right| \ge \mathfrak{c}$. By the same argument as above, we conclude $$\dim C^k(\mathbb{R}) = \mathfrak{c}, \text{ for all } k \in \mathbb{N}_0 \cup \{\infty\}$$
