What is the cardinality of the continuous dual of $C([0,1])$ (the set of continuous functions from $[0,1]\to \mathbb{R}$)?
closed as offtopic by Eric Wofsey, Andrés E. Caicedo, Bill Johnson, user9072, Emil Jeřábek Jul 23 '14 at 10:22
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The polynomial space $P([0,1])$ is dense in $C([0,1])$ by the StoneWeierstrass theorem. Therefore $P$ and $C$ have the same (continuous) dual space. But since $P$ has a countable Hamel basis, its algebraic dual has the same cardinality as $\mathbb R^{\mathbb N}$. Thus $C^*\leq\mathbb R^{\mathbb N}$. The evaluation functionals $\phi_a\in C^*$, $a\in[0,1]$, taking $\phi_a(f)=f(a)$ are continuous, so $[0,1]\leqC^*$. If I'm not mistaken, this implies that $C^*$ has the same cardinality as the real line. (If my memory serves me well, this depends on the choice of axioms, but is true in ZFC.)
Remark: A very similar argument works for the space $C$ itself. A continuous function is determined by its values at rational points, and, on the other hand, constant functions are continuous. This gives $[0,1]\leqC^*([0,1])\leq[0,1]^{\mathbb Q}$.

2$\begingroup$ This is an impeccable answer using appropriate settheoretical techniques but it might be of interest to vlv that much more precise formulations are available in the context of descriptive topology in the sense that a very large class of polish spaces are not only of equal cardinality but even homeomorphic. Thus all separable Fréchet spaces (infinite dimensional, of course) and the space of probability measures on, say, $[0,1]$ are homeomorphic to a countable product of the reals. $\endgroup$ – blackburne Jul 23 '14 at 8:18

$\begingroup$ Sorry, the probability measures on $[0,1]$, being compact, are homeomorphic to the Hilbert cube. $\endgroup$ – blackburne Jul 23 '14 at 8:21
Update: just realized that there is a very simple argument. $C([0,1])$ is separable, i.e. has a dense countable set (choose your favourite, for example polynomials as in the answer of Joonas) and every continuous functional is determined by its values on this set. Hence $C^*([0,1])=\mathbb{R}$.
The previous answer is fine but may be this will be useful.
Answer: $C^*([0,1])=\mathbb{R}$.
Riesz representation theorem states that any continuous functional $F$ on $C(0,1)$ has the form $$F(f)=\int\limits_0^1f(x)d\Phi(x), $$ where $\Phi(x)$ is a function of finite total variation and the integral is Riemann–Stieltjes.
Since $\Phi(x)$ has finite total variation we have $$\Phi(x)=\varphi^+(x)\varphi^(x) $$ for two monotonic functions on $[0,1]$.
Monotonic function $\varphi$ can have only countable many points of discountinuity. Denote them by $a_1,a_2,...$ and denote by $b_n:=\lim\limits_{x\to a_i+0}\varphi(x)\lim\limits_{x\to a_i0}\varphi(x)$. So $$\varphi(x)=\sum\limits_{n=1}^{\infty} b_n \delta(a_i) + \widetilde\varphi(x), $$ where $\delta(a_i)$ is the Dirac delta function at point $a_i$ and $\widetilde\varphi(x)$ is continuous.
So $\Phi(x)$ is determined by the following data: $$a_n^+,b_n^+,\widetilde\varphi^+,a_n^,b_n^,\widetilde\varphi^. $$
So the cardinality of $C^*([0,1])$ is bounded from above by $$\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}\times\mathbb{R}\times\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}\times\mathbb{R}=\mathbb{R} $$ (recall that $\widetilde\varphi^\pm$ are continuous, and that continuous function is determined by its values on some countable dense set, so $C([0,1])=\mathbb{R}$).
Recall that $[0,1]$ is compact and therefore $C([0,1])$ is a Polish space. By standard arguments, there are only $2^{\aleph_0}$ continuous functions between two Polish spaces.
Therefore there are at most $2^{\aleph_0}$ continuous functionals. I'm sure you will have no problem finding witnesses for at least $2^{\aleph_0}$ functionals as well.