Cardinality of $C^*([0,1])$ [closed]

Question

What is the cardinality of the continuous dual of $C([0,1])$ (the set of continuous functions from $[0,1]\to \mathbb{R}$)?

Answer 1

The polynomial space $P([0,1])$ is dense in $C([0,1])$ by the Stone-Weierstrass 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 $|[0,1]|\leq|C^*|$. 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 $|[0,1]|\leq|C^*([0,1])|\leq|[0,1]^{\mathbb Q}|$.

Answer 2

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 $b_n:=\lim\limits_{x\to a_i+0}\varphi(x)-\lim\limits_{x\to a_i-0}\varphi(x)$. 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 $|C([0,1])|=\mathbb{R}$).

Answer 3

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.