The number of Darboux functions

Question

What is the number of real-valued functions of a real variable that have the Darboux property? I know it's at least continuum, because it's a broader class than continuous functions, whose number is continuum; but does it go any further?

Answer 1

Let $X \subset \mathbb{R}$ be an $\mathbb{Q}$-linearly independent subset with cardinality that of $\mathbb{R}$. As $\mathbb{R}$ is equipotent to $\mathbb{R}^2$, there is a partition $X=\cup_{\alpha \in \mathbb{R}}{X_{\alpha}}$ with each $X_{\alpha}$ being in bijection with $\mathbb{R}$.

In particular, by rescaling some vectors in $X_{\alpha}$, we can make sure that every empty bounded open interval with rational endpoints contains some element in $X_{\alpha}$ – thus every nonempty open interval contains points of $X_{\alpha}$.

Now consider any function $f: \mathbb{R}\rightarrow \mathbb{R}$ such that $f(X_{\alpha})=\{\alpha\}$. There are $2^{\mathbb{R}}$ possibilities for $f$ (because there are no constraints on $2X$ and $2X$ is in bijection with $\mathbb{R}$), but by the above construction $f(I)=\mathbb{R}$ for any open interval $I$, so $f$ is Darboux.

Answer 2

Let $I_1,I_2,I_3,\dots$ be an enumeration of the intervals with rational endpoints. Choose pairwise disjoint Cantor sets $C_n\subset I_n$. (This is possible since $C_1\cup C_2\cup\cdots\cup C_{n-1}$ is nowhere dense.) For each $n\in\mathbb N$ define a bijection $f_n:C_n\to\mathbb R$.

If $f:\mathbb R\to\mathbb R$ is such that $f\supseteq\bigcup_{n=2}^\infty f_n$ then $f$ has the Darboux property. The number of such functions $f$ is $2^\mathfrak c$ since $f$ may take arbitrary values on $C_1$.

More generally, given any function $f:\mathbb R\to\mathbb R$ which takes all real values in every interval, we can modify it arbitrarily on a nowhere dense set, and the resulting function will have he same property.