# Is every discontinuous everywhere Darboux function a constant image function?

I've been reading about Darboux functions recently and every example of a everywhere discontinuous Darboux function is also a function such that the image of every non-empty open interval is the image of the entire function. The canonical example being the Conway base-13 function. We will call these functions "constant image" functions for convenience. I was wondering if there were any Darboux functions that are everywhere discontinuous but are not constant image functions.

After discussing this with a colleague we came up with a function that seems to be a Darboux function that does not have this property. Here is a definition of the function we came up with:

Let us define a function $f$. This function will be similar to the Conway base-13 function. We first express the input in base-11 with digits $0-9,A$ such that it does not end in an infinite number of trailing $A$s. If the expansion ends with $A y_1 y_2 y_3 y_4\dots$ where $\forall n:y_n \neq A$ then the function will output $0.y_1 y_2 y_3 y_4\dots$ as a base-10 number.

This function $f$ is a such that the image of every non-empty open interval is the image of the entire function with an image of $[0,1]$, we now make a new function $$g(x)=f(x)+x$$

This discontinuous everywhere function clearly cannot has an unbounded image but every bounded interval on this function has a bounded image. So it cannot satisfy our property but it appears, intuitively, to be a Darboux function. However a proof that this function is Darboux has escaped us.

• @bof Yes that would be a trivial example of a strongly Darboux function. Feb 5, 2018 at 21:01
• @bof It looks like you are correct. I might have the definition of strongly Darboux wrong. It looks like a function that is "strongly Darboux" under my definition is Darboux iff the image is connected. I'll revise the question to avoid using the term strongly Darboux. Feb 5, 2018 at 21:11
• @bof It would seem so. Feel free to write that as an answer. Feb 5, 2018 at 21:19

Start with a function $f:\mathbb R\to\mathbb R$ such that $f(I)=\mathbb R$ for every nonempty open set $I.$ (This can be done very easily: Construct a sequence $P_1,P_2,P_3,\dots$ of pairwise disjoint Cantor sets such that every open interval with rational endpoints contains one of them; choose a surjection $f_n:P_n\to\mathbb R$ for each $n$; and define $f:\mathbb R\to\mathbb R$ so that $f|P_n=f_n$ for each $n.$ No need for any fancy base-$13$ stuff.)
Define $g:\mathbb R\to\mathbb R$ by setting $g(x)=\min\{f(x),|x|\}.$ It's easy to see that $g$ is an everywhere discontinuous Darboux function, but not a constant image function.