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.

  • $\begingroup$ @bof Yes that would be a trivial example of a strongly Darboux function. $\endgroup$ Feb 5, 2018 at 21:01
  • $\begingroup$ @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. $\endgroup$ Feb 5, 2018 at 21:11
  • $\begingroup$ @bof It would seem so. Feel free to write that as an answer. $\endgroup$ Feb 5, 2018 at 21:19

1 Answer 1


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.


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