Darboux function on $[0,1]$ with interesting property I have proved a few years ago the following proposition:

There exists $f: [0,1] \to [0,1]$ with Darboux property such that there exist $A,B \subset[0,1]$ with $A\cap B=\emptyset,\ A \cup B=[0,1]$ with $f(A)\subset B$ and $f(B)\subset A$. (of course $A,B\neq \emptyset$)

A function $f : I\subset \Bbb{R} \to \Bbb{R}$ ($I$ is an interval) has the Darboux property if $f([a,b])$ is an interval forall $[a,b]\subset I$.
The proof resembles the proof of Sierpinski's Therem, that any function $f : \Bbb{R} \to \Bbb{R}$ can be written as the sum of two functions each of them having the Darboux property.
My question is:


*

*have I proved something new, or it is a known fact that such a function exists?

*if the proposition is original can it be useful, I mean, can I submit this as an article?
[edit:] I know I should have done my job and send this to a some magazines to see if it is worth publishing. One of my teachers said that I send the article to JMAA, and of course it got rejected, because it's not that good. I tried at another magazine, but didn't even get an answer if it is rejected or not. I thought then that the proposition is not worthy of an article and proposed it as a problem to AMM. They said its too hard to be published as a problem. As I am a beginner and don't have any paper published until now. I don't know where should I try to send it. 
Could you please name some magazines where I could try and send the paper and recieve an answer to wether the proposition can be published or not?
 A: Just a comment, and a hint. The property of the function $f$ that you   consider, that is, The interval $[0,1]$ admits a partition in two sets that are mapped into each other by $f$, may be restated equivalently as: No iterate of $f$ of odd order has a fixed point (or also, $f$ has no periodic point of odd period). The equivalence is quite obvious, but this puts your result into the category: Darboux property vs Fixed point property, which seems to have a certain interest, looking on Google. A journal that recently had some papers on this topic, and seems suitable for your result, is Real Analysis Exchange. Also, you may consider if your method extends to the construction of a  Darboux function with no periodic points at all.
A: A well knowni application of the Axiom of Choice yields that if A is any subset of the unit interval whose intersection with an interval has size continuum then there is a function $F:A \to [0,1]$ with the following stronger version of the Darboux Property: The image of $[a,b]\cap A$ under $F$ is $[0,1]$ for all $[a,b]$. If A and B are disjoint sets with cardinality continuum in any interval then let $F_A$ and $F_B$ have this strong Darboux property and compose $F_A$ and $F_B$ with bijections from the unit interval to B and the complement of B respectively and then take the union of the resulting functions. This will have the property you want and even have the stronger Darboux property.
The literature on Darboux functions contains quite a few results of this type, but they are not all easily accessible.
