Ryszard2012-08-08 14:05
A Darboux function with a zero-dimensional nowhere dense graph

Classical constructions of Darboux functions with zero-dimensional nowhere dense graphs involve set-theoretic tools like the axiom of choice, the well-ordering of a certain family of subsets of the plane, and cardinality arguments. Here I present a purely topological construction which yields a Borel function of a low class whose graph is a Borel set of a low class. Something for constructivists and perfectionists who love to minimize the set of resources they use.

A Darboux function $f\colon(0,1)\to[0,1]$ with a zero-dimensional nowhere dense graph will be constructed.

If $C\subset(0,1)$ is a Cantor set,
let $C^*=\{x\in C\colon x\in\overline{(0,x)\cap C}\cap\overline{(x,1)\cap C}\}$.
Now, if $C^*\cap[a,b]\not=\emptyset$, $a<b$,
then $(a,b)\cap C^*$ is nonempty and open in $C^*$.

Let $U_n$ be a countable basis of $(0,1)$
and let $C_1\subset U_1$ be a Cantor set.
If $C_1\subset U_1,\ldots,C_n\subset U_n$ are pairwise disjoint Cantor sets,
there is a Cantor set $C_{n+1}\subset U_{n+1}\setminus(C_1\cup\ldots\cup C_n)$.

Fix $n\in{\mathcal N}$. By a similar argument
we obtain a sequence of pairwise disjoint Cantor sets $\{C^k_n\}_{k=1}^\infty\subset C_n^*$
such that each nonempty open subset of $C^*_n$ contains one of them.
Let $f_n^k\colon C_n^k\to[0,1/n]$ be continuous surjections for each $k\in{\mathcal N}$.

Define $f\colon(0,1)\to[0,1]$ by $f(x)=f_n^k(x)$ if $x\in C_n^k$ and $f(x)=0$ otherwise.
In other words, $f=\bigcup\{f|C_n^k\colon n,k\in{\mathcal N}\}\cup (E\times\{0\})$,
where $E=(0,1)\setminus\{C^k_n\colon n,k\in{\mathcal N}\}$.

Since the sets $f|C_n^k$ are homeomorphs of the Cantor set and the set $E$ is zero-dimensional,
the metric space $f$ is a countable union of relatively closed zero-dimensional subsets,
and so $f$ is zero-dimensional.

Notice that $f$ is continuous on the dense set $E$, so $f$ has a nowhere dense graph.

Take any nondegenerate subinterval $A$ of $(0,1)$.
Let $n$ be the least number such that $A\cap C^*_n\not=\emptyset$.
Then $f(A)\subset[0,1/n]$ and
$C^k_n\subset A$ for some $k\in{\mathcal N}$.
$[0,1/n]=f(C^k_n)\subset f(A)$.
In effect, $f(A)=[0,1/n]$ and $f$ is shown to be Darboux.

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