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Let $f,g:\mathbb{R}\rightarrow \mathbb{R}$ and $f(I)\cap g(J)\not=\phi$ for all nonempty open interval $I,J$.

Consider $f_1=\chi_\mathbb{Q}$ and $g_1=\chi_\mathbb{Q^c}$, we know that $f_1$ and $g_1$ satisfy above condition. (some examples can be constructed from this.)

Can we find other examples ?

Can we find $g$ when $f$ is given ?

ps. My question is motivated by this work enter link description here

Pasharapon Pongpattanakit
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2 Answers2

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We can find one $f$ that works for all $g.$ The idea is to construct $f:\mathbb {R}\to \mathbb {R}$ such that $f(I)=\mathbb {R}$ for every nonempty open interval $I.$ Such an $f$ then has the desired property.

The construction makes use of Cantor sets. By a Cantor set I mean here a set of the form $a +bK,$ where $a \in \mathbb R , b > 0,$ and $K$ is "the" Cantor set. We don't need to use any fancy properties of the Cantor set, just that it is compact, is nowhere dense, and has cardinality of $\mathbb {R}.$ All Cantor sets then have this property.

Let $I_n$ be the open intervals with rational end points. We can inductively find Cantor sets $K_n\subset I_n, n=1,2,\dots ,$ that are pairwise disjoint. For each $n$ there exists a bijection $f_n:K_n \to \mathbb {R}.$ Define $f:\mathbb {R}\to \mathbb {R}$ by setting $f = f_n$ on each $K_n,$ and setting $f=0$ on the complement of $\cup K_n.$ If $I$ is any nonempty open interval, then $I$ contains some $I_n,$ hence some $K_n,$ hence $f(I) \supset f(K_n) = \mathbb {R}.$

zhw.
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I know of examples that are very similar to your example. I'm not sure how helpful these examples will be. Basically, two indicator functions on any dense subset of the reals will work. For example $f(x) = 1$ and $g(x)$ is the indicator function of the algebraic numbers.

Also, you can do something like $h(x) = 3$ and $k(x) = 3g(x)$.

When $f$ is given, you won't necessarily be able to uniquely find a corresponding $g$. In your example, your $f(x)$ is the indicator function on the rationals. $g(x) = 1$ works just as well as your other $g(x)$. Also, $g(x)=0$ is another possible pair for your $f(x)$. Thus, in the generic case, it seems that you can't uniquely determine a function from that condition.

https://math.stackexchange.com/a/978359/234911

With the above construction, you should be able to come up with examples where you have functions $f_1,...,f_n$ such that $f_i(I) \cap f_j(J) \neq \emptyset %$ for all nonempty open intervals $I$, $J$ and integers $i$, $j$, such that $1 \leq i \leq j \leq n$.

Patrick Tam
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