# Existence of a strange function

Inspired by A discontinuous construction: Does there exist a function $$a \colon [0,1] \to (0,\infty)$$ and a family $$\{D_x \colon x \in [0,1]\}$$ of countable, dense subsets of $$[0,1]$$ with $$\bigcup_{x \in [0,1]} D_x = [0,1]$$ and $$\sum_{r \in D_x} a(r) < \infty$$ for all $$x \in [0,1]$$,

• Sure, it's not a hard exercise to use a Vitali set to engineer something like this. – Asaf Karagila Feb 23 at 11:54
• Thank you. I shall try it. – Dieter Kadelka Feb 23 at 12:01
• The Vitali set $V$ allows the definition of $\cal{D} := \{(v + \mathbb{Q}) \cap [0,1] \colon v \in V\}$. The $D \in \cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again. – Dieter Kadelka Feb 23 at 12:25

The relation $$x\sim y \quad \iff \quad x-y\in\mathbb{Q}$$ is an equivalence relation that partitions $$[0,1]$$ into countable sets of the form $$[t]=(t+\mathbb{Q})\cap [0,1]$$.
The set $$\{[t]:\,t\in [0,1]\}$$ (the set of all equivalence classes of the relation) has the same cardinality as $$[0,1]$$. Let $$\psi:[0,1]\to\{[t]:\, t\in [0,1]\}$$ be a bijection. For $$x\in [0,1]$$ we define $$D_x=\psi(x)$$. Since the sets $$D_x$$ are precisely equivalence classes of $$\sim$$, we have that they are countable, dense and $$\bigcup_{x\in [0,1]} D_x=[0,1]$$. Each of the sets $$[t]=(t+\mathbb{Q})\cap [0,1]$$ is countable infinite. Let $$\phi_{[t]}:[t]\to\{n^{-2}:\, n\in\mathbb{N}\}$$ be a bijection defined for each of the equivalence classes. Finally we define $$a:[0,1]\to (0,\infty) \quad\text{by}\quad a(x)=\phi_{[x]}(x).$$ It is easy to see that the function $$a$$ has the desired property since $$D_x=[t]$$ for some $$t$$ and hence $$\sum_{r\in D_x} a(r)= \sum_{r\in [t]}\phi_{[r]}(r)= \sum_{r\in [t]}\phi_{[t]}(r)= \sum_{n\in\mathbb{N}} n^{-2}=\frac{\pi^2}{6}.$$ In the second equality we used the fact that $$[r]=[t]$$ for $$r\in [t]$$ which is a property of any equivalence relation.