# example of uncountable family of subsets of a countable set with finite intersections

We wish to give an answer to the following:

Problem. Assume, that $X$ is a countable set. Is there a family $\{X_{i}\}_{i\in I}$ of subsets of $X$ such that $I$ is an uncountable set, but for any $i\neq j\in I$ the intersection  $X_{i}\cap X_{j}$ is finite?

Example. Let $x\in[1,2)$ be a real number. Express $x$ using digits

 $x=1.x_{1}x_{2}x_{3}x_{4}\cdots=1+\sum_{i=1}^{\infty}x_{i}\cdot 10^{-i}$

where each $x_{i}\in\{0,1,2,3,4,5,6,7,8,9\}$. With $x$ we associate the following natural numbers  $\beta_{n}(x)=1x_{1}x_{2}x_{3}\cdots x_{n-1}x_{n}=10^{n+1}+\sum_{i=1}^{n}x_{i}% \cdot 10^{n-i+1}.$

Now define $A:[1,2)\to\mathrm{P}(\mathbb{N})$ (here $\mathrm{P}(X)$ stands for ,,the power set  of $X$”) by

 $A(x)=\{\beta_{1}(x),\beta_{2}(x),\beta_{3}(x),\ldots\}.$

$A$ is injective  . Indeed, note that for any $x,y\in[1,2)$ if $\beta_{i}(x)=\beta_{j}(y)$, then $i=j$ (this is because equal $\beta$ numbers have equal ,,length” and this is because each $\beta$ has $1$ at the begining, zeros are not the problem). Therefore, if $A(x)=A(y)$ for some $x,y$, then it follows, that $\beta_{i}(x)=\beta_{i}(y)$ for each $i$, but this implies that corresponding digits of $x$ and $y$ are equal. Thus $x=y$.

This shows, that $\{A(x)\}_{x\in[1,2)}$ is an uncountable family of subsets of $\mathbb{N}$. Now in order to prove that $A(x)\cap A(y)$ is finite whenever $x\neq y$ it is enough to show that we can uniquely reconstruct $x$ from any infinite   sequence of numbers from $A(x)$. This can be proved by using similar techniques as before and we leave it as a simple exercise.

Title example of uncountable family of subsets of a countable set with finite intersections ExampleOfUncountableFamilyOfSubsetsOfACountableSetWithFiniteIntersections 2013-03-22 19:16:29 2013-03-22 19:16:29 joking (16130) joking (16130) 4 joking (16130) Example msc 03E10