Can a countable set contain uncountably many infinite subsets such that the intersection of any two such distinct subsets is finite?

Can a countable set contain uncountably many infinite subsets such that the intersection of any two such distinct subsets is finite?

• Not sure about $\mathfrak c$-many, but $\omega_1$-many -- yes, I think. – tomasz Jun 24 '12 at 13:47
• @tomasz: continuum many is perfectly doable. – Asaf Karagila Jun 24 '12 at 14:21
• Such system of sets is called almost disjoint family. See e.g. here. – Martin Sleziak Jun 24 '12 at 17:07
• This is problem 13, page 27, of Kaplansky's "Set theory and metric spaces". – Santropedro Apr 19 '17 at 3:32
• Since this is the canonical duplicate I've decided to edit its title to make it easier to find. I invite people to help improve this question in other ways as well. – Asaf Karagila Apr 12 at 16:52

Yes. For every $r\in\mathbb R$ choose a sequence of rational numbers $\{r_n\in\mathbb Q\mid n\in\mathbb N\}$ which converges monotonically to $r$, this sequence is of course a subset of $\mathbb Q$ - a countable set.

If $r\neq s$ are two real numbers then the sequence we chose for them must intersect at a finite subset, otherwise we had a subsequence of the two which would converge to two different limit points.

Since $\mathbb R$ is uncountable (and in fact has cardinality as $\mathcal P(\mathbb Q)$) we have indeed uncountably many subsets of $\mathbb Q$ with the wanted property.

• @pritam: By Cantor's theorem it is not countable. – Asaf Karagila Jun 24 '12 at 13:06
• @pritam he's considering subsets of $\mathbb{Q}$, which is countable. So it meets your conditions exactly. – Dustan Levenstein Jun 24 '12 at 13:07
• @pritam: And $\mathbb Q$ is not a countable set? – Asaf Karagila Jun 24 '12 at 13:09
• @dorsh605: Yes, your previous comment used $X_\alpha$ to denote the sequence converging to $\alpha$. Sorry for not predicting the future change to $S_\alpha$, my crystal ball is not with me right now. ;-) – Asaf Karagila Sep 2 '15 at 9:52
• @Sahiba Arora: For various reasons. Mainly because every convergent sequence in $\Bbb R$ has a monotonous subsequence. You can also enumerate the rational numbers, and proceed by induction to define a sequence which converges monotonically to $r$ from above or from below. – Asaf Karagila Sep 26 '15 at 13:37

Yes. We know $\mathbb{N}$ and $\mathbb{Q}$ are equipotent so we choose a bijection $f:\mathbb{N} \to \mathbb{Q}$. We also know that $\mathbb{R}$ is equipotent to the set of equivalence classes of Cauchy sequences in $\mathbb{Q}$. For every $r \in \mathbb{R}$ choose $(q_{r,n})_n$ a representative from the equivalence class corresponding to $r$. Note that if $r_1\neq r_2 \in \mathbb{R}$ then $q_{r_1,n} = q_{r_2,n}$ for at most finitely many $n$. Since $f$ is a bijection we have that the sequences $(m_{r,n})_n := (f^{-1}(q_{r,n}))_n \subseteq \mathbb{N}$ share the same property. Since $\mathbb{R}$ is uncountable this concludes the proof; just choose the subset $N_\alpha$ to be the range of $m_\alpha$.

Let P be the set of all prime numbers, which is countable. Consider an infinite subset A = {$p_0,p_1,p_2...$} of primes from P arranged in an increasing order. We form $B_A$ = {$p_0,p_0p_1,p_0p_1p_2,...$} corresponding to A.

Now, the prime factorization is unique for integers. Thus, if we choose another infinite subset $A_1$ of P different from A then the corresponding $B_{A_1}$ and $B_A$ has only finitely many elements in common. Since P is countably infinite, the number of different infinite subsets of P is continuum. Hence we are done.

• The definition of $B_A$ is really unclear. – Asaf Karagila Mar 14 '14 at 16:46
• $B_A$ = {${p_0,p_0*p_1,p_0*p_1*p_2,...}$} – Amitayu Banerjee Mar 14 '14 at 16:53
• I asked for a clarification, and you just rewrote the same thing that was unclear. – Asaf Karagila Mar 14 '14 at 17:02
• I still don't understand how do you define $B_A$?? – Asaf Karagila Mar 14 '14 at 18:17
• Given an infinite subset of primes, A = {$p_0,p_1,...$}, I am forming a set $B_A$, the element $p_i$ in A is mapped to ($p_0*p_1*...*p_i$) in B. Can you please elaborate the problem ? – Amitayu Banerjee Mar 14 '14 at 18:35

Another Solution: Since $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ are equivalent, and cardinality of $\mathbb{N}$ is $\omega$, we can work on $\mathbb{N}\times\mathbb{N}$. Fix a $m \in \mathbb{R}$, let $A_m$ be the set of points $(x,y) \in \mathbb{N}\times\mathbb{N}$, that are of distance $\leq 1$ from the line $y = mx$. $A_m$ is infinite as it has a point on every vertical line x=k (k=0,1,2,...) and for any 2 lines $y = mx$ and $y= m_1x$ for $m \neq m_1$ , there are only finite number of points lying of distance $\leq$ 1 from both, so, $\ A_m\cap \ A_{m_1}$ is finite. These sets $\ A_m$ are the required subsets which are continuum many.

Pick $S = \cup_{n \in \mathbb N} \{0;1\}^{\{1;2;\ldots;n\}}$, the set of functions from a finite set to $\{0;1\}$. For any function $f : \mathbb N \to \{0;1\}$, let $g(f) = \{ f|_{\{1;2;\ldots;n\}}, n \in \mathbb N\}$ : $g(f)$ is the subset of $S$ containing all the restrictions of $f$.

Then the set $\{g(f), f : \mathbb N \to \{0;1\}\}$ is an uncountable subset of $\mathcal P(S)$ (because $g$ is injective and there are uncountably many functions $f$) where any two distinct subsets have finite intersection (if $f_1$ and $f_2$ are distinct, they disagree at some integer $n$, from which all their restrictions are different).

Also, this is almost the same as picking $S$ as the set of finite subsets of $\mathbb{N}$, and $g : \mathcal P(\mathbb N) \to \mathcal P( S)$ the injection given by $g(X) = \{X \cap \{1 ; 2 ; \ldots n \}, n \in \mathbb N\}$.

For each $t\in(\frac1{10},1),$ take its decimal expansion (or one of its decimal expansions in case of non-uniqueness), and let $A_t$ be the set of all finite truncations of that decimal expansion, considered as positive integers. For example, $$A_{\frac13}=\{3,\ 33,\ 333,\ 3333,\ 33333,\dots\},$$ $$A_{\pi-3}=\{1,\ 14,\ 141,\ 1415,\ 14159,\dots\}.$$