Countable set having uncountably many infinite subsets

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

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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

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.

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 @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 @Mark: Good idea. Thanks! – Asaf Karagila Jun 24 '12 at 17:13

Consider the directed graph with vertex set $\mathbb{N}$ and edges $(n, 2n)$, $(n, 2n + 1)$ for all $n$. Then infinite paths starting at node 1 in the graph can be considered as subsets of $\mathbb{N}$ and satisfy the condition.

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+1. The graph is a binary tree rooted at node 1 and one can consider injective infinite paths starting at node 1. – Did Jun 24 '12 at 13:59
The question asks the subsets $N_\alpha$ to be indexed by $\mathbb{R}$, perhaps you could mention a bit about how to do this in your answer. – nullUser Jun 24 '12 at 14:00
You reasoning sounds really nice. Perhaps you could enrich it with a little more detail. For example, how do you know that the set of all infinite paths has the required cardinality? I guess that it is easy to construct a bijection to $\{0,1\}^\mathbb{N}$, because at each site $n$, you ca walk to $2n+0$ or $2n+1$. Also, you could elaborate on the reason you believe the intersections are finite. I guess that a good explanation raises in the fact that the graph is acyclic (a tree). – André Caldas Feb 24 at 19:37
@AndréCaldas (and nullUser, Did) Thank you for the comments. Back then I didn't have time enrich the answer, however I guess your explanations does it :) – Serkan Feb 24 at 20:39
I will edit your answer and give it a [+1] if you don't mind. – André Caldas Feb 24 at 23:46
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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 $m_\alpha$.

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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 $g(X) = \{X \cap \{1 ; 2 ; \ldots n \}, n \in \mathbb N\}$.

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