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Suppose A is a family of subsets of R with the property that the intersection of any two sets in A is finite. Show that $|A|\leq 2^{\aleph_0}$.

I was told that choosing a countable $D \subset B$ for all $B \in A$ would be helpful. I'm just really not sure where to go with this. Any hints would be appreciated!

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Here’s an argument that’s a bit closer to the hint. Suppose that $|A|>2^{\aleph_0}$. $\Bbb R$ has only $2^{\aleph_0}$ finite subsets, so without loss of generality we may assume that every member of $A$ is infinite. For each $B\in A$ let $C_B$ be a countably infinite subset of $A$. $\Bbb R$ has $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$ countably infinite subsets, so there must be distinct $B,D\in A$ such that $C_B=C_D$ and hence $B\cap D$ is not finite. In fact there must be some countably infinite $C\subseteq\Bbb R$ such that

$$|\{B\in A:C_B=C\}|>2^{\aleph_0}\;.$$

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Hint: Prove that the set $[\Bbb R]^{<\omega}$, the set of all finite subsets of $\Bbb R$, has size $2^{\aleph_0}$. Fix $C_0\in A$, then prove that for each $C\in [\Bbb R]^{<\omega}$, the set $A_C:=\{B\in A:B\cap C_0=C \}$ has size $\leq2^{\aleph_0}$. Notice that $\{A_C:C\in [\Bbb R]^{<\omega}\}$ is a partition of $A-\{C_0\}$, and thus $|A|\leq 2^{\aleph_0}$.

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