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Here is a theorem:

Let $E$ be an uncountable infinite set. Then there is a collection $\mathcal{A}$ of subsets of $E$ such that $|\mathcal{A}|=|E|^\omega$, $|A|=\omega$ for each $A\in \mathcal{A}$, and the intersection of any two distinct elements of $\mathcal{A}$ is finite.

How to show it? I hope the proof a little simple and clear, because it is too difficult in my textbook. Thanks.

ADD: Thanks Brain. In my textbook, it said, it suffices to construct the desired collection on the set of all finite subset of $\omega \times E$. For each $f\in {}^\omega E$, let $A(f)=\{(f|n):n<\omega\}$. One can easily check $A(f)\cap A(g)$ is finite whenever $f$ and $g$ are distinct elements of ${}^\omega E$, so $\{A(f): f\in {}^\omega E\}$ is the desired collection of sets. I cannot check this sentence: One can easily check $A(f)\cap A(g)$ is finite whenever $f$ and $g$ are distinct elements of ${}^\omega E$.

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For a regular cardinal $\kappa$, by a result of Ulam, it is a disjoint union of $\kappa$ stationary sets. For singular cardinals, use the same argument but on the cofinality $\kappa$. – Eran Mar 12 '13 at 13:03
up vote 5 down vote accepted

Let $T={}^{<\omega}E=\bigcup_{n\in\omega}{}^nE$, the set of functions from finite ordinals into $E$. Then $\langle T,\subseteq\rangle$ is a tree of height $\omega$. Note that for $s\in{}^mE$ and $t\in{}^nE$, $s\subseteq t$ iff $t\upharpoonright m=s$. Each $\sigma\in{}^\omega E$ defines a branch through $T$; for $\sigma\in{}^\omega E$ let $B_\sigma=\{\sigma\upharpoonright n:n\in\omega\}$, the set of nodes on that branch. Each $B_\sigma$ is a countably infinite subset of $T$. $|T|=\omega\cdot|E|=|E|$, so there is a bijection $\varphi:T\to E$; for each $\sigma\in{}^\omega E$ let $A_\sigma=\varphi[B_\sigma]$, and let $\mathscr{A}=\left\{A_\sigma:\sigma\in{}^\omega E\right\}$. I leave it to you to verify that $\mathscr{A}$ has the desired properties.

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Thanks Brain for your answer. I have a small question in the proof of my textbook. Could you help me? see the added part. – Simple Mar 13 '13 at 5:31
@Simple: If $f,g\in{}^\omega E$ and $f\ne g$, then there is an $n\in\omega$ such that $f(n)\ne g(n)$. Then for every $k>n$ we have $f\upharpoonright k\ne g\upharpoonright k$, since $(f\upharpoonright k)(n)\ne(g\upharpoonright k)(n)$. This means that $A(f)\cap A(g)\subseteq\{f\upharpoonright k:k\le n\}$, which is finite. – Brian M. Scott Mar 13 '13 at 8:10
Maybe i'm not clear to the $A(f)$, however, by the definition, $A(f)$ seems that it is not $\subset E$, it is a collection of some subsets of $E$. Is it right? – Simple Mar 13 '13 at 8:58
@Simple: The proof in your book says right up front that it’s constructing subsets of $\omega\times E$, not of $E$, just as I constructed subsets $B_\sigma$ of $T$, not of $E$. But $|\omega\times E|=|E|$, so there is a bijection $\varphi:\omega\times E\to E$, and the sets $\varphi[A(f)]$, which are subsets of $E$, are still almost disjoint. All of the properties that you want $\mathscr{A}$ to have are preserved by bijections, so you can transfer such a family from one set of a given cardinality to any other set of the same cardinality. – Brian M. Scott Mar 13 '13 at 14:52

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