Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ${}^\omega E$.

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
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
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.