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  • 陶哲轩实分析习题 8.5.20

    Let $X$ be a set,and let $\Omega\subset 2^X$ be a collection of subsets of $X$.Assume that $\Omega$ does not contain the empty set $\emptyset$”.Using Zorn's lemma,show that there is a subcollection $\Omega'\subseteq\Omega$ such that all the elements of $\Omega'$ are disjoint from each other,but that all the elements of $\Omega$ intersect at least one element of $\Omega'$.

    Proof:

    Obviously,$\emptyset\not\in\Omega$.Consider  the subsets $A$ of $\Omega$, all the elements of $A$ are disjoint from each other,according to Zorn's lemma,there must exists such $A$ that is maximal with respect to the set inclusion.We denote such $A$ by $\Omega'$.Now I prove that all the elements of $\Omega$ intersect at least one element of $\Omega'$.If there exists $x\in\Omega$,$x$ does not intersect with any elements of $\Omega'$,then we add $x$ into $\Omega'$,then $\Omega'\bigcup\{x\}$ is bigger than $\Omega'$,which is a contradiction.$\Box$

    Conversely,if the above claim is true,show that it implies the claims in Exercise 8.4.2


    Proof:

    First let us review  Exercise 8.4.2.

    Let $I$ be a set,and for each $\alpha\in I$,let $X_{\alpha}$ be a non-empty set.Suppose that all the sets $X_{\alpha}$ is are disjoint from each other.Show that there exists a set $Y$ such that $\#(Y\bigcap X_{\alpha})=1$ for all $\alpha\in I$.


    Now let's prove Exercise 8.4.2. Consider the set of all $\{(0,\alpha),(1,x_\alpha)\}$,denoted $M$,where $\alpha\in I$,$x_{\alpha}\in X_\alpha$.Let $\Omega'$ be the maximal subset of $M$ such that every two distinct elements of $\Omega'$ are disjoint.Then it is easy to prove Exercise 8.4.2.$\Box$


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  • 原文地址:https://www.cnblogs.com/yeluqing/p/3827497.html
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