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.$\Box$