TITLE: A proof of maximality of an antichain in a complete Boolean algebra
QUESTION [1 upvotes]: Suppose $(\mathbf{B},\leqslant)$ is a complete Boolean algebra and let $|\mathbf{B}|=|\gamma|$. Let $C=\langle c_\alpha\mid \alpha<\gamma\rangle$ be a maximal descending chain without a lower bound: $\bigwedge_{\alpha<\gamma}c_\alpha=\mathbf{0}$. Could you please help me to prove the following two things?

For every limit ordinal $\beta<\gamma$, $c_\beta=\bigwedge_{\alpha<\beta}c_\alpha$
$A=\{c_\alpha-c_{\alpha+1}\mid\alpha<\gamma\}$ is a maximal antichain in $\mathbf{B}$.

REPLY [1 votes]: HINT: For the first part, if $c_\beta\ne\bigwedge_{\alpha<\beta}c_\alpha$, then $c_\beta<\bigwedge_{\alpha<\beta}c_\alpha$, and $C\cup\left\{\bigwedge_{\alpha<\beta}c_\alpha\right\}$ is a strictly larger descending chain with no minimum element.
For the second part you must prove two things: that $A$ is an antichain, and that it is maximal. For $\alpha<\gamma$ let $a_\alpha=c_\alpha-c_{\alpha+1}$. Suppose that $\alpha<\beta<\gamma$; you want to show that $a_\alpha\land a_\beta=\mathbf{0}$. To get an intuition for why this must be true (and how to prove it), imagine for a moment that $\mathbf{B}$ is a power set algebra; then $c_\alpha\supsetneqq c_{\alpha+1}\supseteq c_\beta\supsetneqq c_{\beta+1}$, so $a_\beta=c_\beta\setminus c_{\beta+1}\subseteq c_\beta\subseteq c_{\alpha+1}$, while $a_\alpha=c_\alpha\setminus c_{\alpha+1}$ is disjoint from $c_{\alpha+1}$ and hence from $a_\beta$. Now just translate this into abstract Boolean algebra terms.
For maximality, show first that $\bigvee_{\alpha<\gamma}a_\alpha=\mathbf{1}$. Then show that for each non-$\mathbf{0}$ element $b\in\mathbf{B}$, there is an $\alpha<\gamma$ such that $b\land a_\alpha\ne\mathbf{0}$, and conclude that either $A\cup\{b\}=A$, or  is $A\cup\{b\}=A$ is not an antichain.
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