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I am reading Thomas Jech's Axiom of choice in which he gives a concise proof of the Stone Representation theorem using the (Boolean) Prime Ideal theorem. However he states something is trivial, which I am really struggling to see. I shall quote him verbatim:

Stone Representation theorem: Every Boolean algebra is isomorphic to a set algebra. (Set algebra is an algebra on a family of sets with + = union, $\cdot$ = intersection, - = complement)

Proof: Let $B$ be a Boolean algebra, let $$S = \{U:U \textrm{ is an ultrafilter on }B\}$$ for $u\in B$ let $\pi(u) = \{U\in S:u\in U\}$. Then it is easy to see that $$\pi(u+v)=\pi(u)\cup\pi(v)$$ $$\pi(u\cdot v)=\pi(u)\cap\pi(v)$$ $$\pi(-u) = \pi(u)^C$$ Now, I can understand the last line: $U \in \pi(-u)$ iff $-u \in U$ iff $u \notin U$ iff $U \in \pi(u)^C$ (as $U$ an ultrafilter, so xor $u\in U$ or $-u \in U$. But I cannot see how to show the other two lines. Any help?

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Let $U\in S$ and $u,v\in B$. If $u,v\in U$, then $u\cap v\in U$ since $U$ is a filter. Conversely, if $u\cdot v\in U$, then $u\in U$ and $v\in U$, since $u\cdot v\leq u$ and $u\cdot v\leq v$ and $U$ is a filter. Thus $\pi(u\cdot v)=\pi(u)\cap\pi(v)$. The other equation then follows formally: $$\pi(u+v)=\pi(-(-u\cdot-v))=(\pi(u)^C\cap\pi(v)^C)^C=\pi(v)\cup\pi(v).$$

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  • $\begingroup$ In his notes, Jech writes that $u \le v$ iff $u+v = v$. Can you explain why $u\cdot v \le u$? I must be missing something obvious here. $\endgroup$ – Elie Bergman Dec 21 '17 at 21:50
  • $\begingroup$ $u\cdot v+u=u\cdot v+u\cdot 1=u\cdot (v+1)=u\cdot 1=u$. $\endgroup$ – Eric Wofsey Dec 21 '17 at 21:53
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    $\begingroup$ I don't have my copy of Jech's book handy, but I'd expect that, before getting to anything as fancy as ultrafilters and the Stone representation theorem, he'd point out that $u\cdot v$ is the greatest lower bound of $u$ and $v$. That includes the information that $u\cdot v\leq u$. $\endgroup$ – Andreas Blass Dec 21 '17 at 23:20

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