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Let $A$ be a Boolean algebra, and let $A^+$ denote the set of non-zero elements of $A$. A cut $U \subseteq A^+$ is a set such that if $q\in U$, then $p\le q \implies p \in U$, for all $p \in A^+$. A cut $U$ is regular if whenever non-zero $p \not\in U$, there exists $q \le p$ such that $U_q \cap U = \emptyset$, where $U_q := \{ x \in A^+ \mid x \le q\}$.

Since $A^+$ is a regular cut, and any intersection of regular cuts is regular, it follows that any cut $U$ is contained in a least regular cut $\overline{U}$. I want to show that $U_{p+q} = \overline{U_p \cup U_q}$. One direction is obvious, since clearly $U_p, U_q \subseteq U_{p+q} \implies U_p \cup U_q \subseteq U_{p+q} \implies \overline{U_p \cup U_q} \subseteq U_{p+q}$ by leastness. But I am having trouble showing inclusion in the other direction. Any help would be appreciated.

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up vote 2 down vote accepted

Here is one way to see it, obtained after a night's sleep.

Any cut $U$ has the property that $\overline{U} =\{ p \in A^+ \mid (\forall q \le p) \ U_q \cap U \not= \emptyset\}$. This means that $$\overline{U_p \cup U_q} = \{ r \in A^+ \mid (\forall s \le r) \ U_s \cap (U_p \cup U_q ) \not= \emptyset \}.$$

Let $x \in U_{p+q}$, so that $x \le p+q$. Then let $y \le x \le p+q$. We must show that $U_y \cap (U_p \cup U_q) \not= \emptyset$. If $y \in U_q$, then we're done. So suppose $y \not\in U_q$.

It follows that $y\cdot p \not= 0$. This is because $y \le p+q \iff y \cdot (p + q) = y = y\cdot p + y \cdot q $ and so if $y \cdot p =0$, this tells us that $y = y\cdot q \iff y \le q$ which is impossible.

Hence $y \cdot p \in U_y \cap U_p \subseteq U_y \cap (U_p \cup U_q)$, and we're done again.

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Looks okay to me; I think that you could go ahead and accept it. – Brian M. Scott Apr 25 '12 at 19:20
I have to wait 4 hours. Thanks for the confirmation. – Paul Slevin Apr 25 '12 at 21:14

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