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Let $B$ be a complete Boolean algebra. Define 3 subsets of B as follows:

$B_I:= \{ u_{0,i} \mid i \in I \}$

$B_J := \{ u_{1,j} \mid j \in J \}$

$B_{I \times J} := \{ u_{0,i} \cdot u_{1,j} | (i,j) \in I \times J\}$

where $I,J$ are just some indexing sets. I am trying to prove that $\operatorname{sup}B_{I\times J} = \operatorname{sup}B_I \cdot \operatorname{sup} B_J$. (Jech Chapter 7)

It is easy to see that $\operatorname{sup}B_I\cdot\operatorname{sup}B_J$ is an upper bound for $B_{I \times J}$, but I am finding it very hard to show that if $v$ is another upper bound for $B_{I \times J}$ then $\operatorname{sup}B_I\cdot\operatorname{sup}B_J \le v$.

Any help would be appreciated.

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up vote 2 down vote accepted

We use another, simpler, distributive law that holds in all complete Boolean algebras: $$x \cdot \sum_i a_i = \sum_i (x \cdot a_i)$$ which is exercise 17.15(a) in my copy of Jech.

This first law can be shown as follows: For any fixed $i$ we have that $a_i \le \sum_i a_i$, hence $x \cdot a_i \le x \cdot \sum_i a_i$, which makes the right hand side an upper bound for all $x \cdot a_i$, so $\sum_i (x \cdot a_i) \le x \cdot \sum_i a_i$

On the other hand, for all $a$ we have that $a \le x' + a = x' + (a \cdot x)$, and applying this for all $a_i$ we get $\sum_i a_i \le \sum_i (x' + (a_i \cdot x)) \le x' + \sum_i (a_i \cdot x)$, and now we multiply both sides by $x$: $x \cdot (\sum_i a_i) \le x \cdot (x' + \sum_i (a_i \cdot x)) = $ (by finite distributivity!) $ = x\cdot x' + (x \cdot \sum_i (a_i \cdot x)) = x \cdot \sum_i (a_i \cdot x) \le \sum_i (a_i \cdot x)$, which put together gets us the required other inequality.

Now, to the case at hand: for every $i$ we have $$u_{0,i} \cdot B_J = u_{0,i} \cdot \sum_j u_{1,j} = \sum_j (u_{0,i} \cdot u_{1,j}) \le v$$ by applying the above law to the $B_J$ as the infinite sum, and noting that all products in that sum are from $B_{I \times J}$ and thus all bounded by the same $v$ (and so their sum is as well). But as this holds for all $i$: $$\sum_i (u_{0,i} \cdot B_J) \le v$$ and again by applying the above law again, pulling out the $B_J$: $$\sum_i ( u_{0,i} \cdot B_J ) = \sum_i u_{0,i} \cdot B_J = B_I \cdot B_J,$$ finally showing that $B_I \cdot B_J \le v$, as required.

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Thank you. But what is $x'$ ? –  Paul Slevin Feb 10 '12 at 22:05
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The complement of $x$ in the Boolean algebra; I see that Jech uses $-$, which is OK too. –  Henno Brandsma Feb 10 '12 at 22:09
    
Brilliant answer, thanks. –  Paul Slevin Feb 10 '12 at 22:37
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