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Let $B$ be a finite boolean algebra.

Define for $a,b\in B$ $a\leq b$ if $ab=a$

If $x\in B$ and $a_1,\dots,a_k$ are the atoms of B (e.g. $a\neq 0$ and if $b\in B$ such that $0\leq b \leq a$ then $b=0$ or $b=a$) such that $\forall i$ $a_i\leq x$

Could anyone help me prove that $x=a_1+\dots +a_k$

So far I have

$(a_1+\dots +a_k)^2=a_1+\dots +a_k=a_1x+\dots +a_kx=(a_1+\dots +a_k)x$

Therefore $a_1+\dots +a_k\neq 0 \Rightarrow x=a_1+\dots +a_k$

Then I went to consider the case of $a_1+\dots +a_k= 0$ to show that $x=0$. Could anyone help with this second case? Or if i'm completely wrong help show me the correct way?

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    @MichaelHardy I am saying B with the partial order defined as $a,b\in B$ $a\leq b$ if $ab=a$2011-09-26

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Addressing Michael Hardy's comment, note that the Boolean algebra is finite and hence atomic.

Let $a_1,\dots,a_k$ be the atoms below $x$. Note that since the BA is finite, there are only finitely many atoms below $x$. We have $a_1+\dots+a_k\leq x$. Now consider $b=x-(a_1+\dots+a_k)$. If $b=0$, then $x=a_1+\dots+a_k$. Otherwise there is an atom $a\leq b$. Now $a$ is disjoint and in particular different from $a_1,\dots,a_k$. But $a\leq b\leq x$, contradicting the fact that $a_1,\dots,a_k$ lists all atoms below $x$.

Does this help? Adressing your question more specifically, if $x=0$, then there is no atom below $x$ and the sum taken over an empty subset of a BA is usually defined to be $0$. This also follows automatically if you define the sum over a subset of a BA to be the least common upper bound of the set.

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    Thank you, this clears everything up :)2011-09-29