# Atomic Boolean lattice is weakly atomic

The book "Introduction to Lattices and Order" says at Exercise 10.12 that an atomic Boolean lattice is weakly atomic.

Could you tell me why it holds?

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Let $L$ be our atomic Boolean lattice, and suppose $x, y \in L$ with $x < y$. We have to show that there exist $a,b \in L$ such that $x \leq a \prec b \leq y$.
Let $x'$ denote the complement of $x$, and consider the homomorphism \begin{align*} \downarrow x' &\longrightarrow \uparrow x, \\ c &\longmapsto c \vee x \end{align*} which is easily seen to be an isomorphism of lattices with inverse $d \mapsto d \wedge x'$. Now $\downarrow x'$ inherits atomicity from $L$, so $\uparrow x$ is atomic as well. Since $x < y$, there exists an $a \in \uparrow x$ such that $x \prec a \leq y$.
I take it that $\prec$ is the covering (immediate successor) relation? – Brian M. Scott Feb 6 '13 at 23:24
Yes. I use the notation of the book referred to in the question, so $\prec$ is the covering relation, $\downarrow x = \{ a \in L : a \leq x \}$ and similarly for $\uparrow x$. – marlu Feb 6 '13 at 23:26