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Let $\mathbf{B} = \langle B, \wedge, \vee, ', 0, 1 \rangle$ be a Boolean Algebra.
Then $[0,a] = \{ x \in B : 0 \leq x \leq a \}$, and this can be made into a Boolean Algebra $\mathbf{B}_a = \langle [0,a], \wedge, \vee, ^*, 0, a \rangle$, where $x^* = x' \wedge a$.

Suppose $\mathbf{B}_a \cong \mathbf{B}_b$, for some $a,b \in B$ such that $a \wedge b = 0$.
Does it follow that $\mathbf{B}_{a'} \cong \mathbf{B}_{b'}$?
I would like to have an isomorphism $\psi:\mathbf{B}_{a'} \to \mathbf{B}_{b'}$ explicitly defined, and I suppose that if this is true, it can be made such that $\psi(b) = a$ (notice that $b \leq a'$ follows from $a \wedge b = 0$), so that it would extend the inverse of the given isomorphism $\varphi:\mathbf{B}_{a} \to \mathbf{B}_{b}$.

Notice that it is not true if we drop the hypothesis that $a \wedge b = 0$.
For example, if $\mathbf{B}$ is the powerset of $\mathbb{N}$, define $a = \mathbb{N} \setminus \{1\}$ and $b = \mathbb{N}$.
Then $\mathbf{B}_a \cong \mathbf{B}_b = \mathbf{B}$; however, $B_a = \{ \varnothing, \{1\} \}$, and $B_b = \{\varnothing\}$, so they can't give isomorphic Boolean Algebras.

amrsa
  • 12,917

1 Answers1

1

Let us start with two auxiliary results.

  1. If $\mathbf{A}$ and $\mathbf{B}$ are Boolean algebras and $f : A \to B$ is a map, then $f$ is a homomorphism if and only if it is a bounded lattice homomorphism between the bounded lattice reducts of the algebras.
  2. Let $\mathbf{L}$ and $\mathbf{K}$ be lattices and $f:L\to K$ a bijection. Then $f$ is an isomorphism if and only if both $f$ and $f^{-1}$ are isotone if and only if $f$ preserves joins (or equivalently, meets).

Proof:

  1. For $a \in A$, we have $$f(a) \vee f(a') = f(a \vee a') = f(1) = 1,$$ $$f(a) \wedge f(a') = f(a \wedge a') = f(0) = 0,$$ and so $f(a') = (f(a))'$ whence $f$ preserves $'$ and is a Boolean Algebra homomorphism.
  2. For the first equivalence, see, for example, Burris and Sankappanavar, Ch.I, Theorem 2.3. Now suppose that $f$ preserves joins, and let $a, b \in L$ such that $f(a) \leq f(b)$. Then, $$f(a \vee b) = f(a) \vee f(b) = f(b) \Longrightarrow a \vee b = b \Longrightarrow a \leq b,$$ so that $f^{-1}$ is isotone. It is clear that if $f$ preserves joins, then $f$ is isotone. Hence $f$ is an isomorphism.

It follows from 1 and 2 that if a bijection between Boolean Algebras preserves joins, then it is an isomorphism.

Now since $\mathbf{B}_a$ and $\mathbf{B}_b$ are isomorphic, let $\alpha:\mathbf{B}_a \to \mathbf{B}_b$ be an isomorphism and $\beta = \alpha^{-1}$.
Define $\varphi: \mathbf{B}_{a'} \to \mathbf{B}_{b'}$ by $$x \mapsto \beta(x \wedge b) \vee (x \wedge b'),$$ and $\psi: \mathbf{B}_{b'} \to \mathbf{B}_{a'}$ by $$x \mapsto \alpha(x \wedge a) \vee (x \wedge a').$$ For $x \in [0,b']$ we have \begin{align} \varphi( \psi(x) ) &= \varphi( \alpha(x \wedge a) \vee (x \wedge a') )\\ &= \varphi( \alpha(x \wedge a) ) \vee \varphi( x \wedge a' )\\ &= \beta( \alpha(x \wedge a) \wedge b ) \vee ( \alpha(x \wedge a) \wedge b' ) \vee \beta( x \wedge a' \wedge b ) \vee (x \wedge a' \wedge b)\\ &= (\beta( \alpha(x \wedge a) ) \wedge \beta(b)) \vee (\alpha(x \wedge a) \wedge b') \vee 0 \vee (x \wedge a' \wedge b')\tag{$x \wedge b = 0$}\\ &= ((x \wedge a) \wedge a) \vee (\alpha(x \wedge a) \wedge b') \vee (x \wedge a')\tag{$x \leq b'$}\\ &= (x \wedge a) \vee (x \wedge a') \vee ( \alpha(x \wedge a) \wedge b' )\\ &= (x \wedge (a \vee a')) \vee (\alpha(x \wedge a) \wedge b' )\\ &= x \vee (\alpha(x \wedge a) \wedge b' )\\ &= x, \end{align} where $\alpha(x \wedge a) \wedge b' = 0$ because $\alpha(x \wedge a) \leq b$.
Analogously, if $x \leq a'$ then $\psi(\varphi(x)) = x$, and $\varphi, \psi$ are bijections.

Let us see that $\varphi$ preserves joins: \begin{align} \varphi(x \vee y) &= \beta( (x \vee y) \wedge b ) \vee ( (x \vee y) \wedge b' )\\ &= \beta( (x \wedge b) \vee (y \wedge b) ) \vee ( (x \wedge b') \vee (y \wedge b') )\\ &= \beta(x\wedge b) \vee \beta(y \wedge b) \vee (x \wedge b') \vee (y \wedge b')\\ &= \varphi(x) \vee \varphi(y). \end{align} If also follows that $\varphi(b) = a$ and $\varphi(0) = 0$, and therefore, $\varphi$ is an isomorphism.

amrsa
  • 12,917