A Boolean algebra $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is said to be complete if every non-empty subset of $B$ has a greatest lower bound (g.l.b). Prove that a Boolean algebra that is isomorphic to a complete Boolean algebra is complete.
Let us start with a complete Boolean algebra, $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ which is isomorphic to $\mathcal{B'}:=(B',\leq,\lor,\land,^c,0,1)$. Let $h:B\to B'$ be an isomorphism between the two.
For a non-empty subset $X' \subseteq B'$, $\exists X\in B$ such that $h(X)=B$. Is this correct, or does it need more justification? I think this is fine because: for every element $x' \in X'$ we have $x \in X$ such that $h(x) = x'$ since $h$ is an isomorphism. We can construct $X$ such that $h(X)= X'$ by $X = \{x\in B:x = h^{-1}(x') \forall x' \in X'\}$.
Since $\mathcal{B}$ is complete, we know that $X$ has a g.l.b, say $x_g \in X$. This means that for all $x\in X$, $x_g\leq x$. Since isomorphisms between Boolean algebras preserve partial orders, it follows that $h(x_g)\leq h(x)$ for all $x\in X$. Since $X'=\{x'\in B':x'=h(x) \forall x\in X\}$, $h(x_g)$ is in fact the g.l.b of $X'$. Is this fine?
This works for all $X'\subseteq B'$, hence $\mathcal{B}'$ is complete.
I'm new to proof-writing, and I tend to make mistakes or leave out things that need to be stated in order to complete a proof. Hence, it would be helpful if someone could just go through this and let me know if the aforementioned steps are complete, or some gaps need to be filled in? I'd be happy to know about ways to make this proof more rigorous.
Best Answer
What you’ve done is correct, but the fact that $h(x_g)\le x'$ for each $x'\in X'$ isn’t enough to show that $h(x_g)=\bigwedge X'$: it shows that $h(x_g)$ is a lower bound for $X'$, but it doesn’t show that $h(x_g)$ is the greatest lower bound for $X'$. For that you need to verify that $b'\le h(x_g)$ whenever $b'$ is a lower bound for $X'$. But that’s straightforward: if $b'\le x'$ for each $x'\in X'$, then $h^{-1}(b')\le h^{-1}(x')$ for each $x'\in X'$, so $h^{-1}(b')\le x_g$, and therefore $b'\le h(x_g)$.