Lindenbaum algebra complete/incomplete Consider the Lindenbaum algebra for classical logic (defined below). I'm pretty sure that this algebra is not complete, but I can't seem to prove it or even find a proof of it. I can't even seem to find a statement of the incompleteness or completeness of this algebra! Any completeness/incompleteness proof would be appreciated.
Here's the definition. Take a propositional language with countably many propositional variables, and connectives ∧, ∨, and ~. Say that two sentences φ and ψ are equivalent iff the formula (φ ↔ ψ) is a theorem of classical logic. Write |φ| for the equivalence class containing φ. And let A be the set of all equivalences classes. Define ∧, ∨ and ~ on A as follows: |φ| ∧ |ψ| = |φ ∧ ψ|, |φ| ∨ |ψ| = |φ ∨ ψ|, and ~|φ| = |~φ|. (These are well-defined.) Also, define 0 = |P ∨ ~P| and 1 = |P ∧ ~P|, where P is any propositional variable. The Lindenbaum algebra for classical logic is (A, ∧, ∨, ~, 0, 1).
 A: Summary: A "back-and-forth" argument shows that any two countable atomless Boolean algebras (CABAs) are isomorphic. It's not hard to show that the algebra in the OP is a CABA, and so we can make things easier by shifting attention to a simpler to understand CABA. In particular, the algebra of clopen subsets of the Cantor set is clearly an incomplete atomless Boolean algebra, and it's not hard to show that it's countable as well, so we're done.

First step.
Let $\mathbf B(X)$ denote the free Boolean algebra generated by an infinite set $X$.
Let us see that $|\mathbf B(X)|=|X|$.
We can obtain $\mathbf B(X)$ inductively like so: let
$$B_0=X\cup\{0,1\},$$
and once defined $B_n$, define
$$B_{n+1} = B_n \cup \{x':x\in B_n\} \cup \{x\wedge y, x\vee y : x,y \in B_n\}.$$
Then $$\mathbf B(X) = \bigcup_{n\in \omega}B_n.$$
It is clear that $|B_n|=|X|$ for all $n \in \omega$, and so, $|\mathbf B(X)|=|X|$.
In particular, $\mathbf B(\omega)$, the countably generated free Boolean algebra, is countable.

Second step.
Let us see that $\mathbf B(\omega)$ is atomless.
Each element in $\mathbf B(\omega)$ is a term in the language of Boolean algebra whose variables is a finite subset $\{x_1, \ldots, x_n\}$ of $\omega$ (here, we identify $\omega$ with $X=(x_i)_{i\in\omega}$, so we don't have numbers as variables).
So suppose $t(x_1, \ldots, x_n)\neq 0$ (which just means there is no $x\wedge x'$ in a disjunctive normal form of $t$), and let $y \in X$ be a variable not occurring in $t$.
Thus, $$0 < t(x_1,\ldots,x_n)\wedge y < t(x_1, \ldots, x_n),$$
and so $t(x_1, \ldots, x_n)$ isn't an atom.
Since $t$ was any term, $\mathbf B(\omega)$ has no atoms.

Third step.
Let us see that any two countable atomless Boolean algebras are isomorphic.
Suppose that $\mathbf A$ and $\mathbf B$ are two atomless countable Boolean algebras.
We can enumerate its elements
$$A = \{a_n:n\in\omega\}
\quad\text{and}\quad
B = \{b_n:n\in\omega\}.$$
We will construct two sequences of finite algebras
$$\mathbf A_0 \leq \mathbf A_1 \leq \cdots \leq\mathbf A
\quad\text{and}\quad
\mathbf B_0 \leq \mathbf B_1 \leq \cdots \leq\mathbf B,$$
and isomorphisms $\alpha_n:\mathbf A_n\to\mathbf B_n$ such that $\alpha_n$ extends $\alpha_{n-1}$ and
$$A = \bigcup_{n\in\omega}A_n,
\qquad
B = \bigcup_{n\in\omega}B_n,
\quad\text{and}\quad
\alpha = \bigcup_{n\in\omega}\alpha_n,
$$
so that $\alpha:\mathbf A\to\mathbf B$ is an isomorphism.
Start by making $A_0=\{a_0,a_1\}=\{0,1\}$ and $B_0=\{b_0,b_1\}=\{0,1\}$, and $\alpha(0)=0$ and $\alpha(1)=1$.
Now, for $n$ even suppose we already have $\mathbf A_n, \mathbf B_n$ and $\alpha_n$.
We know $\mathbf A_n$ is generated by its atoms $p_0, \ldots, p_k$ and $\mathbf B_n$ by its atoms, $\alpha_n(p_0), \ldots, \alpha_n(p_k)$.
Let $u$ be the first element of $A = \{a_0, a_1, \ldots\}$ such that $u \notin A_n$ and define $\mathbf A_{n+1} = \langle A_n \cup \{u\} \rangle_{\mathbf A}$ (that is, the least subalgebra of $\mathbf A$ containing $A_n \cup \{u\}$).
Each element of $A_{n+1}$ can be written as a join of elements from
$$U=\{ p_0\wedge u, p_0\wedge u', \ldots, p_k\wedge u, p_k\wedge u' \},$$
whence $\mathbf A_{n+1}$ is finite and its atoms are the nonzero elements of $U$.
For $0\leq i\leq k$, let $u_i=u\wedge p_i$ and let $b_i \in B$ such that
\begin{align}
u_i=0 &\implies v_i=0\\
u_i=p_i &\implies v_i=\alpha_n(p_i)\\
0<u_i<p_i &\implies 0<v_i<\alpha_n(p_i),
\end{align}
where, in the last case, the choice can be thus made because $\mathbf B$ has no atoms.
Let $v=\bigvee_{i=0}^k v_i$ e $\mathbf B_{n+1} = \langle B_n \cup \{v\} \rangle_{\mathbf B}$.
Every element of $B_{n+1}$ can be written as a join of elements from
$$V= \{ \alpha_n(p_0)\wedge v, \alpha_n(p_0)\wedge v', \ldots, \alpha_n(p_k)\wedge v, \alpha_n(p_k)\wedge v' \},$$
whence $\mathbf B_{n+1}$ is finite and its atoms are the nonzero elements from $V$.
Now we extend $\alpha_n$ to $\alpha_{n+1}$ by making $\alpha_{n+1}(u)=v$.
If $n$ is odd we do the same, but starting by defining $\mathbf B_{n+1}$, and then $\mathbf A_{n+1}$ and $\alpha_{n+1}$.

Fourth step.
Find a countable atomless Boolean algebra.
By the three steps before, that algebra is, up to isomorphism, the free countable Boolean algebra.
Let $\mathcal C$ be the Cantor set, as a subset of $[0,1]$, and inheriting its topology.
(Here I'll use Stone representation of Boolean algebras.).
It's not difficult to see that $\mathcal C$ is a Boolean space.
Indeed, it has a base of clopen sets given by
$$\{ [0,v], [u,v], [u,1] : u,v\in[0,1]\setminus\mathcal C \}.$$
It's compact because it's the complement in $[0,1]$ of the open set which is the union of the middle third sets removed from the interval.
It's Hausdorff because if $a<b$ in $\mathcal C$, then taking $u \notin \mathcal C$ such that $a<u<b$, we have $a \in [0,u]$, which is clopen, and $b \notin [0,u]$.
Let $\mathbf B(\mathcal C)$ denote the Boolean algebra of the clopen subsets of $\mathcal C$.
Clearly, it is atomless, for if $[u,v]$ is a clopen subset as described above, then there exist $u',v' \in [0,1]\setminus\mathcal C$ such that $u<u'<v'<v$, and thus, $\varnothing < [u',v'] < [u,v]$.
It's also countable.
Notice that the closed intervals that are kept in the construction of $\mathcal C$ are clopen:
$$[0,1/3],\, [2/3,1],\, [0,1/9], [2/9,1/3], \ldots$$
For example, $[0,1/3]\cap \mathcal C$ is closed because $[0,1/3]$ is closed in $[0,1]$; but also $[0,1/3]\cap \mathcal C = [0,1/2)\cap\mathcal C$, and so it's open.
The same reasoning applies to the others.
Moreover, these sets form a basis for the topology of $\mathcal C$.
Indeed, if $c_1<c_2 \in \mathcal C$, then there is one such interval $I$ with $c_1 \in I$ and $c_2 \notin I$ (just get to an iteration $n$, in the construction of $\mathcal C$, such that $c_2-c_1>(1/3)^{n+1}$).
It is clear that the above basis is countable (it's a countable union of finitely many sets).
Now, let $A$ be any clopen set of $\mathcal C$.
Then, there is a family $\{B_i:i\in I\}$ contained in the above defined clopen basis such that
$$A=\bigcup_{i\in I}B_i.$$
All of these sets are clopen.
In particular, $A$ is closed and each of the $B_i$ is open.
By compactness, there is $J\subseteq I$, finite such that
$$A=\bigcup_{i\in J}B_i.$$
Thus, each clopen subset of $\mathcal C$ is a finite union of elements from a countable family, and so there are only countably many clopen subsets.
We conclude that $\mathbf B(\mathcal C) \cong \mathbf B(\omega)$.

Fifth step.
Show that $\mathbf B(\mathcal C)$ is not complete.
For $c \in \mathcal C$, the set $[0,c)$ is open, but is not closed, because it's complement $[c,1]$ is not open.
However,
$$[0,c) = \bigcup_{c>x\notin \mathcal C}[0,x],$$
and these are clopen sets (members of $\mathbf B(\mathcal C)$);
the upper bounds of $[0,c)$ in $\mathbf B(\mathcal C)$ are the sets of the shape $[0,x]$ with $c<x \notin \mathcal C$, whose intersection is $[0,c)$, again, not clopen.
Hence there is no least upper bound, in $\mathbf B(\mathcal C)$, of the family of its elements
$$\{ [0,x] : 0 > c > x \notin \mathcal C \},$$
and therefore, $\mathbf B(\mathcal C)$ is not complete.
