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Let $B$ be an infinite, complete Boolean algebra, and let $\kappa = \operatorname{sat}(B)$. I would like to show that $\kappa$ is uncountable. If we suppose $\kappa$ is countable, that is to say $\kappa = \aleph_0$, then there must exist no descending countable chains in $B$. This means that we can only construct finite descending chains $u_0 > \ldots > u_n$ in $B$.

I think that this tells us $B$ must be finite, but I can't quite see why. Is it to do with the fact that we can construct a partition $W_n$ of $B$ for each $0<n < \aleph_0$, by definition of saturation?

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up vote 4 down vote accepted

Suppose that $\mathbb{B}$ has only finite chains, and that $$0=b_0\lt b_1\lt \cdots\lt b_n=1$$ is a given finite chain. If any of the differences $b_{k+1}-b_k$ is not an atom in $\mathbb{B}$, then we may extend this chain to a strictly longer chain, by inserting an additional node between $b_k$ and $b_{k+1}$, namely, the node $b_k\vee a$, where $0\lt a\lt (b_{k+1}-b_k)$.

Thus, if there is no infinite chain, we must eventually arrive at a finite chain as above, with all the differences $b_{k+1}-b_k$ atoms in $\mathbb{B}$. In this case, $\mathbb{B}$ has a maximal antichain consisting of those atoms, since the differences add up to $1$. It follows that $\mathbb{B}$ is just the power set of $n$ atoms, and has size $2^n$.

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I think I've worked it out. Define $B_u = \{ v \in B \mid v \le u \}$.

Let $u_0 \in B$ be any non-zero, non-unit element. Then one of $B_{u_0}, B_{-u_0}$ must be infinite. Otherwise, say $B_{u_0} = \{ v_1, \ldots v_m \}$, $B_{-u_0} = \{v'_1 , \ldots v'_{m'}\}$. Then given $v \in B$, $v = v\cdot u_0 + v \cdot{-u_0} = v_t + v'_{t'}$ for some $v_t \in B_{u_0}$ and $v'_{t'} \in B_{-u_0}$. Hence every element can be written in this way and $|B| = mm' < \aleph_0$, which is impossible! Choose $w_0$ to be whichever out of $u_0, -u_0$ has $B_{w_0}$ infinite.

Since $B_{w_0}$ is an infinite Boolean algebra with $1 = w_0$ we just repeat the process above to generate a countable decreasing sequence $w_0 > w_1 > \ldots $ which is still in $B$.

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