$\kappa$-saturation implies $\kappa$-chain condition in a complete Boolean algebra

Question

First of all, I am aware of this question. However, I am having difficulties with the other direction of this equivalence. But to make sure I understand all correctly let me prove both directions. So let $B$ be a complete Boolean algebra in the signature $\langle +, \cdot, -, 0_B, 1_B \rangle$. Also $u - v$ stands for $u \cdot (- v)$.

$\kappa$-chain condition $\Rightarrow$ $\kappa$-saturation.

Assume that $\kappa$-saturation fails in $B$. That means we have a $\kappa$-partition $W$ of $B$, that is, $$\left(\forall u, v \in W: u \neq v \rightarrow u\cdot v = 0_B\right) \& \left(\sum_{u \in W}u = 1_B\right) \& \left(|W| = \kappa\right).$$

Let $(u_\alpha)_{\alpha < \kappa}$ be a well-ordering of $W$. Define $v_\alpha = \sum_{\beta \geqslant \alpha} u_\beta$. Clearly, $(v_\alpha)_{\alpha < \kappa}$ is non-ascending sequence. We need to show that it is actually descending. Assume that for some $\alpha < \kappa$ we have $v_\alpha = v_{\alpha + 1}$. That is $v_{\alpha + 1} + u_\alpha = v_{\alpha + 1}$, so (we are using infinite distributivity in the middle)$$u_\alpha = u_\alpha \cdot (v_{\alpha + 1} + u_\alpha) = u_\alpha \cdot v_{\alpha + 1} = u_\alpha \cdot \sum_{\beta \geqslant \alpha + 1} u_b = \sum_{\beta \geqslant \alpha + 1} u_\alpha \cdot u_\beta = \sum_{\beta \geqslant \alpha + 1} 0_B = 0_B,$$ a contradiction. Then $\forall \alpha < \kappa : v_{\alpha} > v_{\alpha + 1}$ and $(v_\alpha)_{\alpha < \kappa}$ is a descending $\kappa$-sequence in $B$. Hence $\kappa$-chain condition implies $\kappa$-saturation.

Question 1. Is this proof valid?

$\kappa$-saturation $\Rightarrow$ $\kappa$-chain condition.

Assume that $\kappa$-chain condition fails in $B$. That means we have a descending $\kappa$-sequence $(v_\alpha)_{\alpha < \kappa}$. Consider the following sequence $u_\alpha = v_\alpha - v_{\alpha + 1}, \alpha < \kappa$. I am trying to show that the set $W = \{u_\alpha \mid \alpha < \kappa\}$ is a $\kappa$-partition of $B$. So assume that $\alpha < \beta < \kappa$, then we have $v_{\beta + 1} < v_\beta \leqslant v_{\alpha + 1} < v_\alpha$. In particular, $v_\beta = v_{\alpha + 1} \cdot v_\beta$. Hence we have $$u_{\alpha} \cdot u_{\beta} = (v_\alpha - v_{\alpha + 1}) \cdot (v_\beta - v_{\beta + 1}) = v_\alpha \cdot (-v_{\alpha + 1}) \cdot v_\beta \cdot (-v_{\beta + 1}) =\\ = v_\alpha \cdot (-v_{\alpha + 1}) \cdot v_{\alpha + 1} \cdot v_\beta \cdot (-v_{\beta + 1}) = 0_B.$$

So $W$ is an antichain of size $\kappa$. It is left to show that $W$ is a partition of $B$, that is, $\sum_{\alpha < \kappa} u_\alpha = 1_B$. This is where I have difficulties. I looked up this question. It is very close to what I'm proving, however, I can't understand how to apply it, so I'm trying to use the same argument. We may assume that $v_0 = 1_B$. Consider the following identity $$v_\beta + \sum_{\alpha < \beta} u_\alpha = 1_B.$$ It holds for $\beta = 0$. If $\beta = \gamma + 1$ then $u_\gamma = v_\gamma - v_{\gamma + 1} = v_\gamma - v_\beta$ and $$v_\beta + \sum_{\alpha < \beta} u_\alpha = v_\beta + u_\gamma + \sum_{\alpha < \gamma} u_\alpha = v_\beta + (v_\gamma - v_\beta) + \sum_{\alpha < \gamma} u_\alpha =\\ = v_\beta + v_\gamma + \sum_{\alpha < \gamma} u_\alpha = v_\beta + 1_B = 1_B.$$ However, I have no ideas about the limit case, because it is unclear how $v_\beta$ is related to $v_\alpha, \alpha < \beta$ and how to deal with the sum. Also our $\kappa$-chain is not maximal as in the mentioned question. Should we extend it to a maximal one with $v_\beta = \prod_{\alpha < \beta} v_\alpha$?

Question 2. Could you give me some hints on how to prove this part in full detail? This equivalence is stated without proof in Jech's "Set Theory", so may be there is an easier way to prove it? I'm thinking about it for several hours with no success.

Answer 1

The first argument is correct, but I’d organize it just a little differently, first noting that $u_\alpha\cdot v_{\alpha+1}=0_B$ and then writing

$$u_\alpha=u_\alpha\cdot v_{\alpha+1}+u_\alpha\cdot u_\alpha=u_\alpha(v_{\alpha+1}+u_\alpha)=u_\alpha\cdot v_{\alpha+1}=0_B\;.$$

For the other direction you want to make $\langle v_\alpha:\alpha<\kappa\rangle$ continuous at limits: for $\alpha\le\kappa$ let

$$w_\alpha=\prod_{\xi<\alpha}v_\xi\;.$$

(Note that this automatically makes $w_0=1_B$.) It’s easy to check that $\langle w_\alpha:\alpha\le\kappa\rangle$ is decreasing. It may not be strictly decreasing, but this happens precisely at limit $\beta$ at which $\langle v_\alpha:\alpha<\kappa\rangle$ was already continuous. That is, if $\beta$ is a limit ordinal, and $v_\beta=\prod_{\xi<\beta}v_\xi$, then $w_{\beta+1}=w_\beta$.

Now let $u_\alpha=w_\alpha-w_{\alpha+1}$ for $\alpha<\kappa$, and let $u_\kappa=w_\kappa$. Let $U=\{u_\alpha:\alpha\le\kappa\}$; then $U\setminus\{0_B\}$ is a partition of $B$. (There are two possible sources of $0_B$ in $U$: because the $u_\alpha$ may not be strictly decreasing, some $w_\alpha$ may be $0_B$, and $w_\kappa$ may be $0_B$.)

You can now handle the limit cases in your argument. Suppose that $\beta$ is a limit; then

$$\begin{align*} w_\beta+\sum_{\alpha<\beta}u_\alpha&=\prod_{\xi<\beta}w_\xi+\sum_{\alpha<\beta}u_\alpha\\ &=\prod_{\xi<\beta}\left(w_\xi+\sum_{\alpha<\beta}u_\alpha\right)\\ &=1_B\;. \end{align*}$$

Answer 2

Here is a short proof for the second part.

As you have already shown, $W = \{u_\alpha \mid \alpha < \kappa\}$ is an antichain of size $\kappa$. If $\sum W=1$, we are done. Otherwise, let $b=\sum W$, which is less than $1$, so $-b\ne 0$. Then $u_\alpha\le b$ for all $\alpha$, each $u_\alpha$ is disjoint from $-b$ and $\sum(W\cup\{-b\})=1$. In other words $W\cup\{-b\}$ is a partition of $B$ (of size $\kappa$).