Saturation of infinite complete Boolean algebra is a regular cardinal

Similar question existed here. However there are still many gaps for stupid persons like me.

A Boolean algebra $$B$$ is called $$\kappa$$-saturated if there is no antichain with supremum $$1$$ (also called partition) of size $$\kappa$$. Let $$\mathrm{sat}(B)$$ to be the least $$\kappa$$ such that $$B$$ is $$\kappa$$-saturated.

Refering to Jech's Set Theory Theorem7.15:

If $$B$$ is an infinite complete Boolean algebra, then $$\mathrm{sat}(B)$$ is a regular uncountable cardinal.

The proof is as follows: Let $$\kappa:=\mathrm{sat}(B)$$. Assume $$\kappa$$ is singular and constuct a partition of size $$\kappa$$ to obtain a contradiction.

Use notation $$\mathrm{sat}(u):=\mathrm{sat}\{v\mid v\le u\}$$ and call $$u\in B$$ stable if $$\mathrm{sat}(v)=\mathrm{sat}(u)$$ for any nonzero $$v\le u$$. Then all stable elements form a dense set $$S$$ : otherwise there will be infinitely descending sequence $$u_0>u_1>\cdots$$ with strictly decreasing saturation, which leads to a contradiction with well-order of Ord. Now one can use Zorn's lemma to choose a maximal set $$T\subset S$$ with elements pairwise disjoint. Then $$T$$ is a partition of $$B$$ and $$|T|<\kappa$$.

Then show that $$\sup\{\mathrm{sat}(u)\mid u\in T\}=\kappa$$. Suffice it to prove for regular $$\lambda<\kappa$$ such that $$\lambda>|T|$$ and partition $$W$$ of $$B$$ of size $$\lambda$$, there is at least one $$u\in T$$ is partitioned by $$W$$ into $$\lambda$$ pieces. Then balabala....

My question is:

1. Why $$T$$ is a partition of $$B$$? It's an antichain actually but why $$\sup T=1$$?

2. Why there is at least one $$u\in T$$ which is "partitioned by $$W$$ into $$\lambda$$ pieces"?

The second question is also asked in previous post. But the given hint is also ambiguous for me to understand.

Thanks in advance and forgive my poor English please.

I've noticed that $$S$$ forms a dense subalgebra so it's regular, hence its maximal antichain $$T$$ is actually also a partition of $$B$$.

But it still confuses me for Question 2.

• Since the set is dense, its supremum is $1$, so any maximal antichain in $T$ is a partition. Commented Apr 10 at 22:43
• @AsafKaragila Thanks, I've learned about that. How about the second question? I couldn't follow the hint of the quoted answer. Commented Apr 11 at 13:12

If $$\sup T<1$$, then let $$u$$ be the complement of $$\sup T$$. Since $$u>0$$, by density of $$T$$ there exists $$v\in T$$ such that $$0. But this is a contradiction, since $$u$$ is disjoint from every element of $$T$$.
Now if $$W$$ is any other partition, for each $$w\in W$$, there exists some $$t\in T$$ such that $$w\wedge t>0$$ (since $$w=w\wedge \sup T=\sup_{t\in T} w\wedge t$$.) So writing $$W_t=\{w\in W:w\wedge t>0\}$$, we have $$W=\bigcup_{t\in T}W_t$$. If $$|W|=\lambda$$ is regular, this implies either $$|T|\geq\lambda$$ or $$|W_t|=\lambda$$ for some $$t$$, i.e. $$t$$ is partitioned by $$W$$ into $$\lambda$$ pieces.