The Amoeba poset has property Knaster.?

Question

The Amoeba poset has property Knaster.?

Amoeba poset $\mathbb{P}=\{p \subseteq \mathbb{R}: p$ $\text{is open} \wedge \mu(p)< \epsilon\}$ for $\epsilon>0$.

Any suggestion. Thanks

Answer 1

You should specify the order on your poset. But with the usual ordering where $p\leq q$ if $p\supseteq q$, yes, amoeba forcing is Knaster.

To see this, fix a countable base for the topology and let $X$ be an uncountable subset of $\mathbb{P}$. By shrinking $X$ a bit we may also assume that there is an $n$ such that $\mu(p)\leq\varepsilon-\frac{1}{n}$ for all $p\in X$.

Now, each $p\in X$ is a countable union of basic opens, so we can associate to each $p$ a $\hat{p}\subseteq p$ which is a finite union of basic opens such that $\mu(p\setminus\hat{p})<\frac{1}{2n}$. But there are only countably many finite unions of basic opens, so there must be an uncountable $Y\subseteq X$ such that $\hat{p}=\hat{q}$ for all $p,q\in Y$.

If we now take any two $p,q\in Y$ we can see that $\mu(p\cup q)<\mu(\hat{p})+\frac{1}{n}\leq\varepsilon$ which shows that $p\cup q$ is a condition and $p$ and $q$ are compatible.