Relation between topological denseness and denseness over poset

Question

In the theory of forcing, the notion of dense set is important. Formally, a subset $D$ of a poset $P$ is dense if, for any $p\in P$ we can find some $q\in D$ with $q\le p$. Intuitively, denseness of $D$ suggests the "property $D$" is easy to get via extension of condition.

I wonder why the denseness called "denseness". It seems to me that the denseness over poset originates from the topological denseness (i.e. whose closure coincides with whole topological space.) More specifically, for any separative forcing poset $P$ we can find the complete boolean algebra $B$ which completely embeds $P$. By Stone duality, such $B$ gives the topological space $S(B)$. I suspect that there is a relation between the denseness over $P$ and the denseness over $S(B)$, but I didn't get it.

I would appreciate any help, thanks!

Answer 1

You already have a good explanation in terms of the topology on $P$ induced by the ordering relation. This topology also accounts for the terminology "dense open", which one often sees in connection with forcing. Nevertheless, your suspicion about the Stone space of the regular completion $B$ of a separative $P$ is also nearly correct. Given a subset $E$ of $P$, think of each element of $E$ as an element of $B$ and thus as a clopen subset of $S(B)$. The union of these clopen sets, over all the elements of $E$, is an open subset of $S(B)$. This open subset is dense in $S(B)$ iff $E$ is pre-dense in $P$, which means that its downward closure in $P$ is dense. In the other direction, an open subset $U$ of $S(B)$ is dense in $S(B)$ iff the set of $p\in P$ that are, when viewed as clopen sets in $S(B)$, subsets of $U$ is dense in $P$.

Answer 2

user40276 implicitly answered your question, but I thought it was a good idea to expand it a little.

Let $(P; \preceq)$ be a poset and define $X \subseteq P$ to be open iff for all $x \in X$ and all $p \in P$ with $p \preceq x$, we have $p \in X$. The family of these open sets defines a topology $\tau$ on $P$.

It immediately follows from this definition that a set $D \subseteq P$ is dense in the sense of the poset $(P; \preceq)$ if and only if it is dense in the topological space $(P;\tau)$. Also note that $A \subseteq P$ is an antichain in the sense of the poset $(P;\preceq)$ if and only if it is an antichain in the topological space $(P; \tau)$.

Answer 3

There's another way to interpret density topologically: given a poset $\mathbb{P}$, let $Fil(\mathbb{P})$ be the set of all maximal filters in $\mathbb{P}$. Then each condition $p\in\mathbb{P}$ yields a subset of $Fil(\mathbb{P})$: $U_p=\{G: p\in G\}$. The sets $U_p$ generate a topology on $Fil(\mathbb{P})$. For example, if $\mathbb{P}=2^{<\omega}$, the resulting topological space is exactly (homeomorphic to) the Cantor set.