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I'm reading Some applications of the method of forcing by Todorchevich and Farah and one of the exercises seems to be wrong to me.


We fix some partially ordered set $(P,\preceq)$.

A filter on the poset is a set $G\subseteq P$ such that

  1. whenever $p\in G$ and $p\preceq q$ then $q\in G$, and
  2. for all $p,q\in G$, there is $r\in G$ such that $r\preceq q$ and $r\preceq p$.

The set $D\subseteq P$ is dense if for all $p\in P$, there is $q\in D$ such that $q\preceq p$.

An atom is an element $p\in P$ such that the set $\{q\in P:q\preceq p\}$ is linearly ordered by $\preceq$.


There is a filter $G$ that intersects every dense set (a generic filter) if and only if there is an atom in $P$.


Let $P=\mathbb{Z}$ and let $p\preceq q$ iff $|p|>|q|$ or $p=q$. Then $(P,\preceq)$ has no atom, but $G=P$ is a filter that intersects every subset and hence every dense subset.

Is there something wrong with my counterexample? If not, is there some version of the exercise that is related to forcing and is correct?

Since there was some speculation what is actually in the text, here is the relevant part of the first page: enter image description here

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I am still trying to figure out what the authors meant. But I don't see anything wrong with the example, indeed I thought of similar examples myself when working through the definitions in the question. – Carl Mummert Sep 12 '12 at 11:15
Michael, I don't have the text, but maybe there was a requirement (to begin with) that the poset is "nontrivial" in the sense that every point has two incompatible extensions (namely, for all $p$ there are $q_1$ and $q_2$ such that $q_1,q_2\preceq p$ and there is no $r$ such that $r\preceq q_1,q_2$)? – Asaf Karagila Sep 12 '12 at 11:32
@Asaf The exercise is right at the beginning of the text, first page, so I'm pretty sure I haven#t missed anything. – Michael Greinecker Sep 12 '12 at 11:33
up vote 4 down vote accepted

The authors are probably assuming that for every $p$ there are $q,r < p$ with $q \perp r$. This is a consequence of a property of the poset known as separativity.

A (nonempty) poset $P$ that is separative has no generic filter. To see this, let $G$ be a generic filter on $P$ and let $p \in P$. Pick $q,r < p$ as above with $q \perp r$. Then at most one of $q$ and $r$ can be in $G$, because they are incompatible. In particular $G \not = P$. Moreover, we see that for every $p \in P$ there is some $s < p $ with $s \not \in G$. Thus the set $P \setminus G$ is dense, and $G$ cannot possibly meet this set. Hence there is no generic filter on $P$.

It appears to me that when the authors defined an atom $p$, they should have said that all elements below $p$ are compatible, not that they are linearly ordered. If $p$ is an atom in this sense, the filter consisting of the upward closure of the set of elements less than $p$ will be generic, and the above argument shows that if there is no atom in this sense then there is no generic filter. The actual content of the text (posted in the question) doesn't appear to be correct, because if $P$ itself is a filter then it will be a generic filter, regardless whether any elements of the poset generate linearly ordered ideals.

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+1 I will live the question open for a day more or so and accept your answer if nothing new comes up. – Michael Greinecker Sep 12 '12 at 11:48
Carl, what you wrote in the first sentence is incorrect. Separativity simply means that any distinct two elements have a point incompatible with at least one. The requirement that every point have two incompatible extensions is somewhat a notion of nontriviality in forcing. – Asaf Karagila Sep 12 '12 at 13:31
@Asaf: I am familiar with the definition that a poset is separative if (it has no minimal elements and) whenever $p \not \leq q$ there is an $r \leq p$ with $r \perp q$. In particular, this implies that for every $p$ there is a $q < p$ and then an $r < p$ with $r \perp q$, so the property I need is a consequence of the definition of separativity I am familiar with. – Carl Mummert Sep 12 '12 at 13:39
Ah, well if it has no minimal elements is an additional assumption and in which case I agree with what you wrote. – Asaf Karagila Sep 12 '12 at 15:15

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