Generic sets in ZFC

I'm reading Shelah's book "Proper and Improper Forcing" (the first two chapters were recommended for learning the basics of forcing)

Given a quasi-order $P$ we say that $\mathcal{I}$ is a dense subset of $P$ if $$(\forall p \in P) (\exists q \in \mathcal{I}) (p\le q)$$

We say that it is open if for any $p,q \in P$ we have that $p \in \mathcal{I} \wedge p\le q$ then $q \in \mathcal{I}$

$G$ is called directed if every two elements in $G$ has an upper bound in $G$.

$G$ is called downward closed if for every $p \in G$ and $q\in P$ if $q \le p$ then $q \in G$

A subset $G$ of $P$ is called generic over $V$ if it is directed, downward closed and for any dense and open subset of $P$ that is in $V$, the intersection with $G$ is non-empty.

Now they give proof that if $P$ has no trivial branches then a generic set cannot exist in the universe, saying that if $G \in V$ then $P\backslash G \in V$ and it is dense and open, how come? I can't figure that out.

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Can you remind me what trivial branch is, please? – Jason DeVito Aug 25 '10 at 14:57
Nevermind - I found it: above any element, there are two elements with no common upper bound. (My edit timer ran out as I was typing this!) – Jason DeVito Aug 25 '10 at 15:14
Also, in Shelah's book, it seems that "directed" has a different meaning: that any two elements have a common upperbound (see page 3, at least on the projecteuclid version) (again the edit timer ran out as I was typing this!). If the mods can (and would like to) combine all these into a single comment, I'd be more than happy with that.) – Jason DeVito Aug 25 '10 at 15:22
I was always taught that a directed set is a set that every two elements has an upper bound. – Asaf Karagila Aug 25 '10 at 15:31
Ah, I see what you're talking about ;) I'll fix that. (man that comments edit timer is really short) – Asaf Karagila Aug 25 '10 at 15:37

1 Answer

I'm going to use the def of "directed" that I found in Shelah's book1: A set $G$ is directed if it is downward closed (if $p$ in G and if $q\leq p$ then $q\in G$) and every two memebers of $G$ have a common upperbound in $G$.

To see that $P-G$ is open is not too hard: Let $p\in P-G$ and suppose $q\in G$ with $p\leq q$. If $q$ were in $G$, then since $G$ is downwardly closed, we'd have $p\in G$, but that contradicts $p\in P-G$. Thus, $q\notin G$ so $q\in P-G$.

Now, why is $P-G$ dense? Well, let $p\in P$. We want to find a $q\in P-G$ with $p\leq q$. Since there are no trivial branches, we know that above $p$ and are two points $r$ and $s$ with no common upperbound. Since $G$ is directed, $r$ and $s$ cannot both be in $G$. Hence, say, $r\notin G$. Then $r\in P-G$, so we're done.

1I'm using chapter 1 and finding the definition on page 3 of the .pdf.

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Just one minor technicality which I think I got but not really sure - why is $P\backslash G \in V$ if $G \in V$? – Asaf Karagila Aug 25 '10 at 15:47
Since $V$ is a model of $ZFC$, it satisfies the axiom of comprehension. That's the axiom (schema) which allows you to form sets of the form $\{x\in A| \phi(x)\}$, where $\phi$ is some first order formula (with parameters). In this case, $P-G = \{p\in P | p\notin G\}$ which uses the parameter $G$. – Jason DeVito Aug 25 '10 at 15:59
Just what I thought myself. I'm all so new to the idea of taking models of ZFC and it's all very confusing. – Asaf Karagila Aug 25 '10 at 16:12
Thank you for fixing my link Grigory! – Jason DeVito Aug 25 '10 at 16:21