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I have a partial answer for the following homework and I was wondering if you could tell me if I have it right and help me with part (c) which I'm currently stuck on. Many thanks for your help!

Let the forcing partial order $\mathbb{U} = ([\omega]^\omega / \mathrm{fin} , \leq)$ be defined as follows:

Define an equivalence relation on $[\omega]^\omega$ by stipulating $ x \sim y \iff x \Delta y$ is finite and let $[\omega]^\omega / \mathrm{fin} := \{ [x] \mid x \in [\omega]^\omega \} $. On $[\omega]^\omega / \mathrm{fin}$ define a partial ordering $\leq$ by stipulating $[x] \leq [y] \iff y \setminus x$ is finite.

Now let $V \models ZFC $ and $G$ be an $\mathbb{U}$-generic filter over V. In the model $V[G]$ let $\mathcal{U} := \bigcup G$


(a) Show that $\mathcal{U}$ is a set of infinite subsets of $\omega$, i.e. $\mathcal{U} \subset [\omega]^\omega$.

My answer:

$\bigcup G = [\bigcup_{g \in G} g ]$ where $g \in [\omega]^\omega$ so for any $y \in \bigcup G$, $y$ has cardinality $\omega$.


(b) Show that $\bigcup G$ is a filter

My answer:

(i) $\bigcup G \neq \emptyset$: If $\bigcup G = \emptyset$ then $G = \emptyset$ which would contradict $G$ being a filter.

(ii) Any two elements in $\bigcup G$ are compatible, i.e. for any $p,q$ in $\bigcup G$ there is an $r$ in $\bigcup G$ such that $p \leq r$ and $q \leq r$:

Let $p,q$ be in $\bigcup G$. Then there exists $g_q , g_p$ in $G$ such that $p \in g_p$ and $q \in g_q$. $g_i$ are sets of ordinals and so $p$ and $q$ are ordinals, too. So either $p \subset q$ or $q \subset p$. Set $r := \max_{\text{w.r.t.} \subset} p, q$, then $p \leq r$ and $q \leq r$.

(iii) If $p,q \in \bigcup G$ and $p \in \bigcup G$ and $q \leq p$ then $q \in \bigcup G$:

Let $p,q \in \bigcup G$ and $p \in \bigcup G$ and $q \leq p$. $q \leq p$ means that $|p \setminus q| < \omega $ which implies that $|p \cap q| = \omega$ which means that $|p \Delta q| < \omega$ and hence $[p] = [q]$ and hence $[q] \in \bigcup G$.


(c) Show that for every set $x \in [\omega]^\omega$ in the ground model $V$ there exists an $y \in \bigcup G$ such that either $y \subset x$ or $y \cap x = \emptyset$.

Here I'm stuck. I thought I could do a proof by contradiction by assuming that there exists $x \in [\omega]^\omega$ such that for all $y \in \bigcup G$, $x \subset y$ but this doesn't lead anywhere.

Many thanks for your help!

Edit

$p \leq q$ reads as $p$ weaker than $q$.

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1  
There’s definitely something wrong in (b)(iii): $|p\setminus q|<\omega$ does not imply that $|p\Delta q|<\omega$, because $q\setminus p$ could still be infinite. You’re essentially saying that if $[x]\le[y]$, then $[x]=[y]$, which is clearly false in general. –  Brian M. Scott Dec 10 '11 at 21:43
    
@BrianM.Scott Thanks Brian! By now I can see it myself : ) –  Rudy the Reindeer Feb 13 '12 at 19:33

2 Answers 2

up vote 1 down vote accepted

Here is future self to past self:

Before we start manipulating symbols blindly let's sit back and take a leisurely look at what the exercise is asking (it's a very nice exercise):

We start with a model $\mathbf V$ of $ZFC$, a forcing partial order $\mathbb U$ and a $\mathbb U$-generic filter $G$. It's important not to confuse the notion of a filter with that of a generic filter: the elements in $\bigcup G$ are sets and not equivalence classes of sets like the elements in $\mathbb U$ and we are talking about a filter on the set $[\omega]^\omega$ and not a generic filter in $\mathbb U$.

We are asked to show that $\bigcup G$ is a set (part (a)) and what's more, that it is an ultrafilter (parts (b) and (c)). Well, to be precise, part (a) asks you to explain why it is a set and also why its elements are infinite sets -- a property which will come in handy in part (b).

That $\bigcup G$ is a set follows from $G\subseteq \mathbf V [G], G \in \mathbf V [G]$ and $\mathbf V [G] \models ZFC$ -- of course $\mathbf V [G]$ has to satisfy "Union" which says exactly that the union of a set is again a set.

As for part (b), we need to show the following three properties of a filter:

(i) $1 = \omega \in \bigcup G$: To this end observe that $D = \{ [\omega]\}$ is dense in $\mathbb U$ and since $G$ is generic, $G \cap D \neq \varnothing$ so that $[\omega] \in G$ and hence $\omega \in \bigcup G$.

(ii) If $a \subseteq b$ and $a$ is in $\bigcup G$ then $b \in \bigcup G$: If $a$ is in $\bigcup G$ we have $[a] \in G$ and since $a \setminus b = \varnothing$ we have $[a] \geq [b]$. Since $G$ is generic it follows that $[b] \in G$.

(iii) If $a,b \in \bigcup G$ then $a \cap b \in \bigcup G$: Let $a,b \in \bigcup G$. Then since $G$ is generic there is $[r] \in G$ such that $[r] \geq [a]$ and $[r] \geq [b]$. Observe that this means that both $r \setminus b$ and $r \setminus a$ are finite. Hence their union is also finite. But $r \setminus (a \cap b) = r \setminus b \cup r \setminus a$ so that $[r] \geq [a \cap b]$ and hence $[a \cap b] \in G$.

For part (c) define $D_x = \{[y] \mid y \in [\omega]^\omega \text{ such that } y \subseteq x \text{ or } y \cap x = \varnothing \}$. Then $D_x$ is dense in $\mathbb U$: Let $z \in [\omega]^\omega$. Then either $z \cap x = \varnothing$ or $z \cap x \neq \varnothing$. If the former, $[z] \in D_x$ and $[z] \geq [z]$. If the latter, let $y = z \setminus x$. Then $y \setminus z = \varnothing$ or in other words, $[y] \geq [z]$. Since $G$ is generic, there is $y \in G \cap D_x$ which proves the claim.

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For part (c) try using a density argument. For each $x\in [\omega]^\omega$, define a dense set $D_x$ in such a way that any $y\in D_x\cap G$ implies that either $y\subset x$ or $y\cap x= \varnothing.$

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