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 $\mathcal{U} := \bigcup G$

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(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 $\omega$.

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(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 $[q] \in \bigcup G$.

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(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 $q$.