**Theorem:** Let $\kappa$ be a regular cardinal in $M$. Let $\mathbb{P} \in M$ be a separative partial order of size $\kappa$, and let $G$ be $\mathbb{P}$-generic over $M$. Then the following are equivalent:

(1) There is some $p \in G$ such that in $M$, $\mathbb{P} \restriction p$ has a dense subset of size $< \kappa$.

(2) For all sets of ordinals $A \in M[G]$ of size $\kappa$, there is $B \subseteq A$ such that $B \in M$ and $|B| = \kappa$.

Per Mohammad's request, I will give a proof. I think this is a nice theorem because it characterizes a certain combinatorial property of partial orders in terms of model comparison with generic extensions, like the distributivity properties. I have found some use of it in comparing submodels of generic extensions.

For a partial order $P$, let $d(P)$ be the least size of a dense subset of $P$. For a given $P$, and $q \leq p$ in $P$, $d(P \restriction q) \leq d(P \restriction p)$. Call an element of $P$ "$d$-stable" if $(\forall q \leq p) d(P \restriction q) = d(P \restriction p)$. By well-foundedness, the $d$-stable elements are dense.

Let $G$ be $P$-generic over $M$. Suppose there is $p \in G$ with $d(P \restriction p) < \kappa$, and let $D \subseteq P \restriction p$ witness this. Then $p \Vdash \kappa$ is regular. Let $A \in M[G]$ be a set of ordinals of size $\kappa$, and let $\dot{A}$ be a name for it. $A = \{ \alpha : (\exists q \in D \cap G) q \Vdash \check{\alpha} \in \dot{A} \}$, so for some $q \in D \cap G$, $B = \{ \alpha : q \Vdash \check{\alpha} \in \dot{A} \}$ has size $\kappa$.

Suppose now that $p \in G$ is $d$-stable, and $d(P \restriction p) = \kappa$. We will find a set of ordinals $A \in M[G]$ unbounded in $\kappa$ that contains no unbounded set from $M$. This suffices because, although $\kappa$ may no longer be regular, the function $f : \kappa \to A$ defined by $f(\alpha) =$ the least $\beta \in A$ above $\alpha$, is an object of size $|\kappa|$, and if it had a size-$\kappa$ subset $g \in M$, then the range of $g$ would be an unbounded subset of $A$ from $M$.

Let $D \subseteq P \restriction p$ be dense, and recursively construct $D' \subseteq D$ also dense, and with the property that for all $q \in D'$, $| \{ r \in D' : r \geq q \}| < \kappa$ (exercise). Let $D' = \langle p_\alpha : \alpha < \kappa \rangle$, and in $M[G]$ consider $A = \{ \alpha : p_\alpha \in D' \cap G \}$. $A$ is unbounded in $\kappa$ (exercise). If we had $q \in D'$ and an unbounded $B \in M$ such that $q \Vdash \check{B} \subseteq \dot{A}$, then $\{ p_\alpha : \alpha \in B \} \subseteq D'$, and by separativity, $q \leq p_\alpha$ for all $\alpha \in B$, which contradicts the property of $D'$.