$\mathbb{1}\Vdash S\subset\kappa$ and $\not\exists\beta<\kappa$ such that $\mathbb{1}\Vdash S\subset\beta$, contradicts $\mathbb{1}\Vdash |S|<\kappa$.

Question

The precise statement would be

If $\kappa$ is regular, $\mathbb{1}\Vdash \mathring{S}\subseteq\check{\kappa}$ and $\not\exists\beta<\kappa$ such that $\mathbb{1}\Vdash \mathring{S}\subseteq\check{\beta}$ contradicts $\mathbb{1}\Vdash |\mathring{S}|<\check{\kappa}$.

This is apparently true as it seems to be assumed in proof for lemma $V.3.9$ in Kunen's Set theory. When I first read it I was quickly convinced it was true but inspecting it more closely I can't give an actual proof.

Of course, if $S\subseteq\kappa$ and $|S|<\kappa$, $S$ must be contained in some $\beta<\kappa$. But it might be true that in every extension $M[G]$ we have $|\mathring{S}_G|<\kappa$ (i.e. $\mathbb{1}\Vdash |\mathring{S}|<\check{\kappa}$) and there still not be 1 $\textit{fixed}$ $\beta<\kappa$ such that in every extension $M[G]$ we have $\mathring{S}_G\subseteq\beta$ (i.e.$\mathbb{1}\Vdash \mathring{S}\subseteq\check{\beta})$ and I can't find an argument for why it isn't.

I include here the statement and proof of the lemma.

Answer 1

Let $(p_i,\beta_i)$ be such that $p_i\Vdash S \subseteq \beta_i$ and $\{p_i\}$ is a maximal anti chain. Then since the poset is $\kappa$-cc, the set $\{\beta_i\}$ has size $<\kappa$ and since $\kappa$ is regular there is $\beta$ such that $\beta_i<\beta$ for all $i$. Now $1\Vdash S\subseteq \beta$.