I have a question about one of the steps in the proof of "$ccc$ preserves cofinalities" in Kunen. The whole proof is below. My question concerns the claim that $\sup Y=\beta$. I am not sure how one gets this. I have managed to show that $X\subseteq Y$, but I am not sure if this is sufficient.
Any help would be greatly appreciated!
The relevant proof:
Theorem IV.3.4 If $\mathbb{P}\in M$ and ($\mathbb{P}$ is $\left.c{{{c}}}c\right)^{M}$ , then $\mathbb{P}$ preserves cofinalities and hence $\mathbb{P}$ preserves cardinals
Proof of Theorem IV.3.4. By Lemma IV.3.3, it is sufficient to fix a that limit ordinal $\beta$ with ${{{\omega}}}<\beta<o(M),$ assume that $({{\beta}}$ is regular) $^M$ , and prove that $({{\beta}}$ is regular) $^{M[G]}$ . If this is false, then fix $X\subseteq\beta$ with $X\in M[G]$ and $\operatorname{sup}(X)\,=\,\beta$ and $\alpha:=\operatorname{type}(X)<\beta$ . Let $f:\alpha\ \to X$ be the unique order preserving bijection. Then $f:\alpha \to \beta,$ so by Lemma IV.3.5，fix $F:\alpha\to{\mathcal{P}}(\beta)$ Let $F\in M$ such that for all $\xi<\alpha$, $f(\xi)\in F(\xi)$ and $(|F(\xi)|\leq\aleph_{0})^{M}$. Let $Y=\cup_{\xi<\alpha}F(\xi).$ Then $Y\subseteq{\beta}$ and $\operatorname{sup}(Y)=\beta.$ Working completely in $\textstyle{M}$: We have an uncountable regular cardinal $|Y|<\beta$ (since $Y$ is a union of fewer than $\beta$ countable sets), so $\operatorname{sup}(Y)=\beta,$ contradicting the definition of “regular”. □
The image of $f$ is unbounded in $\beta$ and for each $\xi \in \alpha$ we have $f(\xi) \in F(\xi)$. Since $\bigcup F(\xi)$ contains each $f(\xi)$, $Y=\bigcup F(\xi)$ must therefore be unbounded in $\beta$, which means (since $Y \subseteq \beta$) that $\sup \bigcup F(\xi) = \beta$.