# $Add(\omega, 1)$ forces $Add(\omega_1$, 1) to not be countably closed.

$$\Vdash_{\mathbb{P}} \check{\mathbb{Q}}$$ is not countably closed. Where $$\mathbb{P} = Add(\omega, 1)$$, $$\mathbb{Q} = Add(\omega_1, 1)$$

I would like to prove this. But I'm sort of lost- I might be drastically overthinking it. Most I can say is that the offending sequence that witnesses failure of countable closure will probably come from the new subset of $$\omega$$ being added, but that doesn't help too much. I have tried some things related to Easton's theorem but I think I may be off target there. Would appreciate some help.

• " the offending sequence that witnesses failure of countable closure will probably come from the new subset of $\omega$ being added" That's exactly right. (Note that trivially $Add(\omega_1,1)^{V[G]}$ is countably closed in $V[G]$; we're crucially looking at the old forcing $Add(\omega_1,1)^V$ in the new model $V[G]$.) Mar 14, 2021 at 1:43
• Ah whoops somehow I forgot that distinction. Then the argument is pretty simple I think; does this work: Let $A$ be the new subset of $\omega$; then the sequence $p_n : n < \omega$ will be increasing finite initial segments of $A$. Then if it was closed the limit $q \leq p_i \forall i$ would describe $A$ and it would be in $V$, thus contradiction? Mar 14, 2021 at 1:51