If $M$ is a model of ZFC set theory, let $I$ be the definable cut of its ordinals, the collection of ordinals that are below an ordinal $\delta$ of $M$ that is definable in $M$ without parameters. This would include all the ordinals of $M$, if the definable ordinals happen to be unbounded in $M$, but one can also construct examples where the definable cut is bounded in $M$.  Let $M_I$ be the corresponding definable cut of $M$ itself, the rank-initial segment of $M$ determined by $I$, or in other words, the collection of all sets $x$ in $M$ of rank below a definable ordinal of $M$. Equivalently, $$M_I=\bigcup _{\delta \in I}{V}_{\delta }^M.$$ It is not difficult to see that this is an elementary substructure $M_I\prec M$, because we can verify the Tarski-Vaught criterion as follows. If $M\models \mathrm{\exists }y\phi (x,y)$, where $x\in M_I$, then let $\delta$ be a definable ordinal above the rank of $x$. In this case, the ordinal $\theta$, which is the supremum over all $a\in V_{\delta }$ of the minimal rank of a set $y$ for which $\phi (a,y)$, if there is such a $y$. This supremum $\theta$ is definable, and so since $x\in V_{\delta }$, the minimal rank of a $y$ such that $\phi (x,y)$ is at most $\theta$. Consequently, since $\theta \in I$, such a $y$ can be found in $M_I$. So we have found the desired witness inside the substructure, and so it is elementary $M_I\prec M$. Note that in the general case, one does not necessarily know that $I$ has a least upper bound in $M$. Under suitable assumptions, it can happen that $I$ is unbounded in $M$, that $I$ is an ordinal of $M$, or that $I$ is bounded in $M$, but has no least upper bound.

What I am interested in for this post is how the definable cut might be affected by forcing. Of course, it is easy to see that if $M$ is definable in $M[G]$, then the definable cut of $M[G]$ is at least as high as the definable cut of $M$, simply because the definable ordinals of $M$ remain definable in $M[G]$.

A second easy observation is that if the definable cut of $M$ is bounded in $M$, then we could perform large collapse forcing, collapsing a cardinal above $I$ to $\omega$, which would of course make every cardinal of $I$ countable in the extension $M[G]$. In this case, since ${\omega }_1^{M[G]}$ is definable, it would change the definable cut. So this kind of very large forcing can change the definable cut, making it larger.

But what about small forcing? Suppose that the forcing notion $\mathbb{P}$ we intend to forcing with is small in the sense that it is in the definable cut $M_I$. This would be true if $\mathbb{P}$ itself were definable, for example, but really we only require that $\mathbb{P}$ has rank less than some definable ordinal of $M$. Can this forcing change the definable cut?

Let me show at least that the definable cut can never go up after small forcing.

Theorem. If $G\subset \mathbb{P}$ is $M$-generic for forcing $\mathbb{P}$ in the definable cut of $M$, then the definable cut of $M[G]$ is below or the same in the ordinals as it was in $M$.

Proof. Suppose that $G\subset \mathbb{P}$ is $M$-generic, and we consider the forcing extension $M[G]$. We have already proved that $M_I\prec M$ is an elementary submodel. I claim that this relation lifts to the forcing extension $M_I[G]\prec M[G]$. Note first that since $\mathbb{P}\in M_I$ and $M_I$ is a rank initial segment of $M$, it follows that $M_I$ has all the subsets of $\mathbb{P}$ in $M$, and so $G$ is $M_I$-generic. So the extension $M_I[G]$ makes sense. Next, suppose that $M[G]\models \phi (a)$ for some $a\in M_I[G]$. If $\dot{a}$ is a name for $a$ in $M_I$, then there is some condition $p\in G$ forcing $\phi (\dot{a})$ over $M$. Since $M_I\prec M$, this is also forced by $p$ over $M_I$, and thus $M_I[G]\models \phi (a)$ as well, as desired. So $M_I[G]\prec M[G]$, and from this it follows that every definable ordinal of $M[G]$ is in the cut $I$. So the definable cut did not get higher. QED

But can it go down? Not if the model $M$ is definable in $M[G]$, by our earlier easy observation. Consequently,

Theorem. If $M$ is definable in $M[G]$, where $G\subset \mathbb{P}$ is $M$-generic for forcing $\mathbb{P}$ below the definable cut of $M$, then the definable cut of $M[G]$ is the same as the definable cut of $M$.

Proof. It didn’t go down, since $M$ is definable in $M[G]$; and it didn’t go up, since $\mathbb{P}$ was small. QED

What if $M$ is not definable in $M[G]$? Can we make the definable cut go down after small forcing? The answer is yes.

Theorem. If ZFC is consistent, then there is a model $M\models \text{ZFC}$ with a definable notion of forcing $\mathbb{P}$ (hence in the definable cut of $M$), such that if $G\subset \mathbb{P}$ is $M$-generic, then the definable cut of the forcing extension $M[G]$ is strictly shorter than the definable cut of $M[G]$.

Proof. Start with a model of $\text{ZFC}+V=L$, whose definable ordinals are bounded by a cardinal $\delta$. Let’s call it $L$, and let $I$ be the definable cut of $L$, which we assume is bounded by $\delta$. Let $M=L[G]$ be the forcing extension of $L$ obtained by performing an Easton product, adding a Cohen subset to every regular cardinal above $\delta$ in $L$. Since this forcing adds no sets below $\delta$, but adds a Cohen set at ${\delta }^{+}$, it follows that $\delta$ becomes definable in $L[G]$. In fact, since the forcing is homogeneous and definable from $\delta$, it follows that the definable ordinals of $L[G]$ are precisely the ordinals that are definable in $L$ with parameter $\delta$. These may be bounded or unbounded in $L[G]$. Now, let $\mathbb{Q}$ be the Easton product forcing at the stages below $\delta$, and suppose that $G\subset \mathbb{Q}$ is $L[G]$-generic. Consider the model $L[G][H]$. Note that the forcing $\mathbb{Q}$ is definable in $L[G]$, since $\delta$ is definable there. This two-step forcing can be combined into one giant Easton product in $L$, the product that simply forces to add a Cohen subset to every regular cardinal. Since this version of the forcing is homogeneous and definable in $L$, it follows that the definable ordinals of $L[G][H]$ are precisely the definable ordinals of $L$, which are bounded by $I$. In summary, the definable cut of $L[G]$ is strictly above $\delta$, since $\delta$ is definable in $L[G]$, and the forcing $\mathbb{Q}$ has size and rank $\delta$; but the forcing extension $L[G][H]$ has definable cut $I$, which is strictly bounded by $\delta$. So the definable cut was made smaller by small forcing, as claimed. QED

This post is an account of some ideas that Alexander Block and I had noted today during the course of our mathematical investigation of another matter.