# How can OD be not transitive

In Kennen's book about the Independence Results, it is said that the class of ordinal definable sets ($$OD$$) need not be necessarily transitive. Therefore, if I'm not mistaken, we would have that $$OD \neq HOD$$. How can $$OD$$ be not transitive? Thank you very much.

• Each $V_\alpha$ is $OD$ (by the formula "$x$ has rank less than $\alpha$"), so to say $OD$ is transitive is to assert $V=OD$. May 10, 2021 at 21:07

The set $$2^\omega$$ is obviously $$\operatorname{OD}$$, since it is just the set of infinite binary sequences. But there may be reals (elements of $$2^\omega$$) which are not definable using only ordinals as parameters. In that case $$\operatorname{OD}$$ is not transitive.
One example where this occurs is when adding a Cohen real. Recall that Cohen forcing is the partial order $$\mathbb{C} = 2^{<\omega}$$ consisting of finite binary strings together with the natural extension relation. Whenever $$G$$ is $$\mathbb{C}$$-generic over $$V$$, $$\bigcup G$$ is an infinite binary sequence (so an element of $$2^\omega$$) which is the so called "Cohen-real".
Theorem. Let $$\varphi(x,\alpha_0, \dots, \alpha_n)$$ be a formula in the language of set theory with one free variable $$x$$ and ordinal parameters $$\alpha_0, \dots, \alpha_n$$. Let $$G$$ be $$\mathbb{C}$$-generic over $$V$$. Then, in $$V[G]$$, $$c := \bigcup G$$ is not uniquely defined by $$\varphi$$.
Proof. Suppose $$V[G] \models \varphi(c,\alpha_0, \dots, \alpha_n)$$. Then, by the forcing theorem, there is a condition $$s \in 2^{<\omega}$$ so that $$s \Vdash \varphi(\dot c, \check \alpha_0, \dots, \check \alpha_n),$$ where $$\dot c$$ is a name for $$c$$. Consider $$c'$$ to be the infinite binary sequence that is equal to $$c$$ except at $$\vert s\vert$$, i.e. $$c'(n) := \begin{cases} c(n) & \text{ if } n \neq \vert s \vert \\ 1-c(n) & \text{ if } n = \vert s \vert \end{cases}.$$ Then you can check that $$c'$$ is also a Cohen real over $$V$$, i.e. there is a $$\mathbb{C}$$-generic $$H$$ over $$V$$ so that $$c' = \bigcup H$$. And in fact you can also see that $$s \in H$$. Thus, by the forcing statement above, $$V[H] \models \varphi(c', \alpha_0, \dots, \alpha_n).$$
But $$V[G]$$ and $$V[H]$$ are exactly the same model since $$H \in V[G]$$ and vice-versa $$G \in V[H]$$ and a forcing extension is the smallest transitive model of ZFC containing the ground model and the generic filter. Thus in fact, $$V[G] \models c' \neq c \wedge \varphi(c, \alpha_0, \dots, \alpha_n) \wedge \varphi(c', \alpha_0, \dots, \alpha_n).$$ So in $$V[G]$$, $$c$$ is not uniquely defined by $$\varphi$$.