# Cohen real without a minimal support

We start with a model of $\sf ZFC$, $V$, for simplicity we can imagine that $V$ satisfies $\sf GCH$ or even $V=L$.

Let $\Bbb P$ be $\operatorname{Add}(\omega,\omega_1)$, and let $G$ be a $V$-generic filter for $\Bbb P$. For $E\subseteq\omega_1$, we will write $G\restriction E$ as the restriction of $G$ to the reals with coordinates in $E$.

By the fact that the Cohen forcing is ccc, if $x\in\Bbb R^{V[G]}$, then there is some $E\in[\omega_1]^{\omega}\cap V$ such that $x\in V[G\restriction E]$. However, I was told there will be such $x$ without a minimal $E$.

What is an example for a real $x$ such that if $x\in V[G\restriction E]$, then there is some $E'\subsetneq E$ for which $x\in V[G\restriction E']$?

• Let $x(n)$ be the first bit of the $n$th Cohen real. Then $x \in V[G \upharpoonright E]$ iff $\omega \setminus E$ is finite. – Ashutosh Dec 12 '16 at 12:42
• Can you perhaps write this as an answer? – Asaf Karagila Dec 12 '16 at 13:00

Define $x \in 2^{\omega} \cap V[G]$ by $x(n) =$ the first bit of the $n$th Cohen real. Then for every $E \subseteq \omega_1$, $E \in V$, we have $x \in V[G \upharpoonright E]$ iff $\omega \setminus E$ is finite. So there is no such minimal $E$ for $x$.