This question is motivated by the "interesting tidbit" in Hamkins' response here: https://mathoverflow.net/a/99025/10671, in which he demonstrates that, after Cohen forcing, there is a perfect set consisting entirely of mutually generic Cohen reals. In particular, this demonstrates that there is a perfect set in the extension containing no ground model reals.

I'm interested in a related question and, while I feel it must have been answered, I can't find a reference.

QUESTION. Consider the reals in the extension $V[c]$ obtained by forcing to add a Cohen real. Does there exist a perfect set in $V[c]$ consisting entirely of ground model reals? Or does every perfect set of reals in $V[c]$ contain a "new" real (added by the forcing)?


2 Answers 2


A very nice question!

The answer is no, there cannot be a perfect set in $V[c]$ consisting entirely of ground-model reals.

Suppose towards contradiction that there is a such a set. So this set consists of the paths through a certain perfect tree $T\subset 2^{<\omega}$ in $V[c]$.

Note, as an easy first case, that we cannot have $T$ in the ground model $V$, for in this case we may consider the branch through $T$ specified by going left-or-right at the splitting nodes of $T$ according to the digits of $c$. From this branch and $T$ we could reconstruct $c$ in the ground model, which is a contradiction.

For the general case, fix a name $\dot T$ for the tree, and suppose that the trivial condition forces $\dot T$ is a perfect tree all of whose branches are in the ground model. For any particular branch $z$ through $T$, there is a finite initial segment of $c$ that forces that $\check z$ is a branch through $\dot T$. Since there are uncountably many branches through $T$ in $V[c]$, there must be some fixed condition $p=c\upharpoonright n$ forcing uncountably many reals $z$ as branches through $\dot T$.

Let $S$ be the tree of finite binary sequences that $p$ forces in $\dot T$. So $S$ is in $V$, and $S$ has uncountably many branches in $V$. So by the Cantor-Bendixson theorem, it follows that $S$ has a perfect subtree $S_0$ in $V$, since every uncountable closed set has a perfect subset. But in this case, we have a perfect tree $S_0$ in $V$ all of whose branches in $V[c]$ are in $V$. And that was the first case we ruled out.

So there can be no such set. QED

Alternative summary of the argument. If $C$ is a perfect set of ground-model reals, let $\dot C$ be a name for it. By pigeon-hole principle, there is a single condition $p$ forcing $\check z\in\dot C$ for uncountably many reals $z$. let $C_p$ be the set of $z$ for which $p\Vdash\check z\in\dot P$. So this is an uncountable closed set in the ground model, which therefore contains a perfect set, the branches through a splitting tree in $V$. But any such tree has branches in $V[c]$ that are not in $V$.

  • 3
    $\begingroup$ Thanks very much, Joel - reducing the argument to a ground-model perfect set is a neat trick! $\endgroup$
    – jonasreitz
    Oct 13, 2016 at 0:11
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    $\begingroup$ That's a lovely pigeonhole argument! $\endgroup$ Oct 13, 2016 at 1:34

There is a more general result proved by Groszek and Slaman which says that if there is a nonconstructible real, then every perfect set has a nonconstructible element. The constructibility can be replaced with any inner models.


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