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I would like to show that by forcing over Add$(\omega, \omega_1)$, one can get the model $HOD(\mathbb{R})^{V[G]}$ where the following weakened perfect set property holds: "for every subset $A$ of $\omega^{\omega}$ either $A$ or its complement (or both ) contains a perfect subset."

The construction is obviously going to parallel the solovay model (which I'm familiar with) but I'm unable to transfer a significant number of details (the Solovay construction I'm familiar with is forcing over Coll$(\omega, <\mu)$ where $\mu$ is an inaccessible) over to this case. The posets seem too different and the lack of an inaccessible hurts.

Does anyone have a reference or hints on how to see this fact?

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2 Answers 2

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You can find a version of this theorem in John Truss' paper, aptly named for your question

Truss, John, Models of set theory containing many perfect sets, Ann. Math. Logic 7, 197-219 (1974). ZBL0302.02024.

The idea behind the proof is fairly simple, you could argue it has parallels to Solovay's model, but it really is just the usual symmetries trick.

  • Suppose that $A$ is a set of reals in $\mathrm{HOD}(\Bbb R)$. Then $A$ is definable from a real, and because we forced with a c.c.c. forcing, it means that there are a name $\dot A$ and a countable initial segment of the forcing such that fixing that initial segment will not change the name $\dot A$. Call this initial segment $\alpha$.

  • Now suppose that $\dot A$ contains a real which is not from $\operatorname{Add}(\omega,\alpha)$. Then by applying permutations we conclude that it contains a perfect set of copies of this real. The key point is that the automorphism group is very wild, it allows us to switch indices, flip bits, apply permutations on different coordinates, etc.

  • Suppose that it doesn't contain such a real, then simply note that $\dot B$ defined to be the name for the complement of $A$ is also fixed by not moving the first $\alpha$ coordinates. So we can apply the previous argument.

In fact, we can observe that a set of reals can be well-ordered if and only if it is a subset of $\Bbb R^{V[G\restriction\alpha]}$ for some $\alpha$. So we really prove that a set of reals can be well-ordered or it contains a perfect subset. (This is generalised in Truss' paper as well.)

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  • $\begingroup$ Could you elaborate a little bit the observation that a subset of $\mathbb{R}^{V[G]}$ has a well-order in $\text{HOD}(\mathbb{R})^{V[G]}$ if and only if it is a subset of $\mathbb{R}^{V[G\upharpoonright\alpha]}$? Why? Thanks! $\endgroup$
    – Lorenzo
    Commented May 6, 2022 at 6:56
  • $\begingroup$ @Lorenzo: Note that one direction is trivial, since $\Bbb R^{V[G\restriction\alpha]}$ can be well ordered there, after all, it is just the same as adding one Cohen real to the ground model in our case here. The other direction is that if it is that if we add reals from unboundedly many coordinates, by homogeneity we have had to have added "pretty much all of them" in some sense, namely a perfect set of them, so your set of reals must have size continuum and is not well orderable. $\endgroup$
    – Asaf Karagila ♦
    Commented May 6, 2022 at 7:12
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    $\begingroup$ @Lorenzo: Because $G\restriction\alpha$ is essentially a single real, $r$ so $V[r]$ is contained in $\rm HOD(\Bbb R)$, and in $V[r]$ choice holds. $\endgroup$
    – Asaf Karagila ♦
    Commented May 6, 2022 at 9:09
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    $\begingroup$ @Lorenzo: That's a different question in a different context. In Truss' work, I presume you mean the models of the form $L(w(\Bbb R))$, which in case we use $\omega_1$ as our uncountable cardinal for the construction, is really just $L(\Bbb R)$. But the idea, in a nutshell, is that any real in $L[G]$, where $G\subseteq\operatorname{Add}(\omega,\omega_1)$ is Cohen for some algebra, and we can look at that quotient algebra and consider its automorphisms and so on. This is why results that are of the form $L(\Bbb R)$ are sometimes harder to formalise in symmetric systems language [...] $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 15, 2022 at 10:57
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    $\begingroup$ [...] because we are actually looking at all automorphisms of the Boolean completion, and so we get a lot of subalgebras and quotients that are "kind of invisible" from the basic partial order standpoint. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 15, 2022 at 10:57
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Here is a relatively permutation-free proof (sorry, Asaf).

Write $\mathbb{C}_{\omega_1}$ for the forcing to add $\omega_1$-many Cohen reals. Let $X$ be a set of reals in $\mathsf{HOD}(\mathbb R)^{V^{\mathbb{C}_{\omega_1}}}$. So there is some formula $\varphi(x,\vec{\alpha},r)$ with free variable $x$ and parameters $\vec{\alpha},r$ which defines $X$ in $V^{\mathbb{C}_{\omega_1}}$.

Without loss of generality (by moving to an intermediate extension if needed), we can assume the parameters are already in the ground model $V$. Furthermore, we break $\mathbb{C}_{\omega_1}$ into the two-step iteration $\mathbb{C}\ast\dot{\mathbb{C}}_{\omega_1}$, where $\mathbb{C}$ is the forcing to add a single Cohen real.

Now if $V[c]$ is the extension by $\mathbb{C}$, then in it, $\varphi(c,\vec{\alpha},r)$ is a statement about ground model elements, so $V[c]$ thinks this is decided by the $1$ of ${\mathbb{C}}_{\omega_1}$. Now we consider cases,

Case 1: in $V[c]$, $1\Vdash_{\mathbb{C}_{\omega_1}} \varphi(c)$.

In this case, there is some condition $p\in V$ which forces that the canonical name of the generic filter names a real which, after forcing with $\mathbb{C}_{\omega_1}$, satisfies $\varphi$. But the forcing to add a perfect set of Cohen reals extending $p$ is the same as adding a single one extending $p$ (see Joel Hamkins's answer here). So if $r$ is a Cohen real extending $p$, in $V[r]$ there is a perfect set of Cohen reals extending $p$. But then in $V[r]$, $1\Vdash_{\mathbb{C}_{\omega_1}} \varphi(r)$. So in the final extension there is a perfect set contained in $X$.

Case 2: in $V[c]$, $1\Vdash_{\mathbb{C}_{\omega_1}} \neg\varphi(c)$.

This is similar to Case 1, except that one argues the perfect set of Cohen reals is disjoint from $X$.

By the way, the property of a set either containing or being disjoint from a perfect set is sometimes called (weakly-) Sacks measurable or Marczewski measurable.

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