Definable Hamel bases and $AC_\omega(R)$

TL;DR

This paper constructs a model of ZF set theory where a definable Hamel basis exists but the countable axiom of choice for reals ($AC_(R)$) fails, highlighting the independence of these concepts.

Contribution

It demonstrates the existence of a ZF model with a definable Hamel basis where $AC_(R)$ does not hold, showing their independence.

Findings

Existence of a $_3$ definable Hamel basis in a ZF model.

Failure of $AC_(R)$ in the same model.

Illustration of the independence between definable bases and choice axioms.

Abstract

There is a model of ZF with a $\Delta^1_3$ definable Hamel basis in which $AC_\omega(R)$ fails.

Equations

$$H={\mathsf{HOD}}_{A\cup \{A\}}^{L[g]}.$$

$$L[g]\models \phi (s^{⌢}c\upharpoonright [n,\omega ),a).$$

$$p{⊩}_L^{\mathbb{C}(\omega )}\phi ({\check{s}}^{⌢}\dot{c}\upharpoonright [\check{n},\omega ),\dot{a}).$$

$$L[g]\models \phi (s^{⌢}{\dot{c}}^{g^{\ast }}\upharpoonright [n,\omega ),a).$$

$$\{(T_1,\dots ,T_N)\in {\mathbb{P}}_{\xi }(N):\exists (T_1^{\prime },\dots ,T_N^{\prime })\in f_{\xi }(n)(T_1,\dots ,T_N)\le (T_1^{\prime },\dots ,T_N^{\prime })\},$$

$$(T_{s_1}^{m_1},\dots ,T_{s_N}^{m_N})\in d_{\xi }(n,N).$$

$$A_{s,\xi }^m=\bigcap _{n\ge \mathrm{lh}(s)}(\bigcup _{\begin{array}{c}t\supset s\\ \mathrm{lh}(t)=n\end{array}}{T}_t^m)=\{\mathrm{stem}(T_{t,\xi }^m)\upharpoonright k:t\supset s,k<\omega \}.$$

$$A_{s,\xi }^m\cap A_{s^{\prime },\xi }^{m^{\prime }}\text{mbox}isfinite,unlessm=m^{\prime }\text{mbox}ands\subset s^{\prime }\text{mbox}or{s}^{\prime }\subset s.$$

$$D=\{(T_1,\dots ,T_N)\in {\mathbb{Q}}_{\xi }(N):\text{mbox}stem(T_i)\perp \text{mbox}stem(T_j)\text{mbox}fori\ne j\}$$

$$n>\max (\{n_0,N-1,m_1,\dots ,m_N,\mathrm{lh}(s_1),\dots ,\mathrm{lh}(s_N)\}),$$

$$(T_{t_1,\eta }^{m_1},\dots ,T_{t_N,\eta }^{m_N})\in d_{\eta }(n,N)\subset D^{\prime }$$

$$(A_{t_1,\eta }^{m_1},\dots ,A_{t_N,\eta }^{m_N})\le (T_{t_1,\eta }^{m_1},\dots ,T_{t_N,\eta }^{m_N}),$$

$$(A_{t_1,\eta }^{m_1},\dots ,A_{t_N,\eta }^{m_N})\le (A_{s_1,\eta }^{m_1},\dots ,A_{s_N,\eta }^{m_N}),$$

$$\{(T_1,\dots ,T_N)\in {\mathbb{Q}}_{\xi }(N):\text{mbox}stem(T_i)\perp \text{mbox}stem(T_j)\text{mbox}fori\ne j\}$$

$$(A_{s_1,\xi }^{m_1},\dots ,A_{s_N,\xi }^{m_N})\in g.$$

$$n>\max (\{N-1,m_1,\dots ,m_N,\mathrm{lh}(s_1),\dots ,\mathrm{lh}(s_N)\}).$$

$$\exists m<\omega (T_{t_1,\xi }^{m_1},\dots ,T_{t_N,\xi }^{m_N})⊩\tau (\check{n})=\check{m}.$$

$$g(n)=\{m<\omega :\exists t_1\supset s_1,\dots \exists t_N\supset t_N(\mathrm{lh}(t_1)=\dots =\mathrm{lh}(t_N)=n+1\wedge$$

$$(T_{t_1,\xi }^{m_1},\dots ,T_{t_N,\xi }^{m_N})⊩\tau (\check{n})=\check{m})\},$$

$$H=H(L)={\mathsf{HOD}}_{A\cup \{A\}}^{L[g]}$$

$$e:\omega \to \omega \times \omega ,$$

$$p{⊩}_L^{\mathbb{P}(\omega )}\phi (\check{x},{\check{\alpha }}_1,\dots {\check{\alpha }}_m,{\dot{c}}_{n_1},\dots {\dot{c}}_{n_k},\dot{A})$$

$$\pi (p){⊩}_L^{\mathbb{P}(\omega )}\phi (\check{x},{\check{\alpha }}_1,\dots {\check{\alpha }}_m,{\dot{c}}_{n_1},\dots {\dot{c}}_{n_k},\dot{A})$$

$$\displaystyle\mbox{If $a$, $b\in[A]^{<\omega}$, then }L[a]\cap L[b]=L[a\cap b].$$

$$\mathbb{R}\cap H=\bigcup \{{\mathbb{R}}_a^{+}:a\in [A{]}^{<\omega }\},$$

$${\mathbb{R}}_a^{+}\cap {\mathbb{R}}_b^{+}=\varnothing \text{mbox}fora,b\in [A{]}^{<\omega },a\ne b.$$

$${ϵ}_n=\frac{ϵ}{n\cdot 2^{2n+2}},$$

$$\{x\in {}^{\omega }2:x\upharpoonright [k_n,k_{n+1})=z\upharpoonright [k_n,k_{n+1})\}\cap P=\varnothing .$$

$${\ell }_n\le k_{n^{\prime }}<k_{n^{\prime }+1}\le {\ell }_{n+1}.$$

$$D=\{x\in {}^{\omega }2:\exists nx\upharpoonright [{\ell }_{e(n)},{\ell }_{e(n+1)})=z^{\prime }\upharpoonright [{\ell }_{e(n)},{\ell }_{e(n+1)})\},$$

$$B=\{x\in \mathbb{R}:\exists \text{mbox}evenn(2^n<\#(x)\le 2^{n+1})\}$$

Full text

Answering a question from [13, p. 433] it was shown in [2] that there is a Hamel basis in the Cohen–Halpern–Lévy model. In this paper we show that in a variant of this model, there is a projective, in fact $\Delta^{1}_{3}$, Hamel basis.

Throughout this paper, by a Hamel basis we always mean a basis for ${\mathbb{R}}$, construed as a vector space over ${\mathbb{Q}}$. We denote by $E_{0}$ the Vitali equivalence relation, $xE_{0}y$ iff $x-y\in{\mathbb{Q}}$ for $x$, $y\in{\mathbb{R}}$. We also write $[x]_{E_{0}}=\{y\colon yE_{0}x\}$ for the $E_{0}$–equivalence class of $x$. A transversal for the set of all $E_{0}$–equivalence classes picks exactly one member from each $[x]_{E_{0}}$. The range of any such transversal is also called a Vitali set. If we identify ${\mathbb{R}}$ with the Cantor space ${}^{\omega}2$, then $xE_{0}y$ iff $\{n\colon x(n)\not=y(n)\}$ is finite.

A set $\Lambda\subset{\mathbb{R}}$ is a Luzin set iff $\Lambda$ is uncountable but $\Lambda\cap M$ is at most countable for every meager set $M\subset{\mathbb{R}}$. A set $S\subset{\mathbb{R}}$ is a Sierpiński set iff $S$ is uncountable but $S\cap N$ is at most countable for every null set $N\subset{\mathbb{R}}$ (“null” in the sense of Lebesgue measure). A set $B\subset{\mathbb{R}}$ is a Bernstein set iff $B\cap P\not=\emptyset\not=P\setminus B$ for every perfect set $P\subset{\mathbb{R}}$. A Burstin basis is a Hamel basis which is also a Bernstein set. It is easy to see that $B\subset{\mathbb{R}}$ is a Burstin basis iff $B$ is a Hamel basis and $B\cap P\not=\emptyset$ for every perfect $P\subset{\mathbb{R}}$.

A set $m\subset{\mathbb{R}}\times{\mathbb{R}}$ is called a Mazurkiewicz set iff $m\cap\ell$ for every straight line $\ell\subset{\mathbb{R}}\times{\mathbb{R}}$.

By ${\sf AC}_{\omega}({\mathbb{R}})$ we mean the statement that for all sequences $(A_{n}\colon n<\omega)$ such that $\emptyset\not=A_{n}\subset{\mathbb{R}}$ for all $n<\omega$ there is some choice function $f\colon\omega\rightarrow{\mathbb{R}}$, i.e., $f(n)\in A_{n}$ for all $n<\omega$.

D. Pincus and K. Prikry study the Cohen-Halpern-Lévy model $H$ in [13]. The model $H$ is obtained by adding a countable set of Cohen reals (say over $L$) without adding their enumeration; $H$ does not satisfy ${\sf AC}_{\omega}({\mathbb{R}})$. It is shown in [13] that there is a Luzin set in $H$, so that in ${\sf ZF}$, the existence of a Luzin set does not even imply ${\sf AC}_{\omega}({\mathbb{R}})$. [2, Theorems 1.7 and 2.1] show that in $H$ there is a Bernstein set as well as a Hamel basis. As in ${\sf ZF}$ the existence of a Hamel basis implies the existence of a Vitali set, the latter also reproves Feferman’s result (see [13]) according to which there is a Vitali set in $H$.

Therefore, in ${\sf ZF}$ the conjunction of the following statements (1), (3), and (5) (which in ${\sf ZF}$ implies (4)) does not yield ${\sf AC}_{\omega}({\mathbb{R}})$.

• (1)

There is a Luzin set.

• (2)

There is a Sierpiński set.

• (3)

There is a Bernstein set.

• (4)

There is a Vitali set.

• (5)

There is a Hamel basis.

• (6)

There is a Burstin basis.

• (7)

There is a Mazurciewicz set.

(2) is false in $H$, see [2, Lemma 1.6]. We neither know if (6) is true in $H$, nor do we know if (7) is true in $H$. We aim to prove that in ${\sf ZF}$, the conjunction of all of these statements does not imply ${\sf AC}_{\omega}({\mathbb{R}})$, even if the respective sets are required to be projective.

The Luzin set which [13, Theorem on p. 429] constructs is $\Delta^{1}_{2}$. In ${\sf ZFC}$, there is no analytic Hamel basis (see [16],[17], [9]), but by a theorem of A. Miller, in $L$ there is a coanalytic Hamel basis, see [12, Theorem 9.26]; see also e.g. [14, Corollary 2 and Lemma 4]. On the other hand, it can be verified that the model from [2] doesn’t have a projective Vitali set.333To display our ignorance: we don’t know if the model from [3] has a definable Hamel basis. For the convenience of the reader as well as to motivate what is to come, we shall sketch the proof of this at the beginning of the first section, see Lemma 1.1.

The papers [3] and [1] produce models of ZF plus DC plus (6) and ZF plus DC plus (7), respectively. By another theorem of A. Miller, see [12, Theorem 7.21], in $L$ there is a coanalytic Mazurkiewicz set. It is not known if there is a Mazurkiewicz set which is Borel.

The result of the current paper is the following.

• (a)

There is a $\Delta^{1}_{2}$ Luzin set.

• (b)

There is a $\Delta^{1}_{2}$ Sierpiński set.

• (c)

There is a $\Delta^{1}_{3}$ Bernstein set.

• (d)

There is a $\Delta^{1}_{3}$ Hamel basis.

1 Jensen’s perfect set forcing, revisited.

In what follows, we shall mostly think of reals as elements of the Cantor space ${}^{\omega}2$. We shall need a variant of the Cohen-Halpern-Lévy model.

In order to get a definable Hamel basis in the absence of ${\sf AC}_{\omega}({\mathbb{R}})$ forces us to indeed work with a model which is different from the original Cohen-Halpern-Lévy model. This follows from the followig folklore result which we include here as a motvation for what is to come. Recall, see [2, Lemma 1.1], that a Hamel basis trivially produces a Vitali set.

Recall that the original Cohen-Halpern-Lévy model is produced as follows. See [13], see also [2, p. 3567]. Let $g$ be ${\mathbb{C}}(\omega)$-generic over $L$, and let $A$ denote the countable set of Cohen reals which ${\mathbb{C}}(\omega)$ adds. Then

[TABLE]

Lemma 1.1

The Cohen-Halpern-Lévy model from (1) does not have a Vitali set which is definable in $H$ from ordinals and reals.444Recall, though, that $H$ does have a Hamel basis which is definable from the set $A$ of Cohen reals, see the proof in [2, section 2].

Proof. Let $g$ be ${\mathbb{C}}(\omega)$-generic over $L$. It suffices to prove that there is no $a\in[A]^{<\omega}$ such that in $L[g]$ there is a Vitali set in ${}^{\omega}2$ which is definable from ordinals and $a$.

Suppose otherwise. By minimizing the ordinal parameters, we may fix $a\in[A]^{<\omega}$ such that in $L[g]$ there is a Vitali set which is definable $a$, say via the formula $\varphi(-,a)$. Let $c\in A\setminus a$, and say $n<\omega$ and $s\in{}^{n}2$ are such that

[TABLE]

Let ${\dot{c}}$ be a canonical ${\mathbb{C}}(\omega)$-name for $c$, in particular, ${\dot{c}}^{g}=c$, let ${\dot{a}}$ be a canonical ${\mathbb{C}}(\omega)$-name for $a$, in particular, ${\dot{a}}^{g}=a$, and pick $p\in g$ such that

[TABLE]

Let $g^{*}$ be ${\mathbb{C}}(\omega)$-generic over $L$ which is identical with $g$ except for that $g^{*}$ incorporates a finite nontrivial variant of $g$ only in the coordinate of ${\mathbb{C}}(\omega)$ which gives rise to $c$ in such a way that ${\dot{c}}^{g^{*}}\upharpoonright[n,\omega)\not=c\upharpoonright[n,\omega)$, but ${\dot{c}}^{g^{*}}$ is $E_{0}$-equivalent with $c$. We have that $L[g^{*}]=L[g]$ and ${\dot{a}}^{g^{*}}=a$, and (3) yields that

[TABLE]

(2) and (4) contradictthe fact that $\varphi(-,a)$ defines a Vitali set in $L[g]$. $\square$

The same argument shows that the model from [3] doesn’t have a Vitali set which is definable from ordinal and real parameters.

In order to construct our model, we now need to introduce a variant of Jensen’s variant of Sacks forcing, see [8] (see also [10, Definition 6.1]), which we shall call ${\mathbb{P}}$. The reason why we can’t work with Jensen’s forcing directly is that it does not seem to have the Sacks property (see e.g. [3, Definition 2.15]).

By way of notation, if ${\mathbb{Q}}$ is a forcing and $N>0$ is any ordinal, then ${\mathbb{Q}}(N)$ denotes the finite support product of $N$ copies of ${\mathbb{Q}}$, ordered component-wise. In this paper, we shall only consider ${\mathbb{Q}}(N)$ for $N\leq\omega$. If $\alpha$ is a limit ordinal, then $<_{J_{\alpha}}$ denotes the canonical well-ordering of $J_{\alpha}$, see [15, Definition 5.14 and p. 79],555The reader unfamiliar with the $J$-hierarchy may read $L_{\alpha}$ instead of $J_{\alpha}$. and $<_{L}=\bigcup\,\{<_{J_{\alpha}}\colon\alpha$ is a limit ordinal $\}$.

Let us work in $L$ until further notice. Let us first define $(\alpha_{\xi},\beta_{\xi}\colon\xi<\omega_{1})$ as follows: $\alpha_{\xi}=$ the least $\alpha>{\rm sup}(\{\beta_{\bar{\xi}}\colon{\bar{\xi}}<\xi\})$ such that $J_{\alpha}\models{\sf ZFC}^{-}$,666Here, ${\sf ZFC}^{-}$ denotes ${\sf ZFC}$ without the power set axiom. Every $J_{\alpha}$ satisfies the strong form of ${\sf AC}$ according to which every set is the surjective image of some ordinal. In the absence of $V=L$, one has to be careful about how to formulate ${\sf ZFC}^{-}$, see [6]. and $\beta_{\xi}=$ the least $\beta>\alpha_{\xi}$ such that $\rho_{\omega}(J_{\beta})=\omega$ (see [15, Definition 11.22]; $\rho_{\omega}(J_{\beta})=\omega$ is equivalent with ${\cal P}(\omega)\cap J_{\beta+\omega}\not\subset J_{\beta}$).

We shall also make use of a sequence $(f_{\xi}\colon\xi<\omega)$ which is defined as follows. Let $({\bar{f}}_{\xi}\colon\xi<\omega)$ be defined by the following trivial recursion: ${\bar{f}}_{\xi}$ be the $<_{L}$-least $f$ such that $f\in({}^{\omega}J_{\omega_{1}}\cap J_{\omega_{1}})\setminus\{{\bar{f}}_{\bar{\xi}}\colon{\bar{\xi}}<\xi\}$. Then if $\pi$ denotes the Gödel pairing function, see [15, p. 35], we let $f_{\pi((\xi_{1},\xi_{2}))}={\bar{f}}_{\xi_{1}}$. We will then have that $f_{\xi}\in J_{\alpha_{\xi}}$ for all $\xi$, and for each $f\in({}^{\omega}J_{\omega_{1}}\cap J_{\omega_{1}})$ the set of $\xi$ such that $f=f_{\xi}$ is cofinal in $\omega_{1}$.

Let us then define $({\mathbb{P}}_{\xi},{\mathbb{Q}}_{\xi}\colon\xi\leq\omega_{1})$. Each ${\mathbb{P}}_{\xi}$ will consist of perfect trees $T\subset{}^{<\omega}2$ such that if $T\in{\mathbb{P}}_{\xi}$ and $s\in T$, then $T_{s}=\{t\in T\colon t\subset s\vee s\subset t\}\in{\mathbb{P}}_{\xi}$ as well.777We denote by $x\subset y$ the fact that $x$ is a (not necessarily proper) subset of $y$. Each ${\mathbb{P}}_{\xi}$ will be construed as a p.o. by stipulating $T\leq T^{\prime}$ ($T$ “is stronger than” $T^{\prime}$) iff $T\subset T^{\prime}$. We will have that ${\mathbb{P}}_{\xi}\in J_{\alpha_{\xi}}$ and ${\mathbb{P}}_{\bar{\xi}}\subset{\mathbb{P}}_{\xi}$ whenever ${\bar{\xi}}\leq\xi\leq\omega_{1}$.

To start with, let ${\mathbb{P}}_{0}$ be the set of all basic clopen sets $U_{s}=\{t\in{}^{<\omega}2\colon t\subset s\vee s\subset t\}$, where $s\in{}^{\omega}2$. If $\lambda\leq\omega_{1}$ is a limit ordinal, then ${\mathbb{P}}_{\lambda}=\bigcup\{{\mathbb{P}}_{\xi}\colon\xi<\lambda\}$.

Now fix $\xi<\omega_{1}$, and suppose that ${\mathbb{P}}_{\xi}$ has already been defined. We shall define ${\mathbb{Q}}_{\xi}$ and ${\mathbb{P}}_{\xi+1}$.

Let $g_{\xi}\in{}^{\omega}J_{\alpha_{\xi}}$ be the following $\omega$-sequence. If there is some $N<\omega$ such that $f_{\xi}$ is an $\omega$-sequence of subsets of ${\mathbb{P}}_{\xi}(N)$, each of which is predense in ${\mathbb{P}}_{\xi}(N)$, then for each $n<\omega$ let $g_{\xi}(n)$ be the open dense set

[TABLE]

and write $N_{\xi}=N$. Otherwise we just set $g_{\xi}(n)={\mathbb{P}}_{\xi}(1)$ for each $n<\omega$, and write $N_{\xi}=1$. Let $d_{\xi}$ be the $<_{J_{\beta_{\xi}+\omega}}$-least $d\in{}^{\omega\times\omega}({\cal P}({\mathbb{P}}_{\xi})\cap J_{\alpha_{\xi}})\cap J_{\beta_{\xi}+\omega}$ such that

1. (i)

for each $(n,N)\in\omega\times\omega$, $d(n,N)$ is an open dense subset of ${\mathbb{P}}_{\xi}(N)$ which exists in $J_{\beta_{\xi}}$, 2. (ii)

for each $N<\omega$ and each open dense subset $D$ of ${\mathbb{P}}_{\xi}(N)$ which exists in $J_{\beta_{\xi}}$ there is some $n<\omega$ with $d(n,N)\subset D$, 3. (iii)

$d(n,N_{\xi})\subset g_{\xi}(n)$ for each $n<\omega$, and 4. (iv)

$d(n+1,N)\subset d(n,N)$ for each $(n,N)\in\omega\times\omega$.

Let us now look at the collection of all systems $(T^{m}_{s}\colon m<\omega,s\in{}^{<\omega}2)$ with the following properties.

1. (a)

$T^{m}_{s}\in{\mathbb{P}}_{\xi}$ for all $m$, $s$, 2. (b)

for each $T\in{\mathbb{P}}_{\xi}$ there are infinitely many $m<\omega$ with $T^{m}_{\emptyset}=T$, 3. (c)

$T^{m}_{t}\leq T^{m}_{s}$ for all $m$, $t\supset s$, 4. (d)

${\rm stem}(T^{m}_{s^{\frown}0})$ and ${\rm stem}(T^{m}_{s^{\frown}1})$ are incompatible elements of $T^{m}_{s}$ for all $m$, $s$, 5. (e)

if $(m,s)\neq(m^{\prime},s^{\prime})$, where $m$,$m^{\prime}<n$ and ${\rm lh}(s)={\rm lh}(s^{\prime})=n+1$ for some $n$, then ${\rm stem}(T^{m}_{s})$ and ${\rm stem}(T^{m^{\prime}}_{s^{\prime}})$ are incompatible, and 6. (f)

for all $N\leq n<\omega$ and all pairwise different $(m_{1},s_{1})$, $\ldots$, $(m_{N},s_{N})$ with $m_{1}$, $\ldots$, $m_{N}<n$ and $s_{1}$, $\ldots$, $s_{N}\in{}^{n+1}2$,

[TABLE]

It is easy to work in $J_{\beta_{\xi}+\omega}$ and construct initial segments $(T^{m}_{s}\colon m<\omega,s\in{}^{<\omega}2,{\rm lh}(s)\leq n)$ of such a system by induction on $n<\omega$. Notice that (f) formulates a constraint only for $m_{1}$, $\ldots$, $m_{N}<{\rm lh}(s_{1})-1=\ldots={\rm lh}(s_{N})-1$, and writing $n={\rm lh}(s_{1})-1$, there are $\sum_{N=1}^{n}\frac{(n\cdot 2^{n+1})!}{(n\cdot 2^{n+1}-N)!}$ (i.e., finitely many) such constraints.

We let $(T^{m}_{s,\xi}\colon m<\omega,s\in{}^{<\omega}2)$ be the $<_{\beta_{\xi}+\omega}$-least such system $(T^{m}_{s}\colon m<\omega,s\in{}^{<\omega}2)$. For every $m<\omega$, $s\in{}^{<\omega}2$, we let

[TABLE]

Notice that (e) implies that

[TABLE]

(5) will imply that $A^{m}_{s,\xi}$ and $A^{m^{\prime}}_{s^{\prime},\xi}$ will be incompatible in every ${\mathbb{P}}_{\eta}$, $\eta>\xi$, unless $m=m^{\prime}$ and $s\subset s^{\prime}$ or $s^{\prime}\subset s$.

We set ${\mathbb{Q}}_{\xi}=\{A^{m}_{s,\xi}\colon m<\omega,s\in{}^{<\omega}2\}$. Finally, we set ${\mathbb{P}}_{\xi+1}={\mathbb{P}}_{\xi}\cup{\mathbb{Q}}_{\xi}$.

Lemma 1.2

Let $N<\omega$, $\xi<\omega_{1}$.

[TABLE]

is dense in ${\mathbb{P}}_{\xi+1}(N)$.888Here, $\mbox{stem}(T_{i})\perp\mbox{stem}(T_{j})$ means that the stem of $T_{i}$ is incompatible with the stem of $T_{j}$.

Proof. Let $(T_{1},\ldots,T_{N})\in{\mathbb{P}}_{\xi+1}(N)$. For $i\in\{1,\ldots,N\}$ such that $T_{i}\in{\mathbb{P}}_{\xi}$ pick some $m_{i}<\omega$ such that $T_{i}=T^{m_{i}}_{\emptyset,\xi}$, and write $s_{i}=\emptyset$. This is possible by (b). If $i\in\{1,\ldots,N\}$ is such that $T_{i}\in{\mathbb{Q}}_{\xi}$, then say $T_{i}=A^{m_{i}}_{s_{i},\xi}$. Now pick $n>{\rm max}(\{m_{1},\ldots,m_{N}\})$ and $t_{1}\supset s_{1}$, $\ldots$, $t_{N}\supset s_{N}$ such that ${\rm lh}(t_{1})=\ldots={\rm lh}(t_{N})=n+1$ and the $(m_{i},t_{i})$ are pairwise different.

Then by (e) the finite sequences ${\rm stem}(T^{m_{i}}_{t_{i},\xi})$ are pairwise incompatible, so that by $A^{m_{i}}_{t_{i},\xi}\leq T^{m_{i}}_{t_{i},\xi}$, the $A^{m_{i}}_{t_{i},\xi}$ are pairwise incompatible. But then $(A^{m_{1}}_{t_{i},\xi},\ldots,A^{m_{N}}_{t_{N},\xi})\in D$ and $(A^{m_{1}}_{t_{i},\xi},\ldots,A^{m_{N}}_{t_{N},\xi})\leq(T_{1},\ldots,T_{N})$. $\square$

Lemma 1.3

(Sealing)* Let $N<\omega$, $\xi<\omega_{1}$. If $D\in J_{\beta_{\xi}}$ is predense in ${\mathbb{P}}_{\xi}(N)$, then $D$ is predense in all ${\mathbb{P}}_{\eta}(N)$, $\eta\geq\xi$, $\eta\leq\omega_{1}$.*

Proof by induction on $\eta$. The cases $\eta=\xi$ and $\eta$ being a limit ordinal are trivial. Suppose $\eta\geq\xi$, $\eta<\omega_{1}$, and $D$ is predense in ${\mathbb{P}}_{\eta}(N)$. Write $D^{\prime}=\{(T_{1},\ldots,T_{N})\in{\mathbb{P}}_{\eta}(N)\colon\exists(T^{\prime}_{1},\ldots,T^{\prime}_{N})\in D\,(T_{1},\ldots,T_{N})\leq(T^{\prime}_{1},\ldots,T^{\prime}_{N})\}$. As $\beta_{\xi}\leq\beta_{\eta}$, $D^{\prime}\in J_{\beta_{\xi}}$ and by (ii) and (iv) there is some $n_{0}<\omega$ with $d_{\eta}(n,N)\subset D^{\prime}$ for every $n>n_{0}$.

To show that $D^{\prime}$ (and hence $D$) is predense in ${\mathbb{P}}_{\eta+1}(N)$, by Lemma 1.2 it suffices to show that for all $(T_{1},\ldots,T_{N})\in{\mathbb{Q}}_{\eta}(N)$ there is some $(T^{\prime}_{1},\ldots,T^{\prime}_{N})\in{\mathbb{Q}}_{\eta}(N)$, $(T^{\prime}_{1},\ldots,T^{\prime}_{N})\leq(T_{1},\ldots,T_{N})$, and $(T^{\prime}_{1},\ldots,T^{\prime}_{N})$ is below some element of $D^{\prime}$.

So let $(A^{m_{1}}_{s_{1},\eta},\ldots,A^{m_{N}}_{s_{N},\eta})\in{\mathbb{Q}}_{\eta}(N)$ be arbitrary. Let

[TABLE]

and let $t_{1}\supset s_{1}$, $\ldots$, $t_{N}\supset s_{N}$ be such that ${\rm lh}(t_{1})=\ldots={\rm lh}(t_{N})=n+1$. By increasing $n$ further if necessary, we may certainly assume that $t_{1}$, $\ldots$, $t_{N}$ are picked in such a way that $(m_{1},t_{1})$, $\ldots$, $(m_{N},t_{N})$ are pairwise different. Then

[TABLE]

by (f). But

[TABLE]

and also

[TABLE]

which means that $(A^{m_{1}}_{s_{1},\eta},\ldots,A^{m_{N}}_{s_{N},\eta})$ is compatible with an element of $D^{\prime}$. $\square$

Corollary 1.4

Let $N<\omega$, $\xi<\omega_{1}$.

[TABLE]

is predense in ${\mathbb{P}}(N)$.

Lemma 1.5

Let $N<\omega$. ${\mathbb{P}}(N)$ has the c.c.c.

Proof. Let $A\subset{\mathbb{P}}(N)$ be a maximal antichain, $A\in L$. Let $j\colon J_{\beta}\rightarrow J_{\omega_{2}}$ be elementary and such that $\beta<\omega_{1}$ and $\{{\mathbb{P}},A\}\subset{\rm ran}(j)$. Write $\xi={\rm crit}(j)$. We have that $j^{-1}({\mathbb{P}}(N))={\mathbb{P}}(N)\cap J_{\xi}={\mathbb{P}}_{\xi}(N)$ and $j^{-1}(A)=A\cap J_{\xi}=A\cap{\mathbb{P}}_{\xi}(N)\in J_{\beta}$ is a maximal antichain in ${\mathbb{P}}_{\xi}(N)$. Moreover, $\beta_{\xi}>\beta$, so that by Lemma 1.4 $A\cap{\mathbb{P}}_{\xi}(N)$ is predense in ${\mathbb{P}}(N)$. This means that $A=A\cap{\mathbb{P}}_{\xi}$ is countable. $\square$

Lemma 1.6

Let $N<\omega$. $(c_{1},\ldots c_{N})\in{}^{N}({}^{\omega}2)$ is ${\mathbb{P}}(N)$-generic over $L$ iff for all $\xi<\omega_{1}$ there is an injection $t\colon\{1,\ldots,N\}\rightarrow{\mathbb{Q}}_{\xi}$ such that for all $i\in\{1,\ldots,N\}$, $c_{i}\in[t(i)]$.

Proof. “$\Longrightarrow$”: This readily follows from Corollary 1.4.

“$\Longleftarrow$”: Let $A\subset{\mathbb{P}}(N)$ be a maximal antichain, $A\in L$. By Lemma 1.5, we may certainly pick some $\xi<\omega_{1}$ with $A\subset{\mathbb{P}}_{\xi}(N)$ and $A\in J_{\alpha_{\xi}}$. Say $n_{0}$ is such that $d_{\xi}(n,N)\subset\{(T_{1},\ldots,T_{N})\in{\mathbb{P}}_{\xi}\colon\exists(T^{\prime}_{1},\ldots,T^{\prime}_{N})\in A\,(T_{1},\ldots,T_{N})\leq(T^{\prime}_{1},\ldots,T^{\prime}_{N})\}$ for all $n\geq n_{0}$. By our hypothesis, we may pick pairwise different $(m_{1},s_{1})$, $\ldots$, $(m_{N},s_{N})$ with ${\rm lh}(s_{1})=\ldots={\rm lh}(s_{N})=n+1$ for some $n\geq n_{0}$ and $c_{i}\in[T^{m_{i}}_{s_{i}},\xi]$ for all $i\in\{1,\ldots,N\}$. But then $(T^{m_{1}}_{s_{i},\xi},\ldots,T^{m_{N}}_{s_{N}})$ is below an element of $A$, which means that the generic filter given by $(c_{1},\ldots,c_{N})$ meets $A$. $\square$

Corollary 1.7

Let $N<\omega$, and let $(c_{1},\ldots c_{N})\in{}^{N}({}^{\omega}2)$ be ${\mathbb{P}}(N)$-generic over $L$. If $x\in L[(c_{1},\ldots c_{N})]$ is ${\mathbb{P}}$-generic over $L$, then $x\in\{c_{1},\ldots c_{N}\}$.

Proof. If $x\in L[(c_{1},\ldots c_{N})]$ is ${\mathbb{P}}$-generic over $L$, then $(c_{1},\ldots c_{N},x)\in{}^{N+1}({}^{\omega}2)$ is ${\mathbb{P}}(N+1)$-generic over $L$, hence $x\notin L[(c_{1},\ldots c_{N})]$. Contradiction! $\square$

Corollary 1.8

Let $N<\omega$, and let $(c_{1},\ldots c_{N})\in{}^{N}({}^{\omega}2)$ be ${\mathbb{P}}(N)$-generic over $L$. Then inside $L[(c_{1},\ldots c_{N})]$, $\{c_{1},\ldots c_{N}\}$ is a (lightface) $\Pi^{1}_{2}$ set.

Proof. Let $\varphi(x)$ express that for all $\xi<\omega_{1}$ there is some $T\in{\mathbb{Q}}_{\xi}$ such that $x\in[T]$. The formula $\varphi(x)$ may be written in a $\Pi^{1}_{2}$ fashion, and it defines $\{c_{1},\ldots c_{N}\}$ inside $L[(c_{1},\ldots c_{N})]$. $\square$

Lemma 1.9

(Sacks property)* Let $N<\omega$, and let $g$ be ${\mathbb{P}}(N)$-generic over $L$. For each $f\colon\omega\rightarrow\omega$, $f\in L[a]$, there is some $g\in L$ with domain $\omega$ such that for each $n<\omega$, $f(n)\in g(n)$ and 999In what follows, the only thing that will matter is that the bound on ${\rm Card}(g(n))$ only depends on $n$ and not on the particular $g$. ${\rm Card}(g(n))\leq(n+1)\cdot 2^{n+1}$.*

Proof. Let $\tau\in L^{{\mathbb{P}}(N)}$, $\tau^{g}=f$. Let $(A_{n}\colon n<\omega)\in L$ be such that for each $n$, $A_{n}$ is a maximal antichain of ${\vec{T}}\in{\mathbb{P}}(N)$ such that $\exists m<\omega\,{\vec{T}}\Vdash\tau({\check{n}})={\check{m}}$. We may pick some $\xi<\omega_{1}$ such that $\bigcup\{A_{n}\colon n<\omega\}\subset{\mathbb{P}}_{\xi}(N)$ and $(A_{n}\colon n<\omega)=f_{\xi}$.

By Lemma 1.6, there are pairwise different $(m_{1},s_{1})$, $\ldots$, $(m_{N},s_{N})$ such that

[TABLE]

Let

[TABLE]

If $t_{1}\supset s_{1}$, $\ldots$, $t_{N}\supset t_{N}$ are such that ${\rm lh}(t_{1})=\ldots={\rm lh}(t_{N})=n+1$, then $(T^{m_{1}}_{t_{1},\xi},\ldots,T^{m_{N}}_{t_{N},\xi})\in d_{\xi}(n,N)\subset A_{n}$, so that also

[TABLE]

Therefore, if we let

[TABLE]

then $(A^{m_{1}}_{s_{1},\xi},\ldots,A^{m_{N}}_{s_{N},\xi})\Vdash\tau({\check{n}})\in(g(n)){\check{}}$ , hence $f(n)\in g(n)$, and ${\rm Card}(g(n))=N\cdot 2^{n+1}\leq(n+1)\cdot 2^{n+1}$ for all but finitely many $n$. $\square$

2 The variant of the Cohen-Helpern-Lévy model.

Let us force with ${\mathbb{P}}(\omega)$ over $L$, and let $g$ be a generic filter. Let $c_{n}$, $n<\omega$, denote the Jensen reals which $g$ adds. Let us write $A=\{c_{n}\colon n<\omega\}$ for the set of those Jensen reals. The model

[TABLE]

of all sets which inside $L[g]$ are hereditarily definable from parameters in ${\rm OR}\cup A\cup\{A\}$ is the variant of the Cohen–Halpern–Lévy model (over $L$) which we shall work with. For the case of Jensen’s original forcing this model was first considered in [5].

For any finite $a\subset A$, we write $L[a]$ for the model constructed from the finitely many reals in $a$.

Lemma 2.1

Inside $H$, $A$ is a (lightface) $\Pi^{1}_{2}$ set.

Proof. Let $\varphi(-)$ be the $\Pi^{1}_{2}$ formula from the proof of Lemma 1.8. If $H\models\varphi(x)$, $x\in L[a]$, $a\in[A]^{<\omega}$, then $L[a]\models\varphi(x)$ by Shoenfield, so $x\in a\subset A$. On the other hand, if $c\in A$, then $L[c]\models\varphi(c)$ and hence $H\models\varphi(c)$ again by Shoenfield. $\square$

Fixing some Gödelization of formulae (or some enumeration of all the rud functions, resp.) at the outset, each $L[a]$, $a\in[A]^{<\omega}$, comes with a unique canonical global well–ordering $<_{a}$ of $L[a]$ by which we mean the one which is induced by the natural order of the elements of $a$ and the fixed Gödelization device in the usual fashion. The assignment $a\mapsto<_{a}$, $a\in[A]^{<\omega}$, is hence in $H$.101010More precisely, the ternary relation consisting of all $(a,x,y)$ such that $x<_{a}y$ is definable over $H$. This is a crucial fact.

Let us fix a bijection

[TABLE]

and let us write $((n)_{0},(n)_{1})=e(n)$.

We shall also make use the following. Cf. [2, Lemma 1.2].

Lemma 2.2

(1) Let $a\in[A]^{<\omega}$ and $X\subset L[a]$, $X\in H$, say $X\in{\sf HOD}^{L[g]}_{b\cup\{A\}}$, where $b\supseteq a$, $b\in[A]^{<\omega}$. Then $X\in L[b]$.

(2) There is no well–ordering of the reals in $H$.

(3) $A$ has no countable subset in $H$.

(4) $[A]^{<\omega}$ has no countable subset in $H$.

Proof sketch. (1) Every permutation $\pi\colon\omega\rightarrow\omega$ induces an automorphism $e_{\pi}$ of ${\mathbb{P}}(\omega)$ by sending $p$ to $q$, where $q(\pi(n))=p(n)$ for all $n<\omega$. It is clear that no $e_{\pi}$ moves the canonical name for $A$, call it ${\dot{A}}$. Let us also write ${\dot{c}}_{n}$ for the canonical name for $c_{n}$, $n<\omega$. Now if $a$, and $b$ are as in the statement of (1), say $b=\{c_{n_{1}},\ldots,c_{n_{k}}\}$, if $p$, $q\in{\mathbb{P}}(\omega)$, if $\pi\upharpoonright\{n_{1},\ldots,n_{k}\}={\rm id}$, $p\upharpoonright\{n_{1},\ldots,n_{k}\}$ is compatible with $q\upharpoonright\{n_{1},\ldots,n_{k}\}$, and ${\rm supp}(\pi(p))\cap{\rm supp}(q)\subseteq\{n_{1},\ldots,n_{k}\}$, if $x\in L$, if $\alpha_{1}$, $\ldots$, $\alpha_{m}$ are ordinals, and if $\varphi$ is a formula, then

[TABLE]

and $\pi(p)$ is compatible with $q$, so that the statement $\varphi({\check{x}},{\check{\alpha}}_{1},\ldots{\check{\alpha}}_{m},{\dot{c}}_{n_{1}},\ldots{\dot{c}}_{n_{k}},{\dot{A}})$ will be decided by conditions $p\in{\mathbb{P}}(\omega)$ with ${\rm supp}(p)\subseteq\{n_{1},\ldots,n_{k}\}$. But every set in $L[b]$ is coded by a set of ordinals, so if $X$ is as in (1), this shows that $X\in L[b]$.

(2) Every real is a subset of $L$. Hence by (1), if $L[g]$ had a well–ordering of the reals in ${\sf HOD}^{L[g]}_{a\cup\{A\}}$, some $a\in[A]^{<\omega}$, then every real of $H$ would be in $L[a]$, which is nonsense.

(3) Assume that $f\colon\omega\rightarrow A$ is injective, $f\in H$. Let $x\in{}^{\omega}\omega$ be defined by $x(n)=f((n)_{0})((n)_{1})$, so that $x\in H$. By (1), $x\in L[a]$ for some $a\in[A]^{<\omega}$. But then ${\rm ran}(f)\subset L[a]$, which is nonsense, as there is some $n<\omega$ such that $c_{n}\in{\rm ran}(f)\setminus a$.

(4) This readily follows from (3). $\square$ (Lemma 2.2)

Let us recall another standard fact.

[TABLE]

To see this, let us assume without loss of generality that $a\setminus b\not=\emptyset\not=b\setminus a$, and say $a\setminus b=\{c_{n}\colon n\in I\}$ and $b\setminus a=\{c_{n}\colon n\in J\}$, where $I$ and $J$ are non–empty disjoint finite subsets of $\omega$. Then $a\setminus b$ and $b\setminus a$ are mutually ${\mathbb{P}}(I)$- and ${\mathbb{P}}(J)$-generic over $L[a\cap b]$. But then $L[a]\cap L[b]=L[a\cap b][a\setminus b]\cap L[a\cap b][b\setminus a]=L[a\cap b]$, cf. [15, Problem 6.12].

For any $a\in[A]^{<\omega}$, we write ${\mathbb{R}}_{a}={\mathbb{R}}\cap L[a]$ and ${\mathbb{R}}^{+}_{a}={\mathbb{R}}_{a}\setminus\bigcup\{{\mathbb{R}}_{b}\colon b\subsetneq a\}$. $({\mathbb{R}}^{+}_{a}\colon a\in[A]^{<\omega})$ is a partition of ${\mathbb{R}}$: By Lemma 2.2 (1),

[TABLE]

and ${\mathbb{R}}_{a}\cap{\mathbb{R}}_{b}={\mathbb{R}}_{a\cap b}$ by (8), so that

[TABLE]

For $x\in{\mathbb{R}}$, we shall also write $a(x)$ for the unique $a\in[A]^{<\omega}$ such that $x\in{\mathbb{R}}^{+}_{a}$, and we shall write $\#(x)={\rm Card}(a(x))$.

Adrian Mathias showed that in the original Cohen–Halpern–Lévy model there is an definable function which assigns to each $x$ an ordering $<_{x}$ such that $<_{x}$ is a well–ordering iff $x$ can be well–ordered, cf. [11, p. 182]. The following is a special simple case of this, adapted to the current model $H$.

Lemma 2.3

(A. Mathias) In $H$, the union of countably many countable sets of reals is countable.

Proof. Let us work inside $H$. Let $(A_{n}\colon n<\omega)$ be such that for each $n<\omega$, $A_{n}\subset{\mathbb{R}}$ and there exists some surjection $f\colon\omega\rightarrow A_{n}$. For each such pair $n$, $f$ let $y_{n,f}\in{}^{\omega}\omega$ be such that $y_{n,f}(m)=f((m)_{0})((m)_{1})$. If $a\in[A]^{<\omega}$ and $y_{n,f}\in{\mathbb{R}}_{a}$, then $A_{n}\in L[a]$. By (8), for each $n$ there is a unique $a_{n}\in[A]^{<\omega}$ such that $A_{n}\in L[a_{n}]$ and $b\supset a_{n}$ for each $b\in[A]^{<\omega}$ such that $A_{n}\in L[b]$. Notice that $A_{n}$ is also countable in $L[a_{n}]$.

Using the function $n\mapsto a_{n}$, an easy recursion yields a surjection $g\colon\omega\rightarrow\bigcup\{a_{n}\colon n<\omega\}$: first enumerate the finitely many elements of $a_{0}$ according to their natural order, then enumerate the finitely many elements of $a_{1}$ according to their natural order, etc. As $A$ has no countable subset, $\bigcup\{a_{n}\colon n<\omega\}$ must be finite, say $a=\bigcup\{a_{n}\colon n<\omega\}\in[A]^{<\omega}$. But then $\{A_{n}\colon n<\omega\}\subset L[a]$. (We don’t claim $(A_{n}\colon n<\omega)\in L[a]$.)

For each $n<\omega$, we may now let $f_{n}$ the $<_{a}$–least surjection $f\colon\omega\rightarrow A_{n}$. Then $f(n)=f_{(n)_{0}}((n)_{1})$ for $n<\omega$ defines a surjection from $\omega$ onto $\bigcup\{A_{n}\colon n<\omega\}$, as desired. $\square$ (Lemma 2.3)

The following is not true in the original Cohen–Halpern–Lévy model. Its proof exploits the Sacks property, Lemma 1.9.

Lemma 2.4

(1) Let $M\in H$ be a null set in $H$. There is then a $G_{\delta}$ null set $M^{\prime}$ with $M^{\prime}\supset M$ whose code is in $L$.

(2) Let $M\in H$ be a meager set in $H$. There is then an $F_{\sigma}$ meager set $M^{\prime}$ with $M^{\prime}\supset M$ whose code is in $L$.

Proof. (1) Let $M\in H$ be a null set in $H$.

Let us work in $H$. Let $(\epsilon_{n}\colon n<\omega)$ be any sequence of positive reals. Let $\bigcup_{s\in X}\,U_{s}\supset H$, where $X\subset{}^{<\omega}2$ and $\mu(\bigcup\{U_{s}\colon s\in X\})\leq\epsilon_{0}$.111111Here, $\mu$ denotes Lebesge measure. Let $e\colon\omega\rightarrow X$ be onto. Let $(k_{n}\colon n<\omega)$ be defined by: $k_{n}=$ the smallest $k$ (strictly bigger than $k_{n-1}$ if $n>0$) such that $\mu(\bigcup\{U_{s}\colon s\in e\mbox{''}\omega\setminus k\})\leq\epsilon_{n}$. Write $k_{-1}=0$. We then have that $\mu(\bigcup\{U_{s}\colon s\in e\mbox{''}[k_{n-1},k_{n})\})\leq\epsilon_{n}$ for every $n<\omega$.

Now fix $\epsilon>0$. Let

[TABLE]

and let $(k_{n}\colon n<\omega)$ and $e\colon\omega\rightarrow{}^{<\omega}2$ be such that $\bigcup_{s\in X}\,U_{s}\supset H$ and $\mu(\bigcup\{U_{s}\colon s\in e\mbox{''}[k_{n-1},k_{n})\})\leq\epsilon_{n}$ for every $n<\omega$. We may now apply Lemma 1.9 inside $L[a]$ for some $a\in[A]^{<\omega}$ such that $\{e,(k_{n}\colon n<\omega)\}\subset L[a]$ and find a function $g\in L$ with domain $\omega$ such that for each $n<\omega$, $g(n)$ is a finite union $U_{n}$ of basic open sets such that $\{U_{s}\colon s\in e\mbox{''}[k_{n-1},k_{n})\}\subset U_{n}$ and $\mu(U_{n})\leq\frac{1}{2^{n+1}}$. But then ${\cal O}=\bigcup\{O_{n}\colon n<\omega\}\supset M$ is open, ${\cal O}$ is coded in $L$ (i.e., there is $Y\in L$, $Y\subset{}^{<\omega}2$, with ${\cal O}=\bigcup\{U_{s}\colon s\in Y\}$), and $\mu({\cal O})\leq\epsilon$.

We may hence for every $n<\omega$ let ${\cal O}_{n}$ be an open set with ${\cal O}_{n}\supset M$, $\mu({\cal O}_{n})\leq\frac{1}{n+1}$, and whose code in $L$ is $<_{L}$-least among all the codes giving such a set. Then $\bigcap\{{\cal O}_{n}\colon n<\omega\}$ is a $G_{\delta}$ null set with code in $L$ and which covers $M$.

(2) Let $M\in H$ be a meager set in $H$, say $M=\bigcup\{N_{n}\colon n<\omega\}$, where each $N_{n}$ is nowhere dense.

Let us again work in $H$. It is easy to verify that a set $P\subset{}^{\omega}2$ is nowhere dense iff there is some $z\in{}^{\omega}2$ and some strictly increasing $(k_{n}\colon n<\omega)$ such that for all $n<\omega$,

[TABLE]

Look at $f\colon\omega\rightarrow\omega$, where $f(m)=k_{n+1}$ for the least $n$ with $m\leq k_{n}$. We may first apply Lemma 1.9 inside $L[a]$ for some $a\in[A]^{<\omega}$ such that $f\in L[a]$ and get a function $g\colon\omega\rightarrow\omega$, $g\in L$, such that $g(m)\geq f(m)$ for all $m<\omega$. Write $\ell_{0}=0$ and $\ell_{n+1}=g(\ell_{n})$, so that for each $n$ there is some $n^{\prime}$ with

[TABLE]

Define $e\colon\omega\rightarrow\omega$ by $e(n)=\sum_{q=0}^{n}\,(q+1)\cdot 2^{q+1}$. We may now apply Lemma 1.9 inside $L[a]$ for some $a\in[A]^{<\omega}$ such that $f\in L[a]$ and get some $n\mapsto(z^{n}_{i}\colon i\leq(n+1)\cdot 2^{n+1})$ inside $L$ such that for all $n$, $i$, $z^{n}_{i}\colon e(n)\rightarrow 2$, and for all $n$ there is some $i$ with $z\upharpoonright e(n)=z^{n}_{i}$. From this we get some $z^{\prime}\colon\omega\rightarrow\omega$, $z^{\prime}\in L$, such that for all $n$ there is some $n^{\prime}$ with $z^{\prime}\upharpoonright[\ell_{n^{\prime}},\ell_{n^{\prime}+1})=z\upharpoonright[\ell_{n^{\prime}},\ell_{n^{\prime}+1})$. But then, writing

[TABLE]

$D\in L$, and $D$ is open and dense.

We may hence for every $n<\omega$ let ${\cal O}_{n}$ be an open dense set with ${\cal O}_{n}\cap N_{n}=\emptyset$, whose code in $L$ is $<_{L}$-least among all the codes giving such a set. Then $\bigcup\{{}^{\omega}2\setminus{\cal O}_{n}\colon n<\omega\}$ is an $F_{\sigma}$ meager set with code in $L$ and which covers $M$. $\square$

Corollary 2.5

In $H$, there is a $\Delta^{1}_{2}$ Sierpiński set as well as a $\Delta^{1}_{2}$ Luzin set.

Proof. There is a $\Delta^{1}_{2}$ Luzin set in $L$. By Lemma 2.4 (2), any such set is still a Luzin set in $H$. The same is true with “Luzin” replaced by “Sierpiński” and Lemma 2.4 (2) replaced by Lemma 2.4 (1). $\square$

Lemma 2.6

In $H$, there is a $\Delta^{1}_{3}$ Bernstein set.

Proof. In this proof, let us think of reals as elements of the Cantor space ${}^{\omega}2$. Let us work in $H$.

We let

[TABLE]

Obviously, $B\cap B^{\prime}=\emptyset$.

Let $P\subset{\mathbb{R}}$ be perfect. We aim to see that $P\cap B\not=\emptyset\not=P\cap B^{\prime}$.

Say $P=[T]=\{x\in{}^{\omega}2\colon\forall n\,x\upharpoonright n\in T\}$, where $T\subseteq{}^{<\omega}2$ is a perfect tree. Modulo some fixed natural bijection ${}^{<\omega}2\leftrightarrow\omega$, we may identify $T$ with a real. By (9), we may pick some $a\in[A]^{<\omega}$ such that $T\in L[a]$. Say ${\rm Card}(a)<2^{n}$, where $n$ is even.

Let $b\in[A]^{2^{n+1}}$, $b\supset a$, and let $x\in{\mathbb{R}}^{+}_{b}$. In particular, $\#(x)=2^{n+1}$. It is easy to work in $L[b]$ and construct some $z\in[T]$ such that $x\leq_{T}z\oplus T$,121212Here, $(x\oplus y)(2n)=x(n)$ and $(x\oplus y)(2n+1)=y(n)$, $n<\omega$. e.g., arrange that if $z\upharpoonright m$ is the $k^{\rm th}$ splitting node of $T$ along $z$, where $k\leq m<\omega$, then $z(m)=0$ if $x(k)=0$ and $z(m)=1$ if $x(k)=1$.

If we had $\#(z)\leq 2^{n}$, then $\#(z\oplus T)\leq\#(z)+\#(T)<2^{n}+2^{n}=2^{n+1}$, so that $\#(x)<2^{n+1}$ by $x\leq_{T}z\oplus T$. Contradiction! Hence $\#(z)>2^{n}$. By $z\in L[b]$, $\#(z)\leq 2^{n+1}$. Therefore, $z\in P\cap B$.

The same argument shows that $P\cap B^{\prime}\not=\emptyset$. $B$ (and also $B^{\prime}$) is thus a Bernstein set.

We have that $x\in B$ iff

[TABLE]

which is true iff

[TABLE]

By Lemma 2.1, this shows that $B$ is $\Delta^{1}_{3}$. $\square$

Recall that for any $a\in[A]^{<\omega}$, we write ${\mathbb{R}}_{a}={\mathbb{R}}\cap L[a]$. Let us now also write ${\mathbb{R}}_{<a}={\rm span}(\bigcup\{{\mathbb{R}}_{b}\colon b\subsetneq a\})$, and ${\mathbb{R}}^{*}_{a}={\mathbb{R}}_{a}\setminus{\mathbb{R}}_{<a}$. In particular, ${\mathbb{R}}_{<\emptyset}=\{0\}$ by our above convention that ${\rm span}(\emptyset)=\{0\}$, and ${\mathbb{R}}^{*}_{\emptyset}=({\mathbb{R}}\cap L)\setminus\{0\}$.

The proof of Claim 2.8 below will show that

[TABLE]

Also, we have that ${\mathbb{R}}_{a}^{*}\subset{\mathbb{R}}_{a}^{+}$, so that by (10),

[TABLE]

Lemma 2.7

In $H$, there is a $\Delta^{1}_{3}$ Hamel basis.

Proof. We call $X\subset{\mathbb{R}}^{*}_{a}$ linearly independent over ${\mathbb{R}}_{<a}$ iff whenever

[TABLE]

where $m\in{\mathbb{N}}$, $m\geq 1$, and $q_{n}\in{\mathbb{Q}}$ and $x_{n}\in X$ for all $n$, $1\leq n\leq m$, then $q_{1}=\ldots=q_{m}=0$. In other words, $X\subset{\mathbb{R}}^{*}_{a}$ is linearly independent over ${\mathbb{R}}_{<a}$ iff

[TABLE]

We call $X\subset{\mathbb{R}}^{*}_{a}$ maximal linearly independent over ${\mathbb{R}}_{<a}$ iff $X$ is linearly independent over ${\mathbb{R}}_{<a}$ and no $Y\supsetneq X$, $Y\subset{\mathbb{R}}_{a}^{*}$ is still linearly independent over ${\mathbb{R}}_{<a}$. In particular, $X\subset{\mathbb{R}}^{*}_{\emptyset}=({\mathbb{R}}\cap L)\setminus\{0\}$ is linearly independent over ${\mathbb{R}}_{<\emptyset}=\{0\}$ iff $X$ is a Hamel basis for ${\mathbb{R}}\cap L$.

For any $a\in[A]^{<\omega}$, we let $b_{a}=\{x_{i}^{a}\colon i<\theta^{a}\}$, some $\theta^{a}\leq\omega_{1}$, be the unique set such that

1. (i)

for each $i<\theta^{a}$, $x_{i}^{a}$ is the $<_{a}$-least $x\in{\mathbb{R}}_{a}^{*}$ such that $\{x_{j}^{a}\colon j<i\}\cup\{x\}$ is linearly independent over ${\mathbb{R}}_{<a}$, and 2. (ii)

$b_{a}$ is maximal linearly independent over ${\mathbb{R}}_{<a}$.

By the above crucial fact, the function $a\mapsto b_{a}$ is well–defined and exists inside $H$. In particular,

[TABLE]

is an element of $H$.

We claim that $B$ is a Hamel basis for the reals of $H$, which will be established by Claims 2.8 and 2.9.

Claim 2.8

${\mathbb{R}}\cap H\subset{\rm span}(B)$.

Proof of Claim 2.8. Assume not, and let $n<\omega$ be the least size of some $a\in[A]^{<\omega}$ such that ${\mathbb{R}}^{*}_{a}\setminus{\rm span}(B)\not=\emptyset$. Pick $x\in{\mathbb{R}}^{*}_{a}\setminus{\rm span}(B)\not=\emptyset$, where ${\rm Card}(a)=n$.

We must have $n>0$, as $b_{\emptyset}$ is a Hamel basis for the reals of $L$. Then, by the maximality of $b_{a}$, while $b_{a}$ is linearly independent over ${\mathbb{R}}_{<a}$, $b_{a}\cup\{x\}$ cannot be linearly independent over ${\mathbb{R}}_{<a}$. This means that there are $q\in{\mathbb{Q}}$, $q\not=0$, $m\in{\mathbb{N}}$, $m\geq 1$, and $q_{n}\in{\mathbb{Q}}\setminus\{0\}$ and $x_{n}\in b_{a}$ for all $n$, $1\leq n\leq m$, such that

[TABLE]

By the definition of ${\mathbb{R}}_{<a}$ and the minimality of $n$, $z\in{\rm span}(\bigcup\{b_{c}\colon c\subsetneq a\})$, which then clearly implies that $x\in{\rm span}(\bigcup\{b_{c}\colon c\subseteq a\})\subset{\rm span}(B)$.

This is a contradiction! $\square$ (Claim 2.8)

Claim 2.9

$B$* is linearly independent.*

Proof of Claim 2.9. Assume not. This means that there are $1\leq k<\omega$, $a_{i}\in[A]^{<\omega}$ pairwise different, $m_{i}\in{\mathbb{N}}$, $m_{i}\geq 1$ for $1\leq i\leq k$, and $q_{n}^{i}\in{\mathbb{Q}}\setminus\{0\}$ and $x_{n}^{i}\in b_{a_{i}}$ for all $i$ and $n$ with $1\leq i\leq k$ and $1\leq n\leq m_{i}$ such that

[TABLE]

By the properties of $b_{a_{i}}$, $\sum_{n=1}^{m_{i}}\,q_{n}^{i}\cdot x_{n}^{i}\in{\mathbb{R}}^{*}_{a_{i}}$, so that (16) buys us that there are $z_{i}\in{\mathbb{R}}_{a_{i}}^{*}$, $z_{i}\not=0$, $1\leq i\leq k$, such that

[TABLE]

There must be some $i$ such that there is no $j$ with $a_{j}\supsetneq a_{i}$, which implies that $a_{j}\cap a_{i}\subsetneq a_{i}$ for all $j\not=i$. Let us assume without loss of generality that $a_{j}\cap a_{1}\subsetneq a_{1}$ for all $j$, $1<j\leq k$.

Let $a_{1}=\{c_{\ell}\colon\ell\in I\}$, where $I\in[\omega]^{<\omega}$, and let $a_{j}\cap a_{1}=\{c_{\ell}\colon\ell\in I_{j}\}$, where $I_{j}\subsetneq I$, for $1<j\leq l$.

In what follows, a nice name $\tau$ for a real is a name of the form

[TABLE]

where each $A_{n,m}$ is a maximal antichain of conditions of the forcing in question deciding that $\tau({\check{n}})={\check{m}}$.

We have that $z_{1}$ is ${\mathbb{P}}(I)$–generic over $L$, so that we may pick a nice name $\tau_{1}\in L^{{\mathbb{P}}(I)}$ for $z_{1}$ with $(\tau_{1})^{g\upharpoonright I}=z_{1}$. Similarly, for $1<j\leq k$, $z_{j}$ is ${\mathbb{P}}(I_{j})$–generic over $L[g\upharpoonright(\omega\setminus I)]$, so that we may pick a nice name $\tau_{j}\in L[g\upharpoonright(\omega\setminus I)]^{{\mathbb{P}}(I_{j})}$ for $z_{j}$ with $(\tau_{j})^{g\upharpoonright I_{j}}=z_{j}$. We may construe each $\tau_{j}$, $1<j\leq k$, as a name in $L[g\upharpoonright(\omega\setminus I)]^{{\mathbb{P}}(I)}$ by replacing each $p\colon I_{j}\rightarrow{\mathbb{P}}$ in an antichain as in (18) by $p^{\prime}\colon I\rightarrow{\mathbb{P}}$, where $p^{\prime}(\ell)=p(\ell)$ for $\ell\in I_{j}$ and $p^{\prime}(\ell)=\emptyset$ otherwise. Let $p\in g\upharpoonright I$ be such that

[TABLE]

We now have that inside $L[g\upharpoonright(\omega\setminus I)]$, there are nice ${{\mathbb{P}}(I)}$–names $\tau_{j}^{\prime}$, $1<j\leq k$ (namey, $\tau_{j}$, $1<j\leq k$), such that still inside $L[g\upharpoonright(\omega\setminus I)]$

1. (1)

$p\Vdash^{{\mathbb{P}}(I)}\tau_{1}+\tau_{2}^{\prime}+\ldots+\tau_{k}^{\prime}=0$, and 2. (2)

for all $j$, $1<j\leq k$ and for all $p$ in one of the antichains of the nice name $\tau_{j}^{\prime}$, ${\rm supp}(p)\subseteq I_{j}$.

By Lemma 1.5, the nice names $\tau_{1}$, $\tau_{2}^{\prime}$, $\ldots$, $\tau_{k}^{\prime}$ may be coded by reals, and both (1) and (2) are arithmetic in such real codes for $\tau_{1}$,$\tau_{2}^{\prime}$, $\ldots$, $\tau_{k}^{\prime}$, so that by $\tau_{1}\in L^{{\mathbb{P}}(I)}$ and $\Sigma^{1}_{1}$–absoluteness between $L$ and $L[g\upharpoonright(\omega\setminus I)]$ there are inside $L$ nice ${{\mathbb{P}}(I)}$–names $\tau_{j}^{\prime}$, $1<j\leq k$, such that in $L$, (1) and (2) hold true. But then, writing $z_{j}^{\prime}=(\tau_{j}^{\prime})^{g\upharpoonright I}$, we have by (2) that $z_{j}^{\prime}\in{\mathbb{R}}_{I_{j}}$ for $1<j\leq k$, and $z_{1}+z_{2}^{\prime}+\ldots+z_{k}^{\prime}=0$ by (1). But then $z_{1}\in{\mathbb{R}}^{*}_{I}\cap{\mathbb{R}}_{<I}$, which is absurd. $\square$ (Claim 2.9)

We now have that $x\in B$ iff

[TABLE]

which is true iff

[TABLE]

By Lemma 2.1, this shows that $B$ is $\Delta^{1}_{3}$. $\square$

3 Open questions.

We finish by stating some open problems.

(1) Is there a model of ${\sf ZF}$ plus $\lnot{\sf AC}_{\omega}({\mathbb{R}})$ where there are sets as in (a)-(d) of Theorem 0.1 of lower projective complexity?

(2) Does the model $H$ from (6) on p. 6 have a Burstin basis? An affirmative answer along the lines of the argument from [3] would require us to show that

[TABLE]

where $s^{0}$ denotes the Marczewski ideal. We don’t know if (19) is true, though, we don’t even know if

[TABLE]

L. Wu and L. Yu have recently shown that (20) is true for ${\rm Card}(a)=2$, but it is not known if (20) holds true for ${\rm Card}(a)=3$. The second author has shown that if $A$ is a countable set of Cohen reals over $L$ (or, for that matter, any countable set of dominating reals over $L$), then (20) is true for $a\in[A]^{<\omega}$ of arbitrary size, i.e., that (20) holds true for ${\mathbb{R}}_{a}^{*}$ as being defied in [2].

(3) Does the model $H$ from (6) on p. 6 have a Mazurkiewicz set?

We may force with the forcings from [3] and [1] to add a Burstin basis and a Mazurkiewicz set, respectively, over $H$ (without adding any reals), but then those sets won’t be definable in that extension.