Transcendence bases, well-orderings of the reals and the axiom of choice

TL;DR

This paper demonstrates the relative consistency of the existence of a transcendence basis for the reals without a well-ordering of the reals, addressing a question in set theory and the axiom of choice.

Contribution

It proves the consistency of having a transcendence basis for the reals while lacking a well-ordering, resolving a question posed by Larson and Zapletal.

Findings

Consistency of $ZF+DC$ with a transcendence basis for reals

Non-existence of a well-ordering of the reals in the same model

Addresses a longstanding open question in set theory

Abstract

We prove that $ZF+DC+"$there exists a transcendence basis for the reals$"+"$there is no well-ordering of the reals$"$ is consistent relative to $ZFC$. This answers a question of Larson and Zapletal.

Full text

Transcendence bases, well-orderings of the reals and the axiom of choice

Haim Horowitz and Saharon Shelah

Abstract

We prove that $ZF+DC+"$there exists a transcendence basis for the reals$"+"$there is no well-ordering of the reals” is consistent relative to $ZFC$. This answers a question of Larson and Zapletal.111Date: January 27, 2019 2010 Mathematics Subject Classification: 03E25, 03E35, 03E40, 12F20 Keywords: transcendence basis, well-ordering, axiom of choice, forcing, amalgamation Publication 1093 of the second author Partially supported by European Research Council grant 338821.

Introduction

It’s well-known that the axiom of choice has far-reaching consequences for the structure of the real line. Among them, to name a few, are the existence of non-measurable sets of reals, nonprincipal ultrafilters on $\omega$, paradoxical decompositions of the unit sphere, mad families and more. As the aforementioned statements are consistently false over $ZF+DC$, it’s natural to study the possible implications between them in the absence of choice. This direction of study has gained considerable interest in recent years, with many consistency results showing mostly the independence over $ZF+DC$ between various properties of the real line implied by the axiom of choice. We mention several such examples:

**Theorem ([Sh:218]): **It’s consistent relative to an inaccessible cardinal that $ZF+DC$ holds, all set of reals are Lebesgue measurable and there is a set of reals without the Baire property.

**Theorem ([HwSh:1113]): **It’s consistent relative to an inaccessible cardinal that $ZF+DC$ holds, all sets of reals are Lebesgue measurable and there is a mad family.

**Theorem ([LaZa1]): **It’s consistent relative to a proper class of Woodin cardinals that there exists a mad family and there are no $\omega_{1}$ sequences of reals, nonatomic measures on $\omega$ and total selectors for $E_{0}$.

Our current paper will focus on two consequences of the axiom of choice for the real line, namely the existence of a transcendence basis for the reals and the existence of a well-ordering of the reals. The following question was asked by Larson and Zapletal in their forthcoming book:

**Question ([LaZa2]): **Does the existence of a transcendence basis for the reals imply the existence of a well-ordering of the reals?

We shall prove that the answer is negative, namely:

**Main result: $ZF+DC+"$**there exists a transcendence basis for the reals$"+"$there is no well-ordering of the reals$"$ is consistent relative to $ZFC$.

The proof strategy will be similar to that of [Sh:218] and [HwSh:1113] (though no inaccessible cardinals will be used in the current proof). Our forcing $\mathbb{P}$ will consist of conditions $p=(u_{p},\mathbb{Q}_{p},\underset{\sim}{R_{p}})$ where $\mathbb{Q}_{p}$ is a ccc forcing from some fixed $H(\lambda)$ that forces $MA_{\aleph_{1}}$ and $\underset{\sim}{R_{p}}$ is a set of $\mathbb{Q}_{p}$-names of reals that’s forced by $\mathbb{Q}_{p}$ to be a transcendence basis for the reals. The order will be defined naturally. The sets of the form $\underset{\sim}{R_{p}}$ will approximate a transcendence basis in the final model, while the forcing notions $\mathbb{Q}_{p}$ will help us to prove the non-existence of a well-ordering of the reals using a standard amalgamation argument. The fact that each $\mathbb{Q}_{p}$ forces $MA_{\aleph_{1}}$ will guarantee that the relevant amalgamation will be ccc.

Acknowledgement: We would like to thank Jindra Zapletal for informing us about a gap in a previous version of this paper.

The rest of the paper will be devoted to the proof of the main result mentioned above. We shall assume basic familiarity with amalgamation of forcing notions (see, e.g., [HwSh:1090]).

Proof of the main result

**Hypothesis 1: **Throughout the paper, we fix infinite regular cardinals $\lambda$ and $\kappa$ and an infinite cardinal $\mu$ such that $\mu=\mu^{\aleph_{1}}<\lambda$, $\kappa=\mu^{+}$ or $\aleph_{2}\leq cf(\kappa)\leq\kappa\leq\lambda$ and $(\forall\alpha<\kappa)([\alpha]^{\aleph_{1}}<\kappa)$.

**Definition 2: **We define the forcing notion $\mathbb{P}$ as follows:

A. $p\in\mathbb{P}$ iff $p=(u,\mathbb{Q},\underset{\sim}{R})=(u_{p},\mathbb{Q}_{p},\underset{\sim}{R}_{p})$ where:

a. $u\in[\lambda]^{<\kappa}$.

b. $\mathbb{Q}\in H(\lambda)$ is a ccc forcing such that $u$ is its underlying set of elements.

c. $\Vdash_{\mathbb{Q}}MA_{\aleph_{1}}$.

d. $\underset{\sim}{R}$ is a set of $\mathbb{Q}$-names of reals that is forced by $\mathbb{Q}$ to be a transcendence basis of the reals.

B. $p\leq_{\mathbb{P}}q$ iff

a. $u_{p}\subseteq u_{q}$.

b. $\mathbb{Q}_{p}\lessdot\mathbb{Q}_{q}$.

c. $\underset{\sim}{R}_{p}\subseteq\underset{\sim}{R}_{q}$.

**Definition 3: **We define the following $\mathbb{P}$ names:

a. $\underset{\sim}{\mathbb{Q}}=\cup\{\mathbb{Q}_{p}:p\in\underset{\sim}{G}_{\mathbb{P}}\}$.

b. $\underset{\sim}{R}=\cup\{\underset{\sim}{R}_{p}:p\in\underset{\sim}{G}_{\mathbb{P}}\}$.

**Claim 4: **a. $\mathbb{P}$ is a forcing notion of cardinality $\lambda^{<\kappa}$, preserving cardinals and cofinalities of cardinals $\leq\kappa$ and $>\lambda^{<\kappa}$.

b. If $\delta<\kappa$ is a limit ordinal and $\bar{p}=(p_{\alpha}:\alpha<\delta)$ is $\leq_{\mathbb{P}}$-increasing and continuous (i.e. $\alpha<\delta\rightarrow\underset{\beta<\alpha}{\cup}\mathbb{Q}_{p_{1+\beta}}\lessdot\mathbb{Q}_{p_{\alpha}}$), then $\bar{p}$ has an upper bound $p_{\delta}$ such that $\bar{p}\hat{}(p_{\delta})$ is $\leq_{\mathbb{P}}$-increasing continuous.

c. In clause (b), if $\aleph_{2}\leq cf(\delta)$, then $p_{\delta}$ can be chosen as the union of the $p_{\alpha}$s.

d. $\Vdash_{\mathbb{P}}"\underset{\sim}{\mathbb{Q}}$ is ccc and $\lambda$ is its underlying set of elements$"$.

e. $\Vdash_{\mathbb{P}}"\Vdash_{\underset{\sim}{\mathbb{Q}}}"\underset{\sim}{R}$ is a transcendence basis for the reals.

f. Every permutation $g$ of $\lambda$ naturally induces an automorphism $\hat{g}$ of $\mathbb{P}$ and $\underset{\sim}{\mathbb{Q}}$ which maps $\underset{\sim}{R}$ to itself.

**Proof: **a. By clause (b), $\mathbb{P}$ is $(<\kappa)$-complete, hence it preserves cardinals and cofinalities $\leq\kappa$. The rest should be straightforward.

b. As $\underset{\alpha<\delta}{\cup}\mathbb{Q}_{p_{\alpha}}$ is ccc, it can be extended to a ccc forcing $\mathbb{Q}_{p_{\delta}}$ such that $\underset{\alpha<\delta}{\cup}\mathbb{Q}_{p_{\alpha}}\lessdot\mathbb{Q}_{p_{\delta}}$ and $\Vdash_{\mathbb{Q}_{p_{\delta}}}MA_{\aleph_{1}}$. As the union of the $\underset{\sim}{R}_{p_{\alpha}}$ is algebraically independent, we can extend it to a transcendence basis for the reals.

c. Letting $\mathbb{Q}_{\delta}=\underset{\alpha<\delta}{\cup}\mathbb{Q}_{p_{\alpha}}$, obviously $\mathbb{Q}_{\delta}$ is ccc. In order to show that $\Vdash_{\mathbb{Q}_{\delta}}MA_{\aleph_{1}}$, it’s enough to show that for forcing notions of cardinality $\aleph_{1}$ in $V^{\mathbb{Q}_{\delta}}$. As $\aleph_{2}\leq cf(\delta)$, the names for a given ccc forcing in $V^{\mathbb{Q}_{\delta}}$ and $\aleph_{1}$-many of its dense subsets are already $\mathbb{Q}_{\alpha}$-names for some $\alpha<\delta$, and as $\Vdash_{\mathbb{Q}_{\alpha}}MA_{\aleph_{1}}$, we’re done. Similarly, every $\mathbb{Q}_{\delta}$-name for a real is already a $\mathbb{Q}_{\alpha}$-name for some $\alpha<\delta$, hence $\underset{\alpha<\delta}{\cup}\underset{\sim}{R}_{p_{\alpha}}$ is a $\mathbb{Q}_{\delta}$-name of a transcendence basis.

d. Let $G\subseteq\mathbb{P}$ be generic over $V$, we shall argue in $V[G]$. Given $I=\{q_{\alpha}:\alpha<\omega_{1}\}\subseteq\mathbb{Q}$, as $\mathbb{P}$ is $(<\kappa)$-complete, it doesn’t add new sequences of ordinals of length $\omega_{1}$, hence $I\in V$. For every $p\in\mathbb{P}$, there is some $q\in\mathbb{P}$ above $p$ such that $I\subseteq\mathbb{Q}_{q}$. Therefore, there is some $p\in G$ such that $I\subseteq\mathbb{Q}_{p}$. As $\mathbb{Q}_{p}$ is ccc, there are two elements of $I$ that are compatible in $\mathbb{Q}_{p}$ and hence they’re compatible in $\mathbb{Q}$. It follows that $\mathbb{Q}$ is ccc. By a similar density argument, for every $\alpha<\lambda$, there is some $p\in G$ such that $\alpha\in\mathbb{Q}_{p}$, hence $\lambda$ is the underlying set of elements of $\mathbb{Q}$.

e. As before, we shall argue in $V[G]$ where $G\subseteq\mathbb{P}$ is generic over $V$. The algebraic independence of $\underset{\sim}{R}$ follows from $G$ being directed. As for the maximality of $\underset{\sim}{R}$, as before, suppose that $\underset{\sim}{r}$ is a $\mathbb{Q}$-name for a real, then by a similar argument as in clause (d), there is $p\in G$ such that $\underset{\sim}{r}$ is a $\mathbb{Q}_{p}$-name. As $\underset{\sim}{R}_{p}$ is a $\mathbb{Q}_{p}$-name of a transcendence basis, we’re done.

f. This is straightforward. $\square$

**Definition/Observation 5: **Let $V_{1}$ be the model $HOD(\mathbb{R}^{<\kappa}\cup\{\underset{\sim}{R}\}\cup V)$ inside $V^{\mathbb{P}*\underset{\sim}{Q}}$, then $V_{1}$ is a model of $ZF+DC_{<\kappa}$ with the same reals as $V^{\mathbb{P}*\underset{\sim}{Q}}$. In particular, $V_{1}$ contains a transcendence basis for the reals (using Claim 4(e)). $\square$

We shall obtain the desired result by proving that there is no well ordering of the reals in $V_{1}$. Before that, we shall prove our main amalgamation claim:

**Main amalgamation claim 6: **(A) implies (B) where:

A. a. $\mathbb{Q}_{0}\lessdot\mathbb{Q}_{l}$ $(l=1,2)$.

b. $\Vdash_{\mathbb{Q}_{l}}"\bar{\underset{\sim}{r}_{l}}=(\underset{\sim}{r_{l,i}}:i<n_{l})$ is algebraically independent over $\mathbb{R}^{V^{\mathbb{Q}_{0}}}$.

c. $\mathbb{Q}=\mathbb{Q}_{1}\times_{\mathbb{Q}_{0}}\mathbb{Q}_{2}$.

B. $\Vdash_{\mathbb{Q}}"\bar{\underset{\sim}{r_{1}}}\hat{}\bar{\underset{\sim}{r_{2}}}$ is algebraically independent over $\mathbb{R}^{V^{\mathbb{Q}_{0}}}$.

**Proof: **Assume towards contradiction that there is a counterexample to the claim. As forcing with $\mathbb{Q}$ is the same as forcing with $\mathbb{Q}_{0}*((\mathbb{Q}_{1}/\mathbb{Q}_{0})\times(\mathbb{Q}_{2}/\mathbb{Q}_{0}))$, if there is a counterexample to the claim, then by working in $V^{\mathbb{Q}_{0}}$ we obtain a counterexample where $\mathbb{Q}_{0}$ is trivial and $\mathbb{Q}=\mathbb{Q}_{1}\times\mathbb{Q}_{2}$. Therefore, we may assume wlog that $\mathbb{Q}=\mathbb{Q}_{1}\times\mathbb{Q}_{2}$ and $\mathbb{Q}_{0}$ is trivial. We may also assume wlog that it’s forced by $\mathbb{Q}$ that $\bar{\underset{\sim}{r_{1}}}$ and $\bar{\underset{\sim}{r_{2}}}$ form a counterexample (if $(q_{1},q_{2})\in\mathbb{Q}_{1}\times\mathbb{Q}_{2}$ forces that $\bar{\underset{\sim}{r_{1}}}$ and $\bar{\underset{\sim}{r_{2}}}$ form a counterexample, then we can replace $\mathbb{Q}_{l}$ by $\mathbb{Q}_{l}\restriction q_{l}$ for $l=1,2$).

**Subclaim: **We may assume wlog that $\mathbb{Q}_{1}$ and $\mathbb{Q}_{2}$ are Cohen forcing.

Proof of Subclaim: Suppose that $\bar{x}=(\mathbb{Q}_{1},\mathbb{Q}_{2},\bar{\underset{\sim}{r_{1}}},\bar{\underset{\sim}{r_{2}}})$ form a counter example to the amalgamation claim, we shall construct a counter example $\bar{x^{\prime}}=(\mathbb{Q}_{1}^{\prime},\mathbb{Q}_{2}^{\prime},\bar{\underset{\sim}{r_{1}^{\prime}}},\bar{\underset{\sim}{r_{2}^{\prime}}})$ where $\mathbb{Q}_{1}^{\prime},\mathbb{Q}_{2}^{\prime}$ are Cohen forcing. As $\bar{x}$ is a counter example to the claim, there is a nontrivial polynomial $P=P(x_{0},...,x_{n_{1}-1},y_{0},...,y_{n_{2}-1})$ with coeficients in $\mathbb{R}^{V}$ and a condition $(p_{1},p_{2})\in\mathbb{Q}_{1}\times\mathbb{Q}_{2}$ such that $(p_{1},p_{2})\Vdash_{\mathbb{Q}_{1}\times\mathbb{Q}_{2}}"P(\bar{\underset{\sim}{r_{1}}},\bar{\underset{\sim}{r_{2}}})=0"$. We shall now choose $(\bar{p_{1,n}},\bar{p_{2,n}},\bar{a_{1,n}},\bar{a_{2,n}})$ by induction on $n<\omega$ such that the following conditions hold:

a. $\bar{p_{l,n}}=(p_{l,n,\nu}:\nu\in\omega^{n})$ $(l=1,2)$.

b. Each $p_{l,n,\nu}$ is a condition in $\mathbb{Q}_{l}$ $(l=1,2)$.

c. If $n=m+1$, $l\in\{1,2\}$ and $\nu\in\omega^{n}$ then $p_{l,m,\nu\restriction m}\leq p_{l,n,\nu}$.

d. $\bar{a_{l,n}}=(a_{l,n,\eta,i}^{-},a_{l,n,\eta,i}^{+}:\eta\in\omega^{n},i<n_{l})$.

e. $a_{l,n,\eta,i}^{-}$ and $a_{l,n,\eta,i}^{+}$ are rationals such that $a_{l,n,\eta,i}^{+}-a_{l,n,\eta,i}^{-}<\frac{1}{2^{n}}$.

f. $p_{l,n,\eta}\Vdash_{\mathbb{Q}_{l}}"\underset{i<n_{l}}{\wedge}a_{l,n,\eta,i}^{-}<\underset{\sim}{r_{l,i}}<a_{l,n,\eta,i}^{+}"$.

g. If $n=m+1$, $\rho\in\omega^{m}$, $l\in\{1,2\}$, $((a_{i},b_{i}):i<n_{l})$ is a sequence of pairs of rationals such that $a_{i}<b_{i}$ for $i<n_{l}$ and $p_{l,m,\rho}\nVdash_{\mathbb{Q}_{l}}"\neg(\underset{i<n_{l}}{\wedge}a_{i}<\underset{\sim}{r_{l,i}}<b_{i})"$, then for some $k<\omega$, $p_{l,n,\rho\hat{}(k)}\Vdash_{\mathbb{Q}_{l}}"\underset{i<n_{l}}{\wedge}a_{i}<\underset{\sim}{r_{l,i}}<b_{i}"$.

h. Moreover, we have $a_{i}<a_{l,n,\rho\hat{}(k),i}^{-}<a_{l,n,\rho\hat{}(k),i}^{+}<b_{i}$.

i. Moreover, if $n=m+1$ and $\nu_{1},\nu_{2}\in\omega^{m}$, then for some $k_{1}$ and $k_{2}$, letting $\rho_{l}=\nu_{l}\hat{}(k_{l})$ $(l=1,2)$ we have: For all $x_{1},...,x_{n_{1}},y_{1},...,y_{n_{2}}$, if $\underset{i<n_{1}}{\wedge}a_{l,n,\rho_{1},i}^{-}<x_{i}<a_{l,n,\rho_{1},i}^{+}$ and $\underset{j<n_{2}}{\wedge}a_{l,n,\rho_{2},j}^{-}<y_{j}<a_{l,n,\rho_{2},j}^{+}$ then $-\frac{1}{2^{n}}<P(x_{1},...,x_{n_{1}-1},y_{1},...,y_{n_{2}-1})<\frac{1}{2^{n}}$.

j. The $a_{l,n,\eta,i}^{-}$ are increasing with $\eta$ and the $a_{l,n,\eta,i}^{+}$ are decreasing with $\eta$.

The induction is straightorward where for clause (i) we use the fact that $(p_{1},p_{2})\Vdash_{\mathbb{Q}_{1}\times\mathbb{Q}_{2}}"P(\bar{\underset{\sim}{r_{1}}},\bar{\underset{\sim}{r_{2}}})=0"$.

For $l=1,2$ we define the following objects:

a. $\mathbb{Q}_{l}^{\prime}=(\omega^{<\omega},\leq)$.

b. $\underset{\sim}{\eta_{l}}$ is the name for the generic real of $\mathbb{Q}_{l}^{\prime}$.

c. For $i<n_{l}$, $\underset{\sim}{r_{l,i}^{\prime}}$ is the unique real in $\underset{n<\omega}{\cap}(a_{l,n,\underset{\sim}{\eta_{l}\restriction n,i}}^{-},a_{l,n,\underset{\sim}{\eta_{l}\restriction n,i}}^{+})$.

Now $\mathbb{Q}_{l}^{\prime}$ are equivalent to Cohen forcing, and by clause (i) of the induction, $\Vdash_{\mathbb{Q}_{1}^{\prime}\times\mathbb{Q}_{2}^{\prime}}"P(\underset{\sim}{r_{1}^{\prime}},\underset{\sim}{r_{2}^{\prime}})=0"$. Therefore, in order to prove the subclaim, it suffices to show that $\Vdash_{\mathbb{Q}_{l}^{\prime}}"\underset{\sim}{r_{l,1}^{\prime}},...,\underset{\sim}{r_{l,n_{1}-1}^{\prime}}$ are algebraically independent over $\mathbb{R}^{V}"$. Assume towards contradiction that there is some $\eta\in\mathbb{Q}_{l}^{\prime}$ and a nontrivial polynomial $P_{l}^{\prime}(x_{0},...,x_{n_{l}-1})$ such that $\eta\Vdash_{\mathbb{Q}_{l}^{\prime}}"P_{l}^{\prime}(\underset{\sim}{r_{l}^{\prime}})=0"$. By the assumption on $(\mathbb{Q}_{l},\underset{\sim}{r_{l}})$, letting $n=lg(\eta)$, $p_{l,n,\eta}\Vdash_{\mathbb{Q}_{l}}"P_{l}^{\prime}(\underset{\sim}{r_{l}})\neq 0"$. Let $G_{l}\subseteq\mathbb{Q}_{l}$ be generic over $V$ such that $p_{l,n,\eta}\in G_{l}$, so wlog $P_{l}^{\prime}(\underset{\sim}{r_{l}}[G_{l}])>0$. By continuity, there are rationals $a_{i}<b_{i}$ $(i<n_{l})$ such that $V[G]\models"$for every $x_{0},...,x_{n_{l}-1}$, $\underset{i<n_{l}}{\wedge}a_{i}<x_{i}<b_{i}\rightarrow P_{l}^{\prime}(x_{0},...,x_{n_{l}-1})>0$ and $\underset{\sim}{r_{l,i}}[G_{l}]\in(a_{i},b_{i})"$. Therefore, the first part of the statement holds in $V$ and there is some $q\in G_{l}$ such that $p_{l,n,\eta}\leq q$ and $q$ forces the second part of the statement. In particular, $p_{l,n,\eta}\nVdash_{\mathbb{Q}_{l}}"\neg(\underset{i<n_{l}}{\wedge}a_{i}<\underset{\sim}{r_{l,i}}<b_{i})"$. By clause (g) of the induction, there is some $k<\omega$ such that $p_{l,n+1,\eta\hat{}(k)}\Vdash_{\mathbb{Q}_{l}}"\underset{i<n_{l}}{\wedge}a_{i}<\underset{\sim}{r_{l,i}}<b_{i}"$ and $a_{i}<a_{l,n+1,\eta\hat{}(k),i}^{-}<a_{l,n+1,\eta\hat{}(k),i}^{+}<b_{i}$. Now $\eta\hat{}(k)$ is a condition in $\mathbb{Q}_{l}^{\prime}$ that forces in $\mathbb{Q}_{l}^{\prime}$ that $\underset{\sim}{r_{l,i}}^{\prime}\in(a_{i},b_{i})$ for all $i<n_{l}$. It follows that $\eta\hat{}(k)$ forces in $\mathbb{Q}_{l}^{\prime}$ that $P_{l}^{\prime}(\underset{\sim}{r_{l,0}}^{\prime},...,\underset{\sim}{r_{l,n_{l}-1}}^{\prime})>0$, contradicting the choise of $\eta$ and $P-l^{\prime}$. It follows that $\Vdash_{\mathbb{Q}_{l}^{\prime}}"\underset{\sim}{r_{l,1}^{\prime}},...,\underset{\sim}{r_{l,n_{1}-1}^{\prime}}$ are algebraically independent over $\mathbb{R}^{V}"$, which completes the proof of the subclaim.

We shall now return to the proof of the main amalgamation claim:

Let $\chi\geq\aleph_{1}$ be large enough and let $N$ be a countable elementary submodel of $(H(\chi),\in)$ such that $\mathbb{Q}_{l},\bar{\underset{\sim}{r_{l}}}\in N$ $(l=1,2)$. As $\mathbb{Q}_{l}$ is Cohen, there is a $\mathbb{Q}_{l}$-name $\underset{\sim}{\eta_{l}}$ for a Cohen real over $V$ that generates the generic for $\mathbb{Q}_{l}$. For each $l<3$ and $i<n_{l}$ there is a Borel function $\mathbb{B}_{l,i}$ such that $\underset{\sim}{r_{l,i}}=\mathbb{B}_{l,i}(\underset{\sim}{\eta_{l}})$, we may assume that the $\mathbb{B}_{l,i}$s belong to $N$ as well. Let $\eta_{1}^{\prime}\in V$ be Cohen over $N$, let $G_{2}\subseteq\mathbb{Q}_{2}$ be generic over $V$ and let $\eta_{2}=\underset{\sim}{\eta_{2}}[G_{2}]$. $\eta_{2}$ is Cohen over $V$ and is also generic over $N[\eta_{1}^{\prime}]$. Therefore, $(\eta_{1}^{\prime},\eta_{2})$ is generic for $\mathbb{Q}_{1}\times\mathbb{Q}_{2}$ over $N$. As it’s forced by $\mathbb{Q}_{1}\times\mathbb{Q}_{2}$ over $V$ that $\bar{\underset{\sim}{r_{1}}}\hat{}\bar{\underset{\sim}{r_{2}}}$ is a counterexample, there is a polynomial $P$ witnessing this, i.e. $V\models"\Vdash_{\mathbb{Q}_{1}\times\mathbb{Q}_{2}}"P(...,\mathbb{B}_{1,l}(\underset{\sim}{\eta_{1}^{\prime}}),...,...,\mathbb{B}_{2,l}(\underset{\sim}{\eta_{2}}),...)=0""$. By absoluteness, the same stetement holds in $N$. By the genericity over $N$ of $(\eta_{1}^{\prime},\eta_{2})$, $N[\eta_{1}^{\prime},\eta_{2}]\models P(...,\mathbb{B}_{1,l}(\eta_{1}^{\prime}),...,...,\mathbb{B}_{2,l}(\eta_{2}),...)=0$. Therefore, there is $p_{2}\in G_{2}\subseteq\mathbb{Q}_{2}$ such that $N[\eta_{1}^{\prime}]\models"p_{2}\Vdash_{\mathbb{Q}_{2}}"\bar{\underset{\sim}{r_{2}}}$ is not algebraically independent over $\mathbb{R}^{V}$, as witnessed by $(\mathbb{B}_{1,l}(\eta_{1}^{\prime}):l<n_{1})""$, and by absoluteness, the same holds in $V$. This contradicts assumption (A)(b) and completes the proof of the claim. $\square$

Before proving the relevant conclusion for $\mathbb{P}$, we need the following algebraic observation:

**Observation 7: **Let $p_{1},p_{2}\in\mathbb{P}$ and suppose that $p_{1}\leq p_{2}$. Denote $\mathbb{Q}_{p_{l}}$ by $\mathbb{Q}_{l}$ and $\underset{\sim}{R_{p_{l}}}$ by $\underset{\sim}{R_{l}}$ $(l=1,2)$. Then $\Vdash_{\mathbb{Q}_{2}}"\underset{\sim}{R_{2}}\setminus\underset{\sim}{R_{1}}$ is algebraically independent over $\mathbb{R}^{V^{\mathbb{Q}_{1}}}"$.

**Proof: **Suppose towards contradiction that there is some $q\in\mathbb{Q}_{2}$ and $\underset{\sim}{r_{0}},...,\underset{\sim}{r_{n_{2}-1}}$ (with no repetition) such that $q\Vdash_{\mathbb{Q}_{2}}"\underset{\sim}{r_{0}},...,\underset{\sim}{r_{n_{2}-1}}\in\underset{\sim}{R_{2}}\setminus\underset{\sim}{R_{1}}$ are not algebraically independent over $\mathbb{R}^{V^{\mathbb{Q}_{1}}}"$. By increasing $q$ if necessary, we may assume wlog that there is a non-trivial polynomial $P(x_{0},...,x_{n_{2}-1})$ over $\mathbb{R}^{V^{\mathbb{Q}_{1}}}$ such that $q\Vdash_{\mathbb{Q}_{2}}"P(\underset{\sim}{r_{0}},...,\underset{\sim}{r_{n_{2}-1}})=0"$. Therefore, there are $\mathbb{Q}_{1}$-names of reals $\underset{\sim}{s_{0}},...,\underset{\sim}{s_{n_{1}-1}}$ and a polynomial $Q(x_{0},...,x_{n_{2}-1},y_{0},...,y_{n_{1}-1})$ over the rationals such that $q\Vdash_{\mathbb{Q}_{2}}"Q(x_{0},...,x_{n_{2}-1},\underset{\sim}{s_{0}},...,\underset{\sim}{s_{n_{1}-1}})=P(x_{0},...,x_{n_{2}-1})"$. Recalling that $\underset{\sim}{R_{1}}$ is a $\mathbb{Q}_{1}$-name of a transcendence basis over the rationals, then by increasing $q$ if necessary, there are $\mathbb{Q}_{1}$-names of reals $\underset{\sim}{t_{0}},...,\underset{\sim}{t_{n_{0}-1}}$ such that $q\Vdash_{\mathbb{Q}_{2}}"\underset{\sim}{t_{0}},...,\underset{\sim}{t_{n_{0}-1}}\in\underset{\sim}{R_{1}}$ (with no repetition)$"$ and $q\Vdash_{\mathbb{Q}_{2}}"\underset{\sim}{s_{0}},...,\underset{\sim}{s_{n_{1}-1}}$ are algebraic over $\mathbb{Q}[\underset{\sim}{t_{0}},...,\underset{\sim}{t_{n_{0}-1}}]"$ (here $\mathbb{Q}$ denotes the field of rational numbers). It follows that $q\Vdash_{\mathbb{Q}_{2}}"\{\underset{\sim}{t_{0}},...,\underset{\sim}{t_{n_{0}-1}},\underset{\sim}{r_{0}},...\underset{\sim}{r_{n_{2}-1}}\}\subseteq\underset{\sim}{R_{2}}$ is not algebraically independent over the rationals$"$. By the choice of the $\underset{\sim}{t_{i}}$s and the $\underset{\sim}{r_{i}}$s, $q\Vdash_{\mathbb{Q}_{2}}"\underset{\sim}{t_{0}},...,\underset{\sim}{t_{n_{0}-1}},\underset{\sim}{r_{0}},...,\underset{\sim}{r_{n_{2}-1}}$ are without repetition”. Together, we get a contradiction to the definition of the conditions in $\mathbb{P}$ and the fact that $p_{2}\in\mathbb{P}$. $\square$

**Conclusion 8: **Suppose that $p_{1},p_{2}\in\mathbb{P}$ such that $p_{1}\leq p_{2}$. Let $g$ be a permutation of $\lambda$ of order $2$ such that $g\restriction u_{p_{1}}=id$ and $g^{\prime\prime}(u_{p_{2}})\cap u_{p_{2}}=u_{p_{1}}$, and let $p_{3}=\hat{g}(p_{2})$. Then there is $q\in\mathbb{P}$ such that $p_{2},p_{3}\leq q$ and $\mathbb{Q}_{p_{2}}\times_{\mathbb{Q}_{p_{1}}}\mathbb{Q}_{p_{3}}\lessdot\mathbb{Q}_{q}$.

**Proof: **Let $\mathbb{Q}=\mathbb{Q}_{p_{2}}\times_{\mathbb{Q}_{p_{1}}}\mathbb{Q}_{p_{3}}$. As $\mathbb{Q}_{1}$ is ccc and $\Vdash_{\mathbb{Q}_{1}}"MA_{\aleph_{1}}+\mathbb{Q}_{2}/\mathbb{Q}_{1}\models ccc+\mathbb{Q}_{3}/\mathbb{Q}_{1}\models ccc"$, it follows that $\mathbb{Q}$ is ccc (see e.g. [HwSh:1090] for details). By the previous observation, for $l=2,3$, $\Vdash_{\mathbb{Q}_{p_{l}}}"\underset{\sim}{R_{p_{l}}}\setminus\underset{\sim}{R_{p_{1}}}$ is algebraically independent over $\mathbb{R}^{V^{\mathbb{Q}_{p_{1}}}}"$. Therefore, by Claim 6, $\Vdash_{\mathbb{Q}}"(\underset{\sim}{R_{p_{2}}}\setminus\underset{\sim}{R_{p_{1}}})\cup(\underset{\sim}{R_{p_{3}}}\setminus\underset{\sim}{R_{p_{1}}})$ is algebraically independent over $\mathbb{R}^{V^{\mathbb{Q}_{1}}}"$. It follows that $\Vdash_{\mathbb{Q}}"\underset{\sim}{R_{p_{2}}}\cup\underset{\sim}{R_{p_{3}}}=\underset{\sim}{R_{p_{1}}}\cup(\underset{\sim}{R_{p_{2}}}\setminus\underset{\sim}{R_{p_{1}}})\cup(\underset{\sim}{R_{p_{3}}}\setminus\underset{\sim}{R_{p_{1}}})$ is algebraically independent over the rationals$"$ (recall that if $\{\alpha_{0},...,\alpha_{n-1}\}$ are algebraically independent over the rationals and $\{\alpha_{n},...,\alpha_{m-1}\}$ are algebraically independent over a field $\mathbb{F}$ containing $\mathbb{Q}\cup\{\alpha_{0},...,\alpha_{n-1}\}$, then $\{\alpha_{0},...,\alpha_{m-1}\}$ are algebraically independent over the rationals). By Hypothesis 1, there is a ccc forcing $\mathbb{Q}_{q}$ such that $\mathbb{Q}\lessdot\mathbb{Q}_{q}$, $\Vdash_{\mathbb{Q}_{q}}MA_{\aleph_{1}}$ and $|\mathbb{Q}_{q}|=u_{q}$ for some $u_{q}\in[\lambda]^{<\kappa}$. As $\Vdash_{\mathbb{Q}_{q}}"\underset{\sim}{R_{p_{2}}}\cup\underset{\sim}{R_{p_{3}}}$ are algebraically independent over the rationals$"$, there is a set $\underset{\sim}{R_{q}}$ of $\mathbb{Q}_{q}$-names of reals such that $\underset{\sim}{R_{p_{2}}}\cup\underset{\sim}{R_{p_{3}}}\subseteq\underset{\sim}{R_{q}}$ and $\Vdash_{\mathbb{Q}_{q}}"\underset{\sim}{R_{q}}$ is a transcendence basis for the reals$"$. Now let $q=(u_{q},\mathbb{Q}_{q},\underset{\sim}{R_{q}})$, it’s easy to verify that $q$ is as required. $\square$

Recalling Observation 5, we shall complete the proof of the main result of the paper by proving the following claim:

**Claim 9: **There is no well-ordering of the reals in $V_{1}$.

**Proof: **Assume towards contradiction that there are $(p_{1},r_{1})\in\mathbb{P}*\underset{\sim}{\mathbb{Q}}$ such that, over $V$, $(p_{1},r_{1})\Vdash_{\mathbb{P}*\underset{\sim}{\mathbb{Q}}}"\underset{\sim}{f}$ is a one-to-one function from $\mathbb{R}$ to $Ord"$ and such that $\underset{\sim}{f}$ is definable from $\underset{\sim}{R}$ and a sequence $(\underset{\sim}{\eta_{\epsilon}}:\epsilon<\epsilon(*))$ where $\epsilon(*)<\kappa$ and wlog each $\underset{\sim}{\eta}_{\epsilon}$ is a $\mathbb{Q}_{p_{1}}$ name for a real (by a similar argument as in claims 4(d) and 4(e), we can always extend $p_{1}$ to make this true). Choose $(p_{2},r_{2})\geq(p_{1},r_{1})$ and a name of a real $\underset{\sim}{r}$ such that $(p_{2},r_{2})\Vdash_{\mathbb{P}*\underset{\sim}{\mathbb{Q}}}"\underset{\sim}{r}\in\mathbb{R}^{V^{\mathbb{Q}_{p_{2}}}}\setminus\mathbb{R}^{V^{\mathbb{Q}_{p_{1}}}}"$, wlog $r_{2}\in\mathbb{Q}_{p_{2}}$, and by extending the condition if necessary, we may assume wlog that $(p_{2},r_{2})$ forces a value $\gamma$ to $\underset{\sim}{f}(\underset{\sim}{r})$.

Let $g$ be a permutation of $\lambda$ of order $2$ such that $g\restriction u_{p_{1}}=id$ and $g^{\prime\prime}(u_{p_{2}})\cap u_{p_{2}}=u_{p_{1}}$. We shall denote both of the induced automorphisms on $\mathbb{P}$ and $\mathbb{Q}$ by $\hat{g}$. Clearly, $\hat{g}(p_{1})=p_{1}$. Let $p_{3}=\hat{g}(p_{2})$ and $r_{3}=\hat{g}(r_{2})$. By the previous claims, there is $q\in\mathbb{P}$ such that $p_{2},p_{3}\leq q$ and $\mathbb{Q}_{p_{2}}\times_{\mathbb{Q}_{p_{1}}}\mathbb{Q}_{p_{3}}\lessdot\mathbb{Q}_{q}$, and by the construction of the amalgamation, there is $r\in\mathbb{Q}_{q}$ above $r_{2}$ and $r_{3}$. As $\Vdash_{\mathbb{P}*\underset{\sim}{\mathbb{Q}}}"\mathbb{R}^{V^{\mathbb{Q}_{p_{2}}}}\cap\mathbb{R}^{V^{\mathbb{Q}_{p_{3}}}}=\mathbb{R}^{V^{\mathbb{Q}_{p_{1}}}}"$, it follows that $(q,r)\Vdash_{\mathbb{P}*\underset{\sim}{\mathbb{Q}}}"\underset{\sim}{r}\neq g(\underset{\sim}{r})"$. As $(p_{2},r_{2})\leq(q,r)$, $(q,r)\Vdash_{\mathbb{P}*\underset{\sim}{\mathbb{Q}}}"\underset{\sim}{f}(\underset{\sim}{r})=\gamma"$. Recalling that $\underset{\sim}{f}$ is forced to be injective, we shall arrive at a contradiction by showing that $(q,r)\Vdash_{\mathbb{P}*\underset{\sim}{Q}}"\underset{\sim}{f}(\hat{g}(\underset{\sim}{r}))=\gamma"$. It’s enough to show that the statement is forced by $(p_{3},r_{3})=(\hat{g}(p_{2}),\hat{g}(r_{2}))$, and in order to show that, it suffices to show that $\underset{\sim}{f}=\hat{g}(\underset{\sim}{f})$. Recalling that each $\underset{\sim}{\eta_{\epsilon}}$ in the definition of $\underset{\sim}{f}$ is a $\mathbb{Q}_{p_{1}}$-name and that $g$ is the identity on $u_{p_{1}}$, it follows that $\hat{g}(\underset{\sim}{\eta_{\epsilon}})=\underset{\sim}{\eta_{\epsilon}}$. By Claim 4(f), $\underset{\sim}{R}$ is preserved by $\hat{g}$. As $\underset{\sim}{f}$ is definable from $\underset{\sim}{R}$ and $(\underset{\sim}{\eta_{\epsilon}}:\epsilon<\epsilon(*))$, it follows that $\hat{g}(\underset{\sim}{f})=\underset{\sim}{f}$. This completes the proof of the claim. $\square$

References

[HwSh:1090] Haim Horowitz and Saharon Shelah, Can you take Toernquist’s inaccessible away?, arXiv:1605.02419

[HwSh:1113] Haim Horowitz and Saharon Shelah, Madness and regularity properties, arXiv:1704.08327

[LaZa1] Paul Larson and Jindrich Zapletal, Canonical models for fragments of the axiom of choice, Journal of Symbolic Logic, 82:489-509, 2017

[LaZa2] Paul Larson and Jindrich Zapletal, Geometric set theory, preprint

[Sh:218] Saharon Shelah, On measure and category, Israel J. Math. 52 (1985) 110-114

(Haim Horowitz) Department of Mathematics

University of Toronto

Bahen Centre, 40 St. George St., Room 6290

Toronto, Ontario, Canada M5S 2E4

E-mail address: haim@math.toronto.edu

(Saharon Shelah) Einstein Institute of Mathematics

Edmond J. Safra Campus,

The Hebrew University of Jerusalem.

Givat Ram, Jerusalem 91904, Israel.

Department of Mathematics

Hill Center - Busch Campus,

Rutgers, The State University of New Jersey.

110 Frelinghuysen Road, Piscataway, NJ 08854-8019 USA

E-mail address: shelah@math.huji.ac.il