USUBA’S PRINCIPLE $\mathrm {UB}_\lambda $ CAN FAIL AT SINGULAR CARDINALS

Abstract

We answer a question of Usuba by showing that the combinatorial principle $\mathrm {UB}_\lambda $ can fail at a singular cardinal. Furthermore, $\lambda $ can be taken to be $\aleph _\omega .$

1 Introduction

In [5], Usuba introduced a new combinatorial principle, denoted $\mathrm {UB}_\lambda .$ ¹ He showed that $\mathrm {UB}_\lambda $ holds for all regular uncountable cardinals and that for singular cardinals, some very weak assumptions like weak square or even $\mathrm {ADS}_\lambda $ imply it. It is known that $\mathrm {ADS}_\lambda $ can fail for singular cardinals, for example if $\kappa $ is supercompact and $\lambda> \kappa $ is such that $\mathrm {cf}(\lambda ) < \kappa $ . Motivated by this results, Usuba asked the following question:

Question 1.1. [5, Question 2.11]

Is it consistent that $\mathrm {UB}_\lambda $ fails for some singular cardinal $\lambda $ ?

In this paper we give a positive answer to the above question by showing that Chang’s transfer principle $(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _1, \aleph _0)$ implies the failure of $\mathrm { UB}_{\aleph _\omega }$ if $\aleph _\omega $ is strong limit (see Theorem 3.1), where a stronger result is proved.

The paper is organized as follows. In Section 2, we present some preliminaries and results and then in Section 3, we prove our main result.

2 Some preliminaries

In this section we present some definitions and results that are needed for the later section of this paper. Let us start by introducing Usuba’s principle.

Definition 2.1. Let $\lambda $ be an uncountable cardinal. The principle $\mathrm {UB}_\lambda $ is the statement: there exists a function $f: [\lambda ^+]^{<\omega } \to \lambda ^+$ such that if $x, y \subseteq \lambda ^+$ are closed under f, $x \cap \lambda =y \cap \lambda $ and $\sup (x \cap \lambda )=\lambda $ , then $x \subseteq y$ or $y \subseteq x.$

It turned out this principle has many equivalent formulations. To state a few of it, let $S=\{x \subseteq \lambda : \sup (x)=\lambda \}$ , $\theta> \lambda $ be large enough regular and let $\lhd $ be a well-ordering of $H(\theta )$ . Then we have the following.

Lemma 2.2. [5]

The following are equivalent $:$

1. (1) $\mathrm {UB}_\lambda $ .

2. (2) If $M, N \prec (H(\theta ), \in , \lhd , \lambda , S, \ldots )$ are such that $M \cap \lambda =N \cap \lambda \in S,$ then either $M \cap \lambda ^+ \subseteq N \cap \lambda ^+$ or $N \cap \lambda ^+ \subseteq M \cap \lambda ^+$ .

3. (3) If $M, N \prec (H(\theta ), \in , \lhd , \lambda , S, \ldots )$ are such that $M \cap \lambda =N \cap \lambda \in S,$ and $\sup (M \cap \lambda ^+) \leq \sup (N \cap \lambda ^+)$ , then $M \cap \lambda ^+$ is an initial segment of $N \cap \lambda ^+.$

The principle $\mathrm {UB}_\lambda $ has many nice implications. Here we only consider its relation with Chang’s transfer principles which is also related to our work.

Definition 2.3. Suppose $\lambda> \mu $ are infinite cardinal. Chang’s transfer principle $(\lambda ^+, \lambda ) \twoheadrightarrow (\mu ^+, \mu )$ is the statement: if $\mathcal L$ is a countable first-order language which contains a unary predicate U, then for any $\mathcal L$ -structure $\mathcal M=(M, U^{\mathcal M}, \ldots )$ with $|M|=\lambda ^+$ and $|U^{\mathcal M}|=\lambda $ , there exists an elementary submodel $\mathcal N= (N, U^{\mathcal N}, \ldots )$ of $\mathcal M$ with $|N|=\mu ^+$ and $|U^{\mathcal N}|=\mu $ .

Given an infinite cardinal $\nu ,$ the transfer principle $(\lambda ^+, \lambda ) \twoheadrightarrow _{\leq \nu } (\mu ^+, \mu )$ is defined similarly, where we allow the language $\mathcal L$ to have size at most $\nu $ .

The next lemma shows the relation between $\mathrm {UB}_{\aleph _\omega }$ and Chang’s transfer principles.

Lemma 2.4. ([5, Corollary 4.2])

Suppose $\mathrm {UB}_{\aleph _{\omega }}$ holds. Then the Chang transfer principles $(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _{n+1}, \aleph _n)$ fail for all $1 \leq n < \omega .$

Remark 2.5. By [4], $(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _{n+1}, \aleph _n)$ fails for all $n \geq 3$ .

Since the consistency of the transfer principle $(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _{n+1}, \aleph _n)$ is open for $n = 1, 2,$ one cannot use the above result to get the consistent failure of $\mathrm {UB}_{\aleph _{\omega }}$ . In the next section we show that if $\aleph _\omega $ is strong limit, then $\mathrm {UB}_{\aleph _{\omega }}$ implies the failure of $(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _{1}, \aleph _0)$ as well, and hence by the results of [3] (see also [1, 2], where the consistency of $\text {GCH}+(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _{1}, \aleph _0)$ is proved using weaker large cardinal assumptions) $\mathrm {UB}_{\aleph _{\omega }}$ can fail. We also need the following notion.

Definition 2.6. An uncountable cardinal $\kappa $ is said to be Jonsson, if for every function $f:[\kappa ]^{<\omega } \to \kappa $ there exists a set $H \subseteq \kappa $ of order type $\kappa $ such that for each n, $f"[H]^n \neq \kappa .$

Notation 2.7. Given a model M and a subset A of M, by $cl(A, M)$ we mean the least substructure of M which includes A as a subset.

Lemma 2.8. Assume $\lambda $ is a singular strong limit cardinal of cofinality $\kappa .$ Then there is a model $M_0$ with vocabulary $\mathcal L_0$ such that $:$

1. (a) $|\mathcal L_0|=\kappa $ and $|M_0|=\lambda ^+.$

2. (b) If M is an $\mathcal L$ -structure which expands $M_0$ , $|\mathcal L|=\kappa $ , and M has Skolem functions, then for $\alpha _1, \alpha _2 < \lambda ^+,$ the following statements are equivalent $:$

   1. (†)_{α1, α2} For some submodels $N_1, N_2$ of M we have:

      • (α) $N_1 \cap \lambda =N_2 \cap \lambda $ is unbounded in $\lambda $ ,

      • (β) $\alpha _1 \in N_1 \setminus N_2$ and $\alpha _2 \in N_2 \setminus N_1$ .

   2. (‡)_{α1, α2} If $V_\ell =cl(\{\alpha _\ell \}, M) \cap \lambda $ , $\ell =1, 2,$ and $V=V_1 \cup V_2$ , then

      $$\begin{align*}\alpha_1 \notin cl(\{\alpha_2\} \cup V, M) \text{~}\& \text{~} \alpha_2 \notin cl(\{\alpha_1\} \cup V, M). \end{align*}$$

Proof Let $\langle \lambda _i: i<\kappa \rangle $ be an increasing sequence cofinal in $\lambda $ such that for all $i<\kappa , 2^{\lambda _i} < \lambda _{i+1}$ . For each $0<n<\omega ,$ let

$$\begin{align*}\langle F_{n, \alpha}: \alpha \in [\lambda_i, 2^{\lambda_i} ) \rangle \end{align*}$$

enumerate all functions from $\lambda _i$ into $\lambda _i.$ Let $M_0$ be defined as follows:

• • the universe of $M_0$ is $\lambda ^+.$

• • $<^{M_0}= \{ (\alpha , \beta ): \alpha < \beta < \lambda ^+ \}$ .

• • $c_i^{M_0}=\lambda _i$ .

• • $P^{M_0}=\{\alpha : \alpha < \lambda \}.$

• • $F_n^{M_0}$ is an $(n+1)$ -ary function such that:

  1. – if $i<\kappa , \alpha \in [\lambda _i, 2^{\lambda _i})$ and $\beta _0, \cdots , \beta _{n-1} < \lambda _i$ , then

     $$\begin{align*}F_n^{M_0}(\beta_0, \ldots, \beta_{n-1}, \alpha)=F_{n, \alpha}(\beta_0, \ldots, \beta_{n-1}), \end{align*}$$

  2. – in all other cases, $F_n^{M_0}(\beta _0, \ldots , \beta _{n-1}, \beta _n)=\beta _n.$

We show that the model $M_0$ is as required. Clause (a) clearly holds. To show that clause (b) is satisfied, let M be an $\mathcal L$ -structure which expands $M_0$ , $|\mathcal L|=\kappa $ , and suppose M has Skolem functions. Let also $\alpha _1, \alpha _2 < \lambda ^+.$

First suppose that $(\dagger )_{\alpha _1, \alpha _2}$ holds, and suppose that the models $N_1, N_2$ witness it. Let also $V_\ell =cl(\{\alpha _\ell \}, M) \cap \lambda $ , $\ell =1, 2$ . Clearly each $V_\ell $ is an unbounded subset of $\lambda $ . Let $V=cl(V_1 \cup V_2, M) \cap \lambda $ and set $N^*_\ell =cl(\{\alpha _\ell \} \cup V, M)$ .

Claim 2.9. $N^*_\ell \subseteq N_\ell ,$ for $\ell =1, 2.$

Proof Fix $\ell .$ Sine $\alpha _\ell \in N_\ell ,$

$$ \begin{align*}V_\ell=cl(\{\alpha_\ell\}, M) \cap \lambda \subseteq N_\ell \cap \lambda.\end{align*} $$

On the other hand, $N_1 \cap \lambda =N_2 \cap \lambda ,$ and hence

$$ \begin{align*}V_{3-\ell}=cl(\{\alpha_{3-\ell}\}, M) \cap \lambda \subseteq N_{3-\ell} \cap \lambda=N_\ell \cap \lambda.\end{align*} $$

It follows that $V_1 \cup V_2 \subseteq N_\ell \cap \lambda ,$ and hence

$$\begin{align*}V=cl(V_1 \cup V_2, M) \cap \lambda \subseteq N_\ell. \end{align*}$$

Thus, as $\{\alpha _\ell \} \cup V \subseteq N_\ell ,$ we have

$$\begin{align*}N^*_\ell=cl(\{\alpha_\ell\} \cup V, M) \subseteq N_\ell. \end{align*}$$

The result follows.

Claim 2.10. $\alpha _1 \in N^*_1\setminus N^*_2$ and $\alpha _2 \in N^*_2\setminus N^*_1$ .

Proof Fix $\ell \in \{1, 2\}.$ Clearly $\alpha _\ell \in N^*_\ell $ . On the other hand, by our assumption, $\alpha _\ell \notin N_{3-\ell }$ , and by Claim 2.9, $N^*_{3-\ell } \subseteq N_{3-\ell }.$ Thus $\alpha _\ell \notin N^*_{3-\ell }$ .

Thus $(\ddagger )_{\alpha _1, \alpha _2}$ is satisfied.

Conversely suppose that $(\ddagger )_{\alpha _1, \alpha _2}$ holds, and for $\ell =1, 2,$ set $N_\ell =cl(\{\alpha _\ell \} \cup V, M)$ . By our assumption, clause ( $\beta $ ) of $(\dagger )_{\alpha _1, \alpha _2}$ holds.

Claim 2.11. For $\ell \in \{1, 2\}$ , $N_\ell \cap \lambda =V.$

Proof Fix $\ell \in \{1, 2\}$ . Clearly $N_\ell \cap \lambda \supseteq V.$ Now suppose towards a contradiction that $N_\ell \cap \lambda \neq V,$ and let $\gamma \in N_\ell \cap \lambda \setminus V.$ As M has Skolem functions, there are $n, \beta _0, \ldots , \beta _{n-1} \in V$ and $(n+1)$ -ary function symbol F in $\mathcal L$ such that

$$\begin{align*}\gamma=F^M(\beta_0, \ldots, \beta_{n-1}, \alpha_\ell). \end{align*}$$

As $\beta _0, \ldots , \beta _{n-1} \in V \subseteq \lambda $ and $\gamma < \lambda $ , there is $i<\kappa $ such that $\beta _0, \ldots , \beta _{n-1}, \gamma < \lambda _i$ . Define an n-ary function $G: \lambda _i \rightarrow \lambda _i$ as follows:

$G(\xi _0, \cdots , \xi _{n-1})=\ \left \{ \begin {array}{l} F^M(\xi _0, \cdots , \xi _{n-1}, \alpha _\ell ), \hspace {0.8cm} \text { if } F^M(\xi _0, \ldots , \xi _{n-1}, \alpha _\ell ) < \lambda _i ,\\ 0, \hspace {3.64cm} \text { otherwise}. \end {array} \right.$

Note that $G \in \{F_{n, \zeta }: \zeta \in [\lambda _i, 2^{\lambda _i}) \}$ . Let

$$\begin{align*}\zeta_*=\min\{\zeta: (\forall \xi_0, \ldots, \xi_{n-1} < c_i) G(\xi_0, \ldots, \xi_{n-1})=F^{M_0}_n(\xi_0, \ldots, \xi_{n-1}, \zeta) \}. \end{align*}$$

$\zeta _*$ is well-defined and is definable in M (even in $M_0$ ) from $\alpha _\ell $ , so clearly $\zeta _* \in cl(\{\alpha _\ell \}, M)$ .

As $\zeta _* \in cl(\{\alpha _\ell \}, M) \cap \lambda = V_\ell \subseteq V$ and $\beta _0, \ldots , \beta _{n-1} \in V,$ so

$$\begin{align*}\gamma=F^M(\beta_0, \ldots, \beta_{n-1}, \alpha_\ell) = F^M_{n, \zeta_*}(\beta_0, \ldots, \beta_{n-1}) \in V. \end{align*}$$

This contradicts our initial assumption that $\gamma \in N_\ell \cap \lambda \setminus V$ . The claim follows.

Claim 2.12. $N_1 \cap \lambda =N_2 \cap \lambda $ .

Proof By Claim 2.11, we have $ N_1 \cap \lambda =V=N_2 \cap \lambda ,$ which concludes the result.

By Claim 2.12, $N_1 \cap \lambda =N_2 \cap \lambda $ , which implies clause ( $\alpha $ ) of $(\dagger )_{\alpha _1, \alpha _2}$ . Thus $N_1$ and $N_2$ are as required in clause $(\dagger )_{\alpha _1, \alpha _2}$ .

This completes the proof of the lemma.

3 $\mathrm {UB}_\lambda $ can fail at singular cardinals

In this section we prove the following theorem which answers Usuba’s Question 1.1.

Theorem 3.1. Assume $\lambda $ is a singular strong limit cardinal. $\mathrm {UB}_\lambda $ fails if at least one of the following holds $:$

1. (a) $\lambda =\aleph _\omega $ and Chang’s transfer principle $(\lambda ^+, \lambda ) \twoheadrightarrow (\aleph _1, \aleph _0)$ holds.

2. (b) $\lambda> \mu \geq \mathrm {cf}(\lambda )$ are such that $(\lambda ^+, \lambda ) \twoheadrightarrow _{\leq \mathrm {cf}(\lambda )} (\mu ^+, \mu )$ holds.

3. (c) $\lambda> \mu \geq \mathrm {cf}(\lambda )$ and for every model M with universe $\lambda ^+$ and vocabulary of cardinality $\mathrm {cf}(\lambda )$ , we can find an increasing sequence $\vec {\alpha }=\langle \alpha _i: i < \mu ^+ \rangle $ of ordinals less than $\lambda ^+$ such that

   $$\begin{align*}S^M_{\vec{\alpha}}=\{i<\mu^+: cl(\{\alpha_i\}, M) \cap \lambda \subseteq cl(\{\alpha_j: j<i \}, M) \} \end{align*}$$
   is stationary in $\mu ^+$ .

4. (d) There exists $\chi $ with $\lambda> \chi =\mathrm {cf}(\chi ) > \mathrm {cf}(\lambda )$ such that for every model M with universe $\lambda ^+$ and vocabulary of cardinality $\mathrm {cf}(\lambda )$ , we can find an increasing sequence $\vec {\alpha }=\langle \alpha _i: i < \chi \rangle $ of ordinals less than $\lambda ^+$ such that

   $$\begin{align*}S^M_{\vec{\alpha}}=\{i<\chi: cl(\{\alpha_i\}, M) \cap \lambda \subseteq cl(\{\alpha_j: j<i \}, M) \} \end{align*}$$
   is stationary in $\chi $ .

5. (e) There is no sequence $\vec {X}=\langle U_i: i<\lambda ^+ \rangle $ such that each $U_i \cap \lambda $ is a cofinal subset of $\lambda $ , $U_i \cap \lambda $ has size $\mathrm {cf}(\lambda )$ , and for every $i<\lambda ^+$ there is a sequence $\vec {X}_i=\langle (\alpha _{i, j}, \beta _{i, j}): j<i \rangle $ such that $:$

   • • $\vec {X}_i$ has no repetition,

   • • $\alpha _{i, j} \in U_i$ ,

   • • $\beta _{i, j} \in U_j \cap \lambda $ .

Furthermore, the statement $(e)$ is equivalent to $\neg \mathrm {UB}_\lambda $ , provided that $\mathrm {cf}(\lambda )$ is not a Jonsson cardinal.

Remark 3.2. The assumption “ $\lambda $ is a strong limit cardinal” is only used in the proof of (e) implies $\neg \mathrm {UB}_\lambda $ .

Proof We prove the theorem by a sequence of claims. First note that:

Claim 3.3. Clause $(a)$ is a special case of clause $(b)$ , and clause $(c) $ implies clause $(d)$ .

Claim 3.4. $(b)$ implies $(c)$ .

Proof Let M be a model with universe $\lambda ^+$ and vocabulary of cardinality at most $\mathrm {cf}(\lambda )$ . By (b), there exists an elementary submodel $N \prec M$ such that $||N||=\mu ^+$ and $|N \cap \lambda |=\mu .$ Let $\vec {\alpha }=\langle \alpha _i: i<\mu ^+ \rangle $ list in increasing order the first $\mu ^+$ elements of $N.$ So for $i<\mu ^+$ we have

$$\begin{align*}cl(\{\alpha_i\}, M) \cap \lambda \subseteq N \cap \lambda, \end{align*}$$

and since $N \cap \lambda $ has size $\mu $ , we can find some $i(*)< \mu ^+$ such that

$$\begin{align*}\forall i<\mu^+,~ cl(\{\alpha_i\}, M) \cap \lambda \subseteq \bigcup\limits_{j<i(*)}cl(\{\alpha_j\}, M). \end{align*}$$

Hence the set $S^M_{\vec {\alpha }}$ includes $[i(*), \mu ^+)$ and so is stationary in $\mu ^+$ , as requested.

Claim 3.5. $(d)$ implies $(e)$ .

Proof Suppose towards a contradiction that (d) holds but (e) fails. As (e) fails, we can find sequences $\vec {X}=\langle U_i: i<\lambda ^+ \rangle $ and $\vec {X}_i=\langle (\alpha _{i, j}, \beta _{i, j}): j<i \rangle $ as in clause (e). Let M be a model in a vocabulary $\mathcal L$ such that:

1. (1) $|\mathcal L|=\mathrm {cf}(\lambda )$ ,

2. (2) M has universe $\lambda ^+$ ,

3. (3) $M=(\lambda ^+, \langle \tau ^M_i: i<\mathrm {cf}(\lambda )\rangle , H^M )$ , where

   1. (a) $\tau ^M_i=i,$

   2. (b) $H^M$ is a 2-place function such that for all i, $U_i \cap \lambda = \{ H^M(i, \alpha ): \alpha < \mathrm {cf}(\lambda ) \}$ .

Now by (d) applied to the model M, we can find a sequence $\vec {\zeta }= \langle \zeta _i: i<\chi \rangle $ of ordinals less than $\lambda ^+$ such that the set $S^M_{\vec {\zeta }}$ is stationary in $\chi $ . Let $\zeta =\sup \limits _{i<\chi }\zeta _i$ . Consider the sequence $\vec {X}_\zeta =\langle (\alpha _{\zeta , \xi }, \beta _{\zeta , \xi }): \xi < \zeta \rangle $ .

For $i<\chi $ , let

$$\begin{align*}W_i=cl(\{\zeta_j: j< i\}, M) \cap \lambda. \end{align*}$$

So $\langle W_i: i<\chi \rangle $ is a $\subseteq $ -increasing continuous sequence of sets each of cardinality $<\chi $ . Note that for each $i \in S^M_{\vec {\zeta }}$ ,

$$\begin{align*}\beta_{\zeta, \zeta_i} \in U_{\zeta_i} \cap \lambda \subseteq cl(\{\zeta_i\}, M) \cap \lambda \subseteq W_i. \end{align*}$$

(The former inclusion $\subseteq $ holds because $\mathrm {cf}(\lambda ) \cup \{\zeta _i \} \subseteq cl(\{\zeta _i\}, M)$ and $cl(\{\zeta _i\}, M)$ is closed under $H^M$ . The latter inclusion $\subseteq $ holds because $i \in S^M_{\vec {\zeta }}$ .) Then since $S^M_{\vec {\zeta }}$ is stationary in $\chi $ , there is $\beta _*$ such that

$$\begin{align*}U=\{i \in S^M_{\vec{\zeta}}: \beta_{\zeta, \zeta_i}=\beta_* \} \end{align*}$$

is stationary. Moreover, since $|U_\zeta |=\mathrm {cf}(\lambda ) < \chi ,$ we get some $i_i < i_2$ in U such that $\alpha _{\zeta , \zeta _{i_1}}=\alpha _{\zeta , \zeta _{i_2}}$ . This contradicts that $\vec {X}_\zeta $ has no repetition.

Claim 3.6. $(e)$ implies $\neg \mathrm {UB}_\lambda $ .

Proof Suppose not. Thus we can assume that both (e) and $\mathrm {UB}_\lambda $ hold. Let $f: [\lambda ^+]^{<\omega } \to \lambda ^+$ witness $\mathrm {UB}_\lambda .$ Choose a vocabulary $\mathcal L$ of size $\mathrm {cf}(\lambda )$ and an $\mathcal L$ -model M such that:

1. (1) M has universe $\lambda ^+.$

2. (2) M expands the model $M_0$ of Lemma 2.8, by expanding $\mathcal L_0$ (the vocabulary of $M_0$ ) using the constant symbols $\langle d^M_i: i< \mathrm {cf}(\lambda ) \rangle $ and the function symbols $( \langle F^M_n: n< \omega \rangle , p^M, G_1^M, G_2^M )$ , where:

   1. (a) $d_i^M=i$ for $i<\mathrm {cf}(\lambda )$ ,

   2. (b) $F^M_n$ is an n-ary function such that

      $$\begin{align*}F^M_n(\alpha_0, \ldots, \alpha_{n-1}) = f(\{\alpha_0, \ldots, \alpha_{n-1} \}), \end{align*}$$

   3. (c) $p^M$ is a pairing function on $\lambda ^+,$ mapping $\lambda \times \lambda $ onto $\lambda $ ,

   4. (d) $G_1^M$ and $G_2^M$ are 2-place functions such that for every $\alpha \in [\lambda , \lambda ^+)$ , $\langle G_1(\beta , \alpha ): \beta < \alpha \rangle $ enumerates $\lambda $ and

      $$\begin{align*}\big( \beta < \alpha ~\& ~ \gamma=G_1(\beta, \alpha) \big) \Rightarrow \beta=G_2(\gamma, \alpha). \end{align*}$$

By expanding M further, let us suppose that

1. (3) M contains Skolem functions.

For $\alpha < \lambda ^+,$ set $N_\alpha = cl(\{\alpha \}, M)$ .

$(*)_1$ $N_\alpha $ belongs to $[\lambda ^+]^{\mathrm {cf}(\lambda )}$ and it contains an unbounded subset of $\lambda $ .

Proof As $\mathcal L$ has size $\mathrm {cf}(\lambda )$ , so $|N_\alpha | \leq \mathrm {cf}(\lambda )$ . On the other hand, by clause (2)(a), $\mathrm {cf}(\lambda ) \subseteq N_\alpha $ and hence $N_\alpha $ belongs to $[\lambda ^+]^{\mathrm {cf}(\lambda )}$ . Also as $\{c_i^{M_0}: i < \mathrm {cf}(\lambda ) \} \subseteq N_\alpha $ (see the proof of Lemma 2.8) and $\langle c_i^{M_0}: i < \mathrm {cf}(\lambda ) \rangle $ is an unbounded sequence in $\lambda $ , we have $N_\alpha $ contains an unbounded subset of $\lambda $ .

Let

$$\begin{align*}E=\{ \delta \in (\lambda, \lambda^+): \delta = cl(\delta, M) \}. \end{align*}$$

E is clearly a club of $\lambda ^+$ and $E \cap \lambda =\emptyset .$ By Lemma 2.8, we have

$(*)_2$ Suppose $\xi < \zeta $ are in E. Then

$$\begin{align*}\xi \in cl\bigg(\{\zeta\} \cup (N_\xi \cap \lambda) \cup (N_\zeta \cap \lambda), M \bigg). \end{align*}$$

Proof Suppose by the way of contradiction that $\xi \notin cl\big (\{\zeta \} \cup (N_\xi \cap \lambda ) \cup (N_\zeta \cap \lambda ), M \big ).$ Let $V_1=N_\xi \cap \lambda $ , $V_2=N_\zeta \cap \lambda $ and $V=V_1 \cup V_2$ . By our assumption,

$$ \begin{align*}\xi \notin cl(\{\zeta\} \cup V, M);\end{align*} $$

also, it is clear that

$$ \begin{align*}\zeta \notin cl(\{\xi\} \cup V, M).\end{align*} $$

Thus by Lemma 2.8, we can find submodels $N^*_1, N^*_2$ of M such that:

1. (1) $N^*_1 \cap \lambda =N^*_2 \cap \lambda $ is unbounded in $\lambda $ .

2. (2) $\xi \in N^*_1 \setminus N^*_2$ and $\zeta \in N^*_2 \setminus N^*_1$ .

The models $N_1^*$ and $N^*_2$ are clearly f-closed, and by clause (1) above and $\mathrm {UB}_\lambda $ , we have $N^*_1 \subseteq N^*_2$ or $N^*_2 \subseteq N^*_1$ , which contradicts clause (2) above.

Let $\langle \sigma _i(x_0, \ldots , x_{n(i)-1}): i<\mathrm {cf}(\lambda ) \rangle $ list all terms of $\mathcal {L}$ . By $(*)_2$ , for each $\xi < \zeta $ from E, we can choose some $i(\xi , \zeta ) < \mathrm {cf}(\lambda )$ together with sequences $\vec {a}_{\xi , \zeta } \in (N_\zeta \cap \lambda )^{<\omega }$ and $\vec {b}_{\xi , \zeta } \in (N_\xi \cap \lambda )^{<\omega }$ such that

$$\begin{align*}(\oplus )_1\hspace {1.0cm}\ \xi = \sigma _{i(\xi , \zeta )}(\zeta , \vec {a}_{\xi , \zeta }, \vec {b}_{\xi , \zeta }). \end{align*}$$

For $\xi \in E$ set $U_\xi =N_\xi =cl(\{\xi \}, M).$ It follows that $U_\xi =cl(U_\xi , M)$ . For $ \xi < \zeta $ use the pairing function $p^M$ to find $\alpha _{\zeta , \xi }$ and $\beta _{\zeta , \xi }$ such that $\alpha _{\zeta , \xi }$ codes $\langle i(\xi , \zeta ) \rangle ^{\frown } \vec {a}_{\xi , \zeta }$ and $\beta _{\zeta , \xi }$ codes $\vec {b}_{\xi , \zeta }$ .

Now the sequences

$$ \begin{align*}\vec{X}=\langle U_\xi: \xi \in E \rangle\end{align*} $$

and

$$ \begin{align*}\langle \langle (\alpha_{\zeta, \xi}, \beta_{\zeta, \xi}): \xi \in \zeta \cap E \rangle: \zeta \in E \rangle\end{align*} $$

witness the failure of (e). We get a contradiction and the claim follows.

Thus so far we have shown that

$$\begin{align*}(a) \implies (b) \implies (c) \implies (d) \implies (e) \implies \neg \mathrm{UB}_\lambda. \end{align*}$$

Claim 3.7. Suppose that $\mathrm {cf}(\lambda )$ is not a Jonsson cardinal. Then $\neg \mathrm {UB}_\lambda $ implies $(e)$ .

Proof Suppose towards a contradiction that (e) fails and let $\vec {X}=\langle U_i: i<\lambda ^+ \rangle $ and $\langle \vec {X}_i: i< \lambda ^+ \rangle $ , where $\vec {X}_i=\langle (\alpha _{i, j}, \beta _{i, j}): j<i \rangle $ as in clause (e) witness this failure. Let $\langle \lambda _i: i<\mathrm {cf}(\lambda ) \rangle $ be an increasing sequence cofinal in $\lambda $ and define the function $c: \lambda \to \mathrm {cf}(\lambda )$ as

$$\begin{align*}c(\alpha) =\min\{i<\mathrm{cf}(\lambda): \alpha < \lambda_i \}. \end{align*}$$

For $\xi < \lambda ^+$ let $\langle \gamma _{\xi , i}: i<\mathrm {cf}(\lambda ) \rangle $ enumerate $U_\xi $ such that each element of $U_\xi $ appears cofinally many often. Let $f: [\lambda ^+]^{<\omega } \to \lambda ^+$ be such that:

1. (1) If $\xi < \zeta < \lambda ^+$ , then

   $$\begin{align*}f(\alpha_{\zeta, \xi}, \beta_{\zeta, \xi}, \zeta)=\xi. \end{align*}$$

2. (2) If $\zeta < \lambda ^+$ and $\alpha < \lambda $ , then for arbitrary large $j< \mathrm {cf}(\lambda )$ , we have

   $$\begin{align*}\sup\limits_{i<j}\lambda_i < \alpha < \lambda_j \implies f(\alpha, \zeta)=\gamma_{\zeta, j}. \end{align*}$$

3. (3) If $A \in [\mathrm {cf}(\lambda )]^{\mathrm {cf}(\lambda )}$ , $c(\alpha _i)=i$ for $i \in A$ and $j<\mathrm {cf}(\lambda ),$ then for some n and some sequence $\vec \xi = \langle \xi _0, \ldots , \xi _{n-1} \rangle \in A^n$ , we have

   $$\begin{align*}j =c (f(\alpha_{\xi_0}, \ldots, \alpha_{\xi_{n-1}})). \end{align*}$$

Since $\mathrm {cf}(\lambda )$ is not a Jonsson cardinal, we can define such a function f. ² Let us show that the pair $(f, c)$ witnesses $\mathrm {UB}_\lambda $ holds, ³ which contradicts our assumption. To see this, suppose $x, y \subseteq \lambda ^+$ are closed under f, $x \cap \lambda =y\cap \lambda $ and $\sup (x\cap \lambda )=\lambda .$ Assume towards a contradiction that $x \nsubseteq y$ and $y \nsubseteq x.$ Let $\xi =\min (x \setminus y)$ and $\zeta =\min (y \setminus x)$ , and let us suppose that $\xi < \zeta .$

By clause (3), $\mathrm {cf}(\lambda ) \subseteq y,$ and then by clause (2), and since $y \cap \lambda $ is cofinal in $\lambda $ , we have $U_\zeta \subseteq y.$ Similarly $U_\xi \subseteq x$ . As $x \cap \lambda =y \cap \lambda $ and $U_\xi \subseteq \lambda ,$ we conclude that $U_\xi \subseteq y$ as well. Thus by item (1), and since $\alpha _{\zeta , \xi }, \beta _{\zeta , \xi }, \zeta \in y$ we have $\xi \in y$ , which contradicts the choice of $\xi \in x \setminus y$ . This completes the proof of the claim.

The theorem follows.

Remark 3.8. The above proof shows that the following are equivalent:

1. (1) clause (e) of Theorem 3.1,

2. (2) for each model M with universe $\lambda ^+$ and vocabulary of cardinality $\mathrm {cf}(\lambda ),$ there are substructures $N_0, N_1$ of M such that $N_0 \cap \lambda =N_1 \cap \lambda $ , $N_0 \nsubseteq N_1$ , and $N_1 \nsubseteq N_0.$

As we noticed earlier, it is consistent relative to the existence of large cardinals that Chang’s transfer principle $(\aleph _{\omega +1}, \aleph _\omega ) \twoheadrightarrow (\aleph _{1}, \aleph _0)$ holds with $\aleph _\omega $ being strong limit. Hence by our main theorem, we have the following corollary.

Corollary 3.9. It is consistent, relative to the existence of large cardinals, that $\mathrm {UB}_{\aleph _\omega }$ fails.

Funding

The first author’s research has been supported by a grant from IPM (No. 1400030417). The second author’s research has been partially supported by Israel Science Foundation (ISF) grant no: 1838/19. This is publication 1216 of the second author.