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 $\alpha _\ell \notin N^*_{3-\ell }$ .

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.

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 $ .

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.

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.Footnote ² Let us show that the pair $(f, c)$ witnesses $\mathrm {UB}_\lambda $ holds,Footnote 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.