Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 144 / 8

Take $f<g$ in $I={\mathbb{Q}}^{\lambda }$ and let $\alpha$ be their first differing coordinate. Choose $r\in \mathbb{Q}$ with $f(\alpha )<r<g(\alpha )$, copy their common initial segment below $\alpha$, put $r$ at $\alpha$, and put zero at every later coordinate. The resulting eventually zero element lies strictly between $f$ and $g$, so $J$ is dense.

For each $\mu <\lambda$, the eventually zero functions supported below $\mu$ number at mostMinimality of $\lambda$ gives $2^{|\mu |}\le \kappa$ for every $\mu <\lambda$. Also $\lambda \le \kappa$, since $2^{\kappa }>\kappa$. Taking the union over $\mu <\lambda$ therefore gives $|J|\le \kappa$.

A formula $\varphi (\bar{x},\bar{y})$ has the order property for $T$ when, for every finite $n$, some model contains tuples $a_i,b_j$ withCompactness realizes this pattern indexed by any linear order.

Fix $\kappa \ge |T|$ and let $\lambda$ be least with $2^{\lambda }>\kappa$. Use the order property along $I={\mathbb{Q}}^{\lambda }$, and take parameterswhere part i gives $|B|\le \kappa$. For distinct $i,i^{\prime }\in I$, choose $j\in J$ strictly between them. Then $\varphi (a_i,b_j)$ and $\varphi (a_{i^{\prime }},b_j)$ have different truth values, so the types $\mathrm{tp}(a_i/B)$ are distinct. There are $|I|=2^{\lambda }>\kappa$ such types over at most $\kappa$ parameters. Enlarging $B$ to size exactly $\kappa$ if necessary preserves them. Hence $T$ is not $\kappa$-stable. This is the order property implies instability argument.