# Definable sets in a $\kappa$-saturated structure.

Exercise 5.2.4. in Ziegler, Tent: A Course in Model Theory states:

If (the $$L$$-structure) $$\mathfrak{A}$$ is $$\kappa$$-saturated, then all definable subsets are either finite or have cardinality at least $$\kappa$$.

This statement seems only correct, if $$|\mathfrak{A}|\geq\kappa$$, as the set $$A$$ defined by $$\top$$ could otherwise have a cardinality $$\aleph_0\leq|A|<\kappa$$ for $$\kappa>\aleph_0$$.

I didn't manage to come up with a viable idea for proofing the statement. I would be thankful for any hint.

A $$\kappa$$-saturated structure has cardinality at least $$\kappa$$: otherwise, consider $$\{x\not=a: a\in\mathfrak{A}\}$$ ... (Fine, unless it's finite.)
• @Dalhamun You could view Noah's answer as a significant hint to proving the statement in general. The partial type $\{x\neq a : a\in \mathfrak{A}\}$ says that $x$ is not equal to any element of the definable set defined by $\top$... – Alex Kruckman Oct 15 '18 at 22:36
• Small correction: a $\kappa$-saturated structure has cardinality at least $\kappa$, unless it is finite. – Levon Haykazyan Oct 16 '18 at 14:16
• @Dalhamun What you wrote makes almost no sense to me. Why should $[\varphi(x)]$ consist only of isolated types? – Alex Kruckman Oct 16 '18 at 23:54
• @Dalhamun just look at the partial type $\{\varphi(x)\}\cup\{x\neq a \mid a\in \varphi(\mathfrak{A})\}$! Is it realized in $\mathfrak{A}$? – Alex Kruckman Oct 16 '18 at 23:56