I've been given the following problem (note $T_{ord}$ is the theory of dense linear orders without endpoints):
Show that $\langle \mathbb{R}, < \rangle$ (the reals with their usual ordering) is not a saturated model of $T_{ord}$ of cardinality $\aleph_1$.
I don't think this is too terribly hard, but I'm thrown by the phrasing in light of a previous problem:
Let $M$ be a saturated model of $T_{ord}$. Show that for every $A \subseteq |M|$ such that $|A| < ||M||$ there exists $b \in |M|$ such that $M \models a < b$ for every $a \in A$.
So it seems to me that we don't need the "of cardinality $\aleph_1$" clause in the first question at all; if the reals were a saturated model of $T_{ord}$, then, eg, setting $A = \mathbb{N}$ would give us a real number $b$ larger than every natural number. So I am wondering if there's a flaw in this argument (maybe a snafu with the subtleties of $\models$ that I'm missing?) or if I'm correct that "of cardinality $\aleph_1$" is redundant here.
Thanks!
Best Answer
Yes, the "of cardinality $\aleph_1$" here is not really relevant (it simply gives a much shorter proof in the event that $\mathbb{R}$ does not have cardinality $\aleph_1$ at all, i.e. if the continuum hypothesis is false). It is possible that the intended statement was instead that $\langle \mathbb{R}, < \rangle$ is not an $\aleph_1$-saturated model (which is equivalent to saturated if $\mathbb{R}$ has cardinality $\aleph_1$, but is stronger otherwise; in any case your proof still works thoough since $A=\mathbb{N}$ is countable).