# Saturation of the reals as a dense linear order

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!

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).