# Dense and complete linear orderings

Let $$(X,\leq)$$ be a totally (or linearly) ordered set with $$|X|>1$$ and with the following properties:

1. it is dense, that is, for any $$a there is $$x\in X$$ with $$a, and
2. it is complete, that is every non-empty subset $$S\subseteq X$$ has a (unique) least upper bound (= supremum) and a largest lower bound (= infimum).

The prime example for such a total order is the real interval $$[0,1]$$.

Two questions:

1. Is it possible in $${\sf ZFC}$$ that $$|X|<2^{\aleph_0}$$?
2. Given any cardinal $$\lambda > 2^{\aleph_0}$$, is there a total ordering relation $$\leq$$ making $$(\lambda,\leq)$$ complete and dense?

1. No. Taking $$a in $$X$$, we can find a countable dense linear order without endpoints as a sub order of $$X$$ between $$a$$ and $$b$$ by repeatedly applying density. Such an order is isomorphic to the rational order, which has continuum many cuts, each of which must be filled in $$X$$, by completeness.
2. Yes. The order $$\lambda\times [0,1)$$, with the lexicographic order, and with a top element added, is dense and complete of cardinality $$\lambda$$ when $$\lambda>2^{\aleph_0}$$.