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 $a<b \in X$ there is $x\in X$ with $a<x<b$, 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 $a<b$ 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 $\lambda>2^{\aleph_0}$.

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