# Can a Proper Partial Order have a Totally-Ordered 'Spine'?

Suppose $P$ is an arbitrary poset.

Define a subset $Q \subset P$ to be dense to mean that for every $a,b \in P$ with $a < b$ there is some $q \in Q$ such that $a \le q \le b$. This may not be the standard notion for a dense subset of a poset.

Question (1): Suppose some dense subset $Q \subset P$ is totally-ordered under the induced ordering. Does it follow that $P$ is totally-ordered?

Answer (1): No. For a counterexample let $P = \mathbb R \cup \{5'\}$ where $5'$ is greater than every element of $(-\infty, 5)$; less than every element of $(5,\infty)$; and incomparable to $5$. Then $\mathbb R$ is dense but $P$ is not totally-ordered.

Question (2): Suppose some dense subset $Q \subset P$ with dense complement is totally-ordered under the induced ordering. Does it follow that $P$ is totally-ordered?

Edits: I never meant to say 'well-ordered' instead of 'totally ordered'. The definition of dense has been changed to require $a < b$.
• For question 2, the $P$ from question 1 and $Q = \mathbb{R}\setminus \mathbb{Q}$ say the answer is "no". – Daniel Fischer Oct 17 '15 at 21:07
• Take $P$ to be any set with at least two elements and impose on it a trivial partial order, i.e. $a\leq b\Leftrightarrow a=b$. Then whole $P$ satisfies (2) but is not totally ordered. – Wojowu Oct 17 '15 at 21:08
The answer to $(2)$ is no even if we rule out the trivial partial order of equality noted in the comments by Wojowu, and your example $P$ for $(1)$ can be used to show this. Let $Q$ be the set of rational numbers in $\Bbb R\setminus\{5\}$; clearly $Q$ is dense in $P$ and linearly ordered. $P\setminus Q$ includes the irrationals, so it also is dense in $P$, but $P$ is not linearly ordered.