Exercise 32 of chapter 2 of Kunen (1980) tells me to show that there exists a total ordering with no $\omega_1$ strictly increasing/decreasing sequencies such that every separable subspace is nowhere dense (that is, the interior of it's closure is empty).
It tells me to try to use a Aronszajn Tree to construct an Aronszajn line. However, the book doesn't tell me what is a Aronszajn line.
Wikipedia and the Handbook of Set Theoretic Topology tells me that a Aronszajn line is a total order of cardinality $\omega_1$ that contains no $\omega_1$ strictly increasing/decreasing sequences and no uncountable subsets order-isomorphic to a subset of the reals. I have also seen that every lexicographic ordering of an Aronszajn line is an Aronszajn tree. My question is: Is it true that every Aronszajn line satisfies what Kunen's exercise tells me to do? Is there an counter-example? I am supposed to use an Aronszajn tree to construct an Aronszajn line and then use the line to build another line that satisfies what Kunen wants or I am supposed to build an Aronszajn line and then proof that every Aronszajn line fits?
With a little extra work an Aronszajn line does what’s wanted. If $\langle X,\le\rangle$ is an Aronszajn line, define a relation $\sim$ on $X$ by setting $x\sim y$ iff the interval between $x$ and $y$ is countable. It’s easy to check that $\sim$ is an equivalence relation with order-convex equivalence classes, so that $X/\!\!\sim$ naturally inherits a linear order $\preceq$ from $X$. It’s also straightforward to check that $\langle X/\!\!\sim,\preceq\rangle$ is a densely ordered Aronszajn line. The final step is to verify that every separable subset of $X/\!\!\sim$ is nowhere dense, and dividing out the equivalence relation ensures that.