dense in terms of order and in terms of the order topology In a densely and totally ordered set, induce a order topology from the order.
Are the dense in terms of the order and the dense in terms of the order topology equivalent?
Thanks and regards!
 A: Yes.
Let X be dense in terms of order.  Let U be an open subset.  We need to show X intersects U.  Since U is open, it contains a subset of the form (a,b) with a < b, since these sets form a basis for the topology (by definition of order topology).  Since X is dense in terms of order, there is an x in X such that a < x < b.  Then x is in X and (a,b), and hence x is in X and U.
Conversely, suppose X is dense in terms of topology, meaning it intersects every open set.  Choose two elements a and b and suppose a < b.  Then (a,b) is an open set.  Since X is dense in terms of topology, there is an x in (a,b), and thus this x will satisfy a < x < b.  Hence, X is dense in terms of order.
A: No.
Consider $\mathbb{R}$ with the usual order/topology and consider $S = \mathbb{Q} \cap (0,1) \subset \mathbb{R}$. $S$ is dense as an order (it is order isomorphic to $\mathbb{Q}$), but not (topologically) dense in $\mathbb{R}$, since, e.g., $2 \in \mathbb{R}$ is clearly neither in $S$ nor a limit point of $S$.
(If you are looking for a density-related parallel between orders and order topologies, one can indeed verify that dense-in-itself as a topology does coincide with dense-in-itself as an order; both are strictly weaker than the corresponding notion of "dense".)
