Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
1  
As you can infer from the two conflicting answers, "dense in terms of the order" can be understood in two different ways. One is that the subset has elements between any two elements of the original set, also expressed as "dense in the original set"; the other is that it has elements between any two elements of the subset, also expressible as "dense as an ordered set in its own right." I suspect that you meant the first of these, but @Justaskin assumed you meant the second. – Andreas Blass Oct 25 '15 at 17:23
up vote 3 down vote accepted

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.

share|cite|improve this answer
    
Thanks! Are you finished? – Tim Aug 19 '10 at 18:00
1  
"Since X is dense in terms of order, there is an x in X such that a < x < b." The definition of a dense order only guarantees this is true for $a,b \in X$. Here we only have $a,b \in T$ where $T$ is the topological space. – barron Oct 25 '15 at 15:54
    
@Justaskin_: On the off chance you haven't yet seen Andreas Blass's comment on the question, it seems as though the expression "dense as an order" is ambiguous. – Jason DeVito Oct 26 '15 at 1:29

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".)

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.