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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?

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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.

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Thanks! Are you finished? –  Tim Aug 19 '10 at 18:00
    
Yes, sorry. It kept interpreting "<" as an html tag, I think. –  Jason DeVito Aug 19 '10 at 18:21
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