Editing Order topology (section)

== Example of a subspace of a linearly ordered space whose topology is not an order topology ==
Though the subspace topology of ''Y'' = {−1} &cup; {1/''n''}<sub>''n''&isin;'''N'''</sub> in the section above is shown not to be generated by the induced order on ''Y'', it is nonetheless an order topology on ''Y''; indeed, in the subspace topology every point is isolated (i.e., singleton {''y''} is open in ''Y'' for every ''y'' in ''Y''), so the subspace topology is the [[discrete topology]] on ''Y'' (the topology in which every subset of ''Y'' is open), and the discrete topology on any set is an order topology. To define a total order on ''Y'' that generates the discrete topology on ''Y'', simply modify the induced order on ''Y'' by defining −1 to be the greatest element of ''Y'' and otherwise keeping the same order for the other points, so that in this new order (call it say <<sub>1</sub>) we have 1/''n'' <<sub>1</sub> −1 for all ''n''&nbsp;&isin;&nbsp;'''N'''. Then, in the order topology on ''Y'' generated by <<sub>1</sub>, every point of ''Y'' is isolated in ''Y''.

We wish to define here a subset ''Z'' of a linearly ordered topological space ''X'' such that no total order on ''Z'' generates the subspace topology on ''Z'', so that the subspace topology will not be an order topology even though it is the subspace topology of a space whose topology is an order topology.

Let <math>Z = \{-1\} \cup (0,1)</math> in the [[real line]]. The same argument as before shows that the subspace topology on ''Z'' is not equal to the induced order topology on ''Z'', but one can show that the subspace topology on ''Z'' cannot be equal to any order topology on ''Z''.

An argument follows. Suppose by way of contradiction that there is some [[Totally ordered set#Strict total order|strict total order]] < on ''Z'' such that the order topology generated by < is equal to the subspace topology on ''Z'' (note that we are not assuming that < is the induced order on ''Z'', but rather an arbitrarily given total order on ''Z'' that generates the subspace topology).

Let ''M''&nbsp;=&nbsp;''Z''&nbsp;\&nbsp;{−1}&nbsp;=&nbsp;(0,1), then ''M'' is [[connected space|connected]], so ''M'' is dense on itself and has no gaps, in regards to <. If −1 is not the smallest or the largest element of ''Z'', then <math>(-\infty,-1)</math> and <math>(-1,\infty)</math> separate ''M'', a contradiction. Assume without loss of generality that −1 is the smallest element of ''Z''. Since {−1} is open in ''Z'', there is some point ''p'' in ''M'' such that the interval (−1,''p'') is [[empty set|empty]], so ''p'' is the minimum of ''M''. Then ''M''&nbsp;\&nbsp;{''p''}&nbsp;=&nbsp;(0,''p'')&nbsp;∪&nbsp;(''p'',1) is not connected with respect to the subspace topology inherited from {{math|'''R'''}}. On the other hand, the subspace topology of ''M''&nbsp;\&nbsp;{''p''} inherited from the order topology of ''Z'' coincides with the order topology of ''M''&nbsp;\&nbsp;{''p''} induced by <, which is connected since there are no gaps in ''M''&nbsp;\&nbsp;{''p''} and it is dense. This is a contradiction.