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For a linearly (totally) ordered set $A$, one can define its order topology: that is the smallest topology containing the set B of all the intervals of the form $\{x\mid x < a\}, \{x\mid x > a\}$ or $\{x\mid a < x < b\}$ where $x$, $a$ and $b$ are elements of $A$.

The order topology can be shown to be generated by the set $B$ of all intervals mentioned above (that is any element of the order topology can be written as the union of intervals). But to prove this, isn't AC required?

Otherwise how can we prove that: if we call $U$ the set of all union of indexed families of elements of $B$, then any union of indexed families of elements of $U$ are also elements of $U$? This is required for $U$ to be the order topology.

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    The Axiom of Choice does tell us that every set has a topology induced by a well-ordering, but is not necessary here. Indeed, most order topologies that people are interested in do not come from a well-ordering (the metric topology on $\mathbb{R}$, for example, is isomorphic to the topology induced by the dense linear order).2012-03-05

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