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Give a non-discrete #topology on the real numbers (or y'know, whatever) closed under arbitrary *intersections*. Yeah, it won't be T1. (Why?)
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A T_1 #topology on the real numbers closed under arbitrary *intersections* is discrete because the arbitrary union of singletons is closed.
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Let {(x,∞):-∞≤x≤∞}∪{[x,∞):-∞<x<∞} give a T_0 #topology on the real numbers. Then the arbitrary union *or intersection* of open sets is open.