As seen in a preceding entry of this blog, topologies on a set can be defined by the means of closure operators. In this note, we prove that topologies can also be constructed using interior operators. In fact, the note is a solution of an exercise formulated previously. We repeat the statement of the exercise in question here, as a reminder.
Exercise.
Let be a set. If
is an operator which carries subsets of
into subsets of
, and
the family of all subsets such that
, under what conditions will
define a topology on
and
be the interior operator relative to this topology?
Solution of the exercise as PDF file
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