prove:
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So you mean like?
1) Supposewere a "point of adherence" as you guys call them apparently. Then for any open U containing x, you need
but this is clearly not possible. So the set of all points of adherence of
is
.
2) Similarly supposewere a "point of adherence." Then for any open U containing x,
. So every point in X is a point of adherence, so yeah, you are done.