letbe a metric space and
a subset of X then
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Supposebut
. Then consider
this is an open neighborhood containing x, but does not intersect A, thus
, contradiction. d(x,A)=0 as desired.
Let d(x,A)=0. Suppose
then there exists an open set U containing x, but does not intersect A. By definition of open set, this means there is a basis element (a ball) containing x and contained in U. That is
. But this means
contradiction. thus
as desired.
QED