Theorem
For a nowhere dense subset
of a metric space
and an open set
an open subset of
exists with no intersection with![]()
Proof
Let
the intersection of
with the closure of![]()
Then
and![]()
since
is open and
is dense in![]()
Hence there exists![]()
is open as the intersection of two open sets.
Hence
exists such that![]()
Hence![]()