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boundary of an open set is nowhere
dense∗
CWoo†
2013-03-22 0:44:53
This entry provides another example of a nowhere dense set.
Proposition 1. If A is an open set in a topological space X, then ∂A, the
boundary of A is nowhere dense.
Proof. Let B = ∂A. Since B = A∩A{ , it is closed, so all we need to show is that
B has empty interior int(B) = ∅. First notice that B = A ∩ A{ , since A is open.
Now, we invoke one of the interior axioms, namely int(U ∩ V ) = int(U ) ∩ int(V ).
So, by direct computation, we have
{
{
int(B) = int(A) ∩ int(A{ ) = int(A) ∩ A ⊆ A ∩ A = ∅.
The second equality and the inclusion follow from the general properties of the
interior operation, the proofs of which can be found here.
Remark. The fact that A is open is essential. Otherwise, the proposition
fails in general. For example, the rationals Q, as a subset of the reals R under
the usual order topology, is not open, and its boundary is not nowhere dense,
as Q ∩ Q{ = R ∩ R = R, whose interior is R itself, and thus not empty.
∗ hBoundaryOfAnOpenSetIsNowhereDensei
created: h2013-03-2i by: hCWooi version:
h40422i Privacy setting: h1i hDerivationi h54A99i
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