2.3 Closure Of A Set
Next is the topic of the closure of a set.
Definition 2.3.1 Let A be a nonempty subset of
a metric space. Then
A is defined to be the intersection of all closed sets which contain A. Note the whole
space, X is one such closed set which contains A. The whole space X is closed because
its complement is open, its complement being ∅. It is certainly true that every point of
the empty set is an interior point because there are no points of ∅.
Lemma 2.3.2 Let A be a nonempty set in
. Then A is a closed set and
where A′ denotes the set of limit points of A.
Proof: First of all, denote by C the set of closed sets which contain A. Then
and this will be closed if its complement is open. However,
Each HC is open and so the union of all these open sets must also be open. This is because if
x is in this union, then it is in at least one of them. Hence it is an interior point of that one.
But this implies it is an interior point of the union of them all which is an even larger set.
Thus A is closed.
The interesting part is the next claim. First note that from the definition, A ⊆A so
if x ∈ A, then x ∈A. Now consider y ∈ A′ but y
a closed set, then
there exists B
cannot be a limit point of A,
Next suppose x ∈A and suppose x
Then if B
contains no points of
itself is not in A,
it would follow that B
and so recalling that
open balls are open, B
is a closed set containing A
so from the definition, it also
which is contrary to the assertion that x ∈A.
Hence if x
then x ∈ A′