Theorem 6.6 Let S be a subset of T. A necessary and sufficient condition for S be a
closed set is that S contains all the accumulation points of S.
Proof From Theorem 6.2, that S is a closed set is equivalent to S ¼ S . From
Theorem 6.5 this is equivalent to S ¼ S
d
[ S. Obviously, this is equivalent to S
d
⊂ S.
In other words, S contains all the accumulation points of S. This completes the
proof.
∎
As another proof, taking account of (6.25), we have
S
d
À Á À ¼ ∅ ⟺ S ¼ S ⟺ S is a closed set from Theorem 6:2
ð
Þ :
This relation is equivalent to the statement that S contains all the accumulation
points of S.
In Fig. 6.6 we show the relationship among S, S, S
d , and S
dsc
. From Fig. 6.6, we
can readily show that
S
dsc
¼ S À S
d
¼ S À S
d
:
We have a further example of direct sum decomposition. Using Lemma 6.5, we
have
S
b
¼ S \ S
c
¼ S \ S
∘
ð Þ
c ¼ S À S
∘
:
ð6:27Þ
Since S ⊃ S
∘ , we get
S ¼ S
∘
[ S
b ; S
∘
\ S
b
¼ ∅:
Notice that (6.27) is a succinct expression of Theorem 6.3. That is,
S
b
¼ ∅ ⟺ S ¼ S
∘
⟺ S is a clopen set:
̅
( )
( )
̅ =
∪ =
∪
,
= ( ) ∪ ( )
Fig. 6.6 Relationship
among S, S, S
d
, and S
dsc .
Regarding the symbols and
notations, see text
6.1 Set and Topology
193