This means that S and S
c have the same boundary. Since both S and S
c are closed
sets and the boundary is defined as their intersection, from Axiom (C3) the boundary
is a closed set. From Definition 6.1 a closed set is defined as a complement of an
open set, and vice versa. Thus, the open set and closed set do not make a confrontation concept but a complementation concept. In this respect we have the following
lemma and theorem.
Lemma 6.5 Let S be an arbitrary subset of T. Then, we have
S
c
¼ S
∘
ð Þ
c :
Proof Since S
∘
⊂ S, (S
∘
)
c
⊃ S
c
. As S
∘ is open, (S
∘
)
c is closed. Hence, S
∘
ð Þ
c ⊃ S
c ,
because S
c is the smallest closed set containing S
c
. Next let C be an arbitrary closed set
that contains S
c
. Then C
c is an open set that is contained in S, and so we have C
c
⊂ S
∘
,
because S
∘ is the largest open set contained in S. Therefore, C ⊃ (S
∘
)
c
. Then, if we
choose S
c for C, we must have S
c
⊃ (S
∘
)
c
. Combining this relation with S
∘
ð Þ
c ⊃ S
c
obtained above, we get S
c
¼ S
∘
ð Þ
c .
∎
Theorem 6.3 [4] A necessary and sufficient condition for S to be both open and
closed at once is S
b
¼ ∅.
Proof
(i) Necessary condition: Let S be open and closed. Then, S ¼ S and S ¼ S
∘ .
Suppose that S
b
6 ¼ ∅. Then, from Definition 6.5 we would have
∃ a 2 S \ S
c
for some a. Meanwhile, from Lemma 6.5 we have S
c
¼ S
∘
ð Þ
c . This leads to
a 2 S \ S
∘
ð Þ
c
. But, from the assumption we would have a 2 S \ S
c . We have no
such a, however. Thus, using the proof by contradiction we must have S
b
¼ ∅.
(ii) Sufficient condition: Suppose that S
b
¼ ∅. Starting with S ⊃ S
∘ and S 6 ¼ S
∘ , let
us assume that there is a such that a 2 S and a =
2 S
∘ . That is, we have a 2
S
∘
ð Þ
c ) a 2 S \ S
∘
ð Þ
c ⊂ S \ S
∘
ð Þ
c ¼ S \ S
c
¼ S
b , in contradiction to the supposition. Thus, we must not have such a, implying that S ¼ S
∘ . Next, starting
with S ⊃ S and S 6 ¼ S, we assume that there is a such that a 2 S and a =
2 S. That
is, we have a 2 S
c
) a 2 S \ S
c
⊂ S \ S
c
¼ S
b , in contradiction to the supposition. Thus, we must not have such a, implying that S ¼ S. Combining this
relation with S ¼ S
∘ obtained above, we get S ¼ S ¼ S
∘ . In other words, S is
both open and closed at once. These complete the proof.
∎
The abovementioned set is sometimes referred to as a closed-open set or a clopen
set as a portmanteau word. An interesting example can be seen in Chap. 20 in
relation to topological groups (or continuous groups).
190
6 Theory of Analytic Functions
c have the same boundary. Since both S and S
c are closed
sets and the boundary is defined as their intersection, from Axiom (C3) the boundary
is a closed set. From Definition 6.1 a closed set is defined as a complement of an
open set, and vice versa. Thus, the open set and closed set do not make a confrontation concept but a complementation concept. In this respect we have the following
lemma and theorem.
Lemma 6.5 Let S be an arbitrary subset of T. Then, we have
S
c
¼ S
∘
ð Þ
c :
Proof Since S
∘
⊂ S, (S
∘
)
c
⊃ S
c
. As S
∘ is open, (S
∘
)
c is closed. Hence, S
∘
ð Þ
c ⊃ S
c ,
because S
c is the smallest closed set containing S
c
. Next let C be an arbitrary closed set
that contains S
c
. Then C
c is an open set that is contained in S, and so we have C
c
⊂ S
∘
,
because S
∘ is the largest open set contained in S. Therefore, C ⊃ (S
∘
)
c
. Then, if we
choose S
c for C, we must have S
c
⊃ (S
∘
)
c
. Combining this relation with S
∘
ð Þ
c ⊃ S
c
obtained above, we get S
c
¼ S
∘
ð Þ
c .
∎
Theorem 6.3 [4] A necessary and sufficient condition for S to be both open and
closed at once is S
b
¼ ∅.
Proof
(i) Necessary condition: Let S be open and closed. Then, S ¼ S and S ¼ S
∘ .
Suppose that S
b
6 ¼ ∅. Then, from Definition 6.5 we would have
∃ a 2 S \ S
c
for some a. Meanwhile, from Lemma 6.5 we have S
c
¼ S
∘
ð Þ
c . This leads to
a 2 S \ S
∘
ð Þ
c
. But, from the assumption we would have a 2 S \ S
c . We have no
such a, however. Thus, using the proof by contradiction we must have S
b
¼ ∅.
(ii) Sufficient condition: Suppose that S
b
¼ ∅. Starting with S ⊃ S
∘ and S 6 ¼ S
∘ , let
us assume that there is a such that a 2 S and a =
2 S
∘ . That is, we have a 2
S
∘
ð Þ
c ) a 2 S \ S
∘
ð Þ
c ⊂ S \ S
∘
ð Þ
c ¼ S \ S
c
¼ S
b , in contradiction to the supposition. Thus, we must not have such a, implying that S ¼ S
∘ . Next, starting
with S ⊃ S and S 6 ¼ S, we assume that there is a such that a 2 S and a =
2 S. That
is, we have a 2 S
c
) a 2 S \ S
c
⊂ S \ S
c
¼ S
b , in contradiction to the supposition. Thus, we must not have such a, implying that S ¼ S. Combining this
relation with S ¼ S
∘ obtained above, we get S ¼ S ¼ S
∘ . In other words, S is
both open and closed at once. These complete the proof.
∎
The abovementioned set is sometimes referred to as a closed-open set or a clopen
set as a portmanteau word. An interesting example can be seen in Chap. 20 in
relation to topological groups (or continuous groups).
190
6 Theory of Analytic Functions