According to the above definition, with the adherent point x of S we explicitly
write x 2 S. Think of negation of the statement of Definition 6.4. That is, with some
neighborhood
∃
N of x we have (i) N \ S ¼ ∅ , x =
2 S . Meanwhile,
(ii) N \ S ¼ ∅ ) x =
2 S. Combining (i) and (ii), we have x =
2 S ) x =
2 S. Similarly
to the above argument about the interior, we get
S ⊂ S:
ð6:10Þ
Combining (6.9) and (6.10), we get
S
∘
⊂ S ⊂ S:
ð6:11Þ
In relation to the interior and closure, we have following important lemmas.
Lemma 6.1 Let S be a subset of T and O be any open set contained in S. Then, we
have O ⊂ S
∘ .
Proof Suppose that with an arbitrary point x, x 2 O. Since we have x 2 O ⊂ S, S is
a neighborhood of x (due to Definition 6.2). Then, from Definition 6.3 we have
x 2 S
∘ . Then, we have x 2 O ) x 2 S
∘ . This implies O ⊂ S
∘ . This completes the
proof.
∎
Lemma 6.2 Let S be a subset of T. If x 2 S
∘ , then there is some open set
∃
O that
satisfies x 2 O ⊂ S.
Proof Suppose that with a point x, x 2 S
∘ . Then by Definition 6.3, S is a neighborhood of x. Meanwhile, by Definition 6.2 there is some open set
∃ O that satisfies
x 2 O ⊂ S. This completes the proof.
∎
Lemmas 6.1 and 6.2 teach us further implications. From Lemma 6.1 and (6.11)
we have
O ⊂ S
∘
⊂ S
ð6:12Þ
with an arbitrary open set
8 O contained in S
∘ . From Lemma 6.2 we have
x 2 S
∘ ) x 2 O; i.e., S
∘ ⊂ O with some open set
∃ O containing S
∘ . Expressing
this specific O as e
O, we have S
∘
⊂ e
O. Meanwhile, using Lemma 6.1 once again we
must get
e
O ⊂ S
∘
⊂ S:
ð6:13Þ
That is, we have S
∘
⊂ e
O and e
O ⊂ S
∘ at once. This implies that with this specific
open set e
O, we get e
O ¼ S
∘ . That is,
6.1 Set and Topology
187
write x 2 S. Think of negation of the statement of Definition 6.4. That is, with some
neighborhood
∃
N of x we have (i) N \ S ¼ ∅ , x =
2 S . Meanwhile,
(ii) N \ S ¼ ∅ ) x =
2 S. Combining (i) and (ii), we have x =
2 S ) x =
2 S. Similarly
to the above argument about the interior, we get
S ⊂ S:
ð6:10Þ
Combining (6.9) and (6.10), we get
S
∘
⊂ S ⊂ S:
ð6:11Þ
In relation to the interior and closure, we have following important lemmas.
Lemma 6.1 Let S be a subset of T and O be any open set contained in S. Then, we
have O ⊂ S
∘ .
Proof Suppose that with an arbitrary point x, x 2 O. Since we have x 2 O ⊂ S, S is
a neighborhood of x (due to Definition 6.2). Then, from Definition 6.3 we have
x 2 S
∘ . Then, we have x 2 O ) x 2 S
∘ . This implies O ⊂ S
∘ . This completes the
proof.
∎
Lemma 6.2 Let S be a subset of T. If x 2 S
∘ , then there is some open set
∃
O that
satisfies x 2 O ⊂ S.
Proof Suppose that with a point x, x 2 S
∘ . Then by Definition 6.3, S is a neighborhood of x. Meanwhile, by Definition 6.2 there is some open set
∃ O that satisfies
x 2 O ⊂ S. This completes the proof.
∎
Lemmas 6.1 and 6.2 teach us further implications. From Lemma 6.1 and (6.11)
we have
O ⊂ S
∘
⊂ S
ð6:12Þ
with an arbitrary open set
8 O contained in S
∘ . From Lemma 6.2 we have
x 2 S
∘ ) x 2 O; i.e., S
∘ ⊂ O with some open set
∃ O containing S
∘ . Expressing
this specific O as e
O, we have S
∘
⊂ e
O. Meanwhile, using Lemma 6.1 once again we
must get
e
O ⊂ S
∘
⊂ S:
ð6:13Þ
That is, we have S
∘
⊂ e
O and e
O ⊂ S
∘ at once. This implies that with this specific
open set e
O, we get e
O ¼ S
∘ . That is,
6.1 Set and Topology
187