where T À {x} is an open set, from Definition 6.2 N ¼ T À {x} is a neighborhood of y, and moreover N does not contain x. Then, from Definition 6.8 (T, τ) is
a T 1 -space. These complete the proof.
∎
A set comprising a unit element is called a singleton or a unit set.
Theorem 6.8 Let S be a subset of a T 1 -space. Then S
d is a closed set in the T 1 -space.
Proof Let p be a point of S. Suppose p 2 S
d . Suppose, furthermore, p =
2 S
d . Then,
from Definition 6.6 there is some neighborhood
∃
N of p such that N \ (S À {p}) ¼ ∅.
Since p 2 S
d , from Definition 6.4 we must have N \ S
d
6 ¼ ∅ at once for this special
neighborhood N of p. Then, as we have p =
2 S
d , we can take a point q such that
q 2 N \ S
d and q 6 ¼ p. Meanwhile, we may choose an open set N for the above
neighborhood of p (i.e., an open neighborhood), because p 2 N (¼N
∘
) ⊂ N. Since
N \ (T À {p}) ¼ N À {p} is an open set from Theorem 6.7 as well as Axiom
(O2) and Definition 6.1, N À {p} is an open neighborhood of q which does not
contain p. Namely, we have q 2 N À {p}[¼(N À {p})
∘ ] ⊂ N À {p}. Note that
q 2 N À {p} and p =
2 N À {p}, in agreement with Definition 6.8. As q 2 S
d , again
from Definition 6.6 we have
N À p
f g
ð
Þ\ S À q
f g
ð
Þ6 ¼ ∅:
ð6:30Þ
This implies that N \ (S À {q}) 6 ¼ ∅ as well. But, it is in contradiction to the
original relation of N \ (S À {p}) ¼ ∅, which is obtained from p =
2 S
d . Then, we
must have p 2 S
d . Thus, from the original supposition we get S
d
⊂ S
d . Meanwhile,
we have S
d
⊃ S
d from (6.11). We have S
d
¼ S
d accordingly. From Theorem 6.2, this
means that S
d is a closed set. Hence, we have proven the theorem.
∎
The above two theorems are intuitively acceptable. For, in a one-dimensional
Euclidean space ℝ we express a singleton {x} as [x, x] and a closed interval as [x, y]
(x 6 ¼ y). With the latter, [x, y] might well be expressed as (x, y)
d . Such subsets are well
known as closed sets. According to the separation axioms, various topological
spaces can be obtained by imposing stronger constraints upon the T 1 -space [2]. A
typical example for this is a metric space. In this context, the T 1 -space has acquired
primitive notions of metric that can be shared with the metric space. Definitions of
the metric (or distance function) and metric space will briefly be summarized in
Chap. 13 in reference to an inner product space.
In the above discussion, we have described a brief outline of the set theory and
topology. We use the results in this chapter and in Part IV as well.
196
6 Theory of Analytic Functions
a T 1 -space. These complete the proof.
∎
A set comprising a unit element is called a singleton or a unit set.
Theorem 6.8 Let S be a subset of a T 1 -space. Then S
d is a closed set in the T 1 -space.
Proof Let p be a point of S. Suppose p 2 S
d . Suppose, furthermore, p =
2 S
d . Then,
from Definition 6.6 there is some neighborhood
∃
N of p such that N \ (S À {p}) ¼ ∅.
Since p 2 S
d , from Definition 6.4 we must have N \ S
d
6 ¼ ∅ at once for this special
neighborhood N of p. Then, as we have p =
2 S
d , we can take a point q such that
q 2 N \ S
d and q 6 ¼ p. Meanwhile, we may choose an open set N for the above
neighborhood of p (i.e., an open neighborhood), because p 2 N (¼N
∘
) ⊂ N. Since
N \ (T À {p}) ¼ N À {p} is an open set from Theorem 6.7 as well as Axiom
(O2) and Definition 6.1, N À {p} is an open neighborhood of q which does not
contain p. Namely, we have q 2 N À {p}[¼(N À {p})
∘ ] ⊂ N À {p}. Note that
q 2 N À {p} and p =
2 N À {p}, in agreement with Definition 6.8. As q 2 S
d , again
from Definition 6.6 we have
N À p
f g
ð
Þ\ S À q
f g
ð
Þ6 ¼ ∅:
ð6:30Þ
This implies that N \ (S À {q}) 6 ¼ ∅ as well. But, it is in contradiction to the
original relation of N \ (S À {p}) ¼ ∅, which is obtained from p =
2 S
d . Then, we
must have p 2 S
d . Thus, from the original supposition we get S
d
⊂ S
d . Meanwhile,
we have S
d
⊃ S
d from (6.11). We have S
d
¼ S
d accordingly. From Theorem 6.2, this
means that S
d is a closed set. Hence, we have proven the theorem.
∎
The above two theorems are intuitively acceptable. For, in a one-dimensional
Euclidean space ℝ we express a singleton {x} as [x, x] and a closed interval as [x, y]
(x 6 ¼ y). With the latter, [x, y] might well be expressed as (x, y)
d . Such subsets are well
known as closed sets. According to the separation axioms, various topological
spaces can be obtained by imposing stronger constraints upon the T 1 -space [2]. A
typical example for this is a metric space. In this context, the T 1 -space has acquired
primitive notions of metric that can be shared with the metric space. Definitions of
the metric (or distance function) and metric space will briefly be summarized in
Chap. 13 in reference to an inner product space.
In the above discussion, we have described a brief outline of the set theory and
topology. We use the results in this chapter and in Part IV as well.
196
6 Theory of Analytic Functions