6.1.3 T 1 -Space
We can “enter” a variety of topologies τ into a set T to get a topological space (T, τ).
We have two extremes of them. One is an indiscrete topology (or trivial topology)
and the other is a discrete topology [2]. For the former the topology is characterized
as
τ Ã ¼ ∅, T
f
g
ð6:28Þ
and the latter case is
τ
Ã
¼ ∅, all the subsets of T, T
f
g :
ð6:29Þ
With τ
à all the subsets are open and complements to the individual subsets are
closed by Definition 6.1. But, such closed subsets are contained in τ
à , and so they are
again open. Thus, all the subsets are clopen sets. Practically speaking, the aforementioned two extremes are of less interest and value. For this reason, we need some
moderate separation conditions with topological spaces. These conditions are well
established as the separation axioms. We will not get into details about the discussion [2], but we mention the first separation axiom (Fréchet axiom) that produces the
T 1 -space.
Definition 6.8 Let (T, τ) be a topological space. Suppose that with respect to
8 x, y (x 6 ¼ y) 2 T there is a neighborhood N in such a way that
x 2 N and y =
2 N:
The topological space that satisfies the above separation condition is called a
T 1 -space.
We mention two important theorems of the T 1 -space.
Theorem 6.7 A necessary and sufficient condition for (T, τ) to be a T 1 -space is that
for each point x 2 T, {x} is a closed set of T.
Proof
(i) Necessary condition: Let (T, τ) to be a T 1 -space. Choose any pair of elements
x, y (x 6 ¼ y) 2 T. Then, from Definition 6.8 there is a neighborhood
∃ N of y (6 ¼x)
that does not contain x; y 2 N and N \ {x} ¼ ∅. From Definition 6.4 this implies
that y =
2 x
f g. Thus, we have y 6 ¼ x ) y =
2 x
f g. This means that y 2 x
f g ) y ¼ x.
Namely, x
f g ⊂ x
f g, but from (6.11) we have x
f g ⊂ x
f g. Thus, x
f g ¼ x
f g.
From Theorem 6.2 this shows that {x} is a closed set of T.
(ii) Sufficient condition: Suppose that {x} is a closed set of T. Choose any x and
y such that y 6 ¼ x. Since {x} is a closed set, N ¼ T À {x} is an open set that
contains y; i.e., y 2 T À {x} and x =
2 T À {x}. Since y 2 T À {x} ⊂ T À {x}
6.1 Set and Topology
195
We can “enter” a variety of topologies τ into a set T to get a topological space (T, τ).
We have two extremes of them. One is an indiscrete topology (or trivial topology)
and the other is a discrete topology [2]. For the former the topology is characterized
as
τ Ã ¼ ∅, T
f
g
ð6:28Þ
and the latter case is
τ
Ã
¼ ∅, all the subsets of T, T
f
g :
ð6:29Þ
With τ
à all the subsets are open and complements to the individual subsets are
closed by Definition 6.1. But, such closed subsets are contained in τ
à , and so they are
again open. Thus, all the subsets are clopen sets. Practically speaking, the aforementioned two extremes are of less interest and value. For this reason, we need some
moderate separation conditions with topological spaces. These conditions are well
established as the separation axioms. We will not get into details about the discussion [2], but we mention the first separation axiom (Fréchet axiom) that produces the
T 1 -space.
Definition 6.8 Let (T, τ) be a topological space. Suppose that with respect to
8 x, y (x 6 ¼ y) 2 T there is a neighborhood N in such a way that
x 2 N and y =
2 N:
The topological space that satisfies the above separation condition is called a
T 1 -space.
We mention two important theorems of the T 1 -space.
Theorem 6.7 A necessary and sufficient condition for (T, τ) to be a T 1 -space is that
for each point x 2 T, {x} is a closed set of T.
Proof
(i) Necessary condition: Let (T, τ) to be a T 1 -space. Choose any pair of elements
x, y (x 6 ¼ y) 2 T. Then, from Definition 6.8 there is a neighborhood
∃ N of y (6 ¼x)
that does not contain x; y 2 N and N \ {x} ¼ ∅. From Definition 6.4 this implies
that y =
2 x
f g. Thus, we have y 6 ¼ x ) y =
2 x
f g. This means that y 2 x
f g ) y ¼ x.
Namely, x
f g ⊂ x
f g, but from (6.11) we have x
f g ⊂ x
f g. Thus, x
f g ¼ x
f g.
From Theorem 6.2 this shows that {x} is a closed set of T.
(ii) Sufficient condition: Suppose that {x} is a closed set of T. Choose any x and
y such that y 6 ¼ x. Since {x} is a closed set, N ¼ T À {x} is an open set that
contains y; i.e., y 2 T À {x} and x =
2 T À {x}. Since y 2 T À {x} ⊂ T À {x}
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195