Introduction
Today, I will write about General Topology. General topology is very impossible for studying mathematics. General topology defines phase to define continuity of function. Today, I will explain phase.
Also, This post is my review of General Topology.
Overview
Distance space
Firstly, Distance space is defined.
Let X is set, d:X\times X ->\mathbb{R},
(X,d) is distance space \iff
this condition is called axiom of distance.
d is called distance function.
Topology space
Let (X,d) is distance space.
\mathbb{O} \in 2^X is phase in (X,d) \iff
here, 2^X := \{A | A \subset X \}
Its condition is called axiom of phase.
We must watch out for \Lambda.
\Lambda is any Suffix set.
then (X,d,\mathbb{O}) is phase space.
Open set
A \subset X is open set\iff
\forall x \in A,~~\exists \epsilon > 0, ~~s.t.~~ B(x,\epsilon) \subset A
here, B(x,\epsilon):= \{y\in A| d(x,y) < \epsilon\}
This definition of the open set is a satisfied condition of the axiom of phase.
Let X:set , d:X \times X -> \mathbb{R}: distance function.
Let
\mathbb{O} := \{A \subset X|A :open set\}.
then \mathbb{O} is satisfyed axiom of phase.
proof.
This is obvious, because \phi do not have element, so \phi is open set, and \forall x \in X, \exists \epsilon > 0 ~~s.t.~~ B(x,\epsilon) \subset X. This is satisfyed becaue X is whole set.
about \forall O_1,O_2 \in \mathbb{O},\forall x \in O_1 \cap O_2
because x \in O_1and x \in O_2, \exists \epsilon_1 ~~s.t.~~ B(x,\epsilon_1) \in O_1 and \exists \epsilon_2 ~~s.t.~~ B(x,\epsilon_2) \in O_2
So,Let \epsilon := \min\{\epsilon_1,\epsilon_2\}, following is satisfyed.
B(x,\epsilon) \subset B(x,\epsilon_1)
B(x,\epsilon) \subset B(x,\epsilon_2)
Thus, B(x,\epsilon) \subset O_1 \cap O_2
about,\forall \{O_\lambda\}_{\lambda \in \Lambda},
\forall x \in \bigcup_{\lambda \in \Lambda} \{O_\lambda\}_{\lambda \in \Lambda},
because \forall x \in \{O_\lambda\}_{\lambda \in \Lambda}, following is satisfed.
\exists \lambda_0 \in \Lambda ~~.st.~~ x \in O_{\lambda_0}
Thus, \exists \epsilon ~~s.t.~~ B(x,\epsilon) \subset O_{\lambda_0}
So, B(x,\epsilon) \subset \{O_\lambda\}_{\lambda \in \Lambda}
Conclusion
We often define phase by open set.
Continuity of function is defined by this phase.
I will explain the definition of continuity of function by phase.
Reference
https://ja.wikipedia.org/wiki/%E8%B7%9D%E9%9B%A2%E7%A9%BA%E9%96%93
https://ja.wikipedia.org/wiki/%E4%BD%8D%E7%9B%B8%E7%A9%BA%E9%96%93
Also, This post is my review of General Topology.
Overview
- Distance space
- Axiom of phase
- Topology space
- Open set
Distance space
Firstly, Distance space is defined.
Let X is set, d:X\times X ->\mathbb{R},
(X,d) is distance space \iff
- \forall x,y \in X,~~~~~d(x,y) \geq 0
- \forall x,y \in X,~~~~~x = y \implies d(x,y) = 0
- \forall x,y \in X,~~~~~d(x,y) = d(y,x)
- \forall x,y,z \in X, ~~~~d(x,y) + d(y,z) \geq d(x,z)
this condition is called axiom of distance.
d is called distance function.
Topology space
Let (X,d) is distance space.
\mathbb{O} \in 2^X is phase in (X,d) \iff
- \phi,X \in \mathbb{O}
- \forall O_1,O_2 \in \mathbb{O} \implies O_1 \cap O_2 \in \mathbb{O}
- \forall \Lambda ,~~\forall \{O_\lambda \}_{\lambda \in \Lambda} \in O \implies \bigcup_{\lambda \in \Lambda} O_\lambda \in \mathbb{O}
here, 2^X := \{A | A \subset X \}
Its condition is called axiom of phase.
We must watch out for \Lambda.
\Lambda is any Suffix set.
then (X,d,\mathbb{O}) is phase space.
Open set
A \subset X is open set\iff
\forall x \in A,~~\exists \epsilon > 0, ~~s.t.~~ B(x,\epsilon) \subset A
here, B(x,\epsilon):= \{y\in A| d(x,y) < \epsilon\}
This definition of the open set is a satisfied condition of the axiom of phase.
Let X:set , d:X \times X -> \mathbb{R}: distance function.
Let
\mathbb{O} := \{A \subset X|A :open set\}.
then \mathbb{O} is satisfyed axiom of phase.
proof.
- \phi,X \in \mathbb{O}
This is obvious, because \phi do not have element, so \phi is open set, and \forall x \in X, \exists \epsilon > 0 ~~s.t.~~ B(x,\epsilon) \subset X. This is satisfyed becaue X is whole set.
- \forall O_1,O_2 \in \mathbb{O} \implies O_1 \cap O_2 \in \mathbb{O}
about \forall O_1,O_2 \in \mathbb{O},\forall x \in O_1 \cap O_2
because x \in O_1and x \in O_2, \exists \epsilon_1 ~~s.t.~~ B(x,\epsilon_1) \in O_1 and \exists \epsilon_2 ~~s.t.~~ B(x,\epsilon_2) \in O_2
So,Let \epsilon := \min\{\epsilon_1,\epsilon_2\}, following is satisfyed.
B(x,\epsilon) \subset B(x,\epsilon_1)
B(x,\epsilon) \subset B(x,\epsilon_2)
Thus, B(x,\epsilon) \subset O_1 \cap O_2
- \forall \Lambda ,~~\forall \{O_\lambda \}_{\lambda \in \Lambda} \in O \implies \bigcup_{\lambda \in \Lambda} O_\lambda \in \mathbb{O}
about,\forall \{O_\lambda\}_{\lambda \in \Lambda},
\forall x \in \bigcup_{\lambda \in \Lambda} \{O_\lambda\}_{\lambda \in \Lambda},
because \forall x \in \{O_\lambda\}_{\lambda \in \Lambda}, following is satisfed.
\exists \lambda_0 \in \Lambda ~~.st.~~ x \in O_{\lambda_0}
Thus, \exists \epsilon ~~s.t.~~ B(x,\epsilon) \subset O_{\lambda_0}
So, B(x,\epsilon) \subset \{O_\lambda\}_{\lambda \in \Lambda}
Q.E.D
Conclusion
We often define phase by open set.
Continuity of function is defined by this phase.
I will explain the definition of continuity of function by phase.
Reference
https://ja.wikipedia.org/wiki/%E8%B7%9D%E9%9B%A2%E7%A9%BA%E9%96%93
https://ja.wikipedia.org/wiki/%E4%BD%8D%E7%9B%B8%E7%A9%BA%E9%96%93
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