Definition of Topology space

Introduction

日本語 ver

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
• 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 $x \in O_1$and $x \in O_2$, $\exists \epsilon_1 ~~s.t.~~ B(x,\epsilon_1) \in O_1$ 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