definition of continuity by Topology

Introduction

日本語 ver

Today, I will write definition of continuity of function by definition of an open set.
I wrote this post about the definition of Topology space, open set, and a thing that open set satisfy the axiom of Topology, but I did not write about continuity of function by definition of an open set. Actually, It is very important.

Overview

•  Open set
• $\epsilon-\delta$ reasoning
• definition of continuity of function by an open set
• Equivalence

Open set

Let (X,d) is distance space.
$A \subset X$ 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 open set satisfies Axim of Topology. It written the last time post.

$\epsilon-\delta$ reasoning

I will explain the $\epsilon-\delta$ reasoning. This reasoning is learned in bachelor third student at Univ.

Let f:X-> Y: map
f is countinous where $x=x_0$
$$\iff$$
$$\lim_{x \rightarrow x_0} f(x) = f(x_0)$$
$$\iff$$
$$\forall \epsilon > 0,~~\exists \delta >0 ~~s.t~~ d(x,x_0) < \delta \implies d(f(x),f(x_0)) < \epsilon $$

definition of continuity of function by an open set

Define1.0

Let $(X,\mathbb{O}_X),(Y,\mathbb{O}_Y)$ is Topology space, and
$f:X \rightarrow Y$ is countinous where $x=x_0$
$$\iff$$
$$f(x_0) \in \forall V:\textrm{open set} \subset Y~~,f^{-1} (V) \subset X ~~\textrm{is open set}$$
here,$\mathbb{O_X}$ and $\mathbb{O_Y}$ is open set family in X,Y, and  $f^{-1} (V) := \{a \in X| f(a) \in V \}$.

This definition is equivalence with $\epsilon-\delta$ reasoning.

Equivalence

- Define 1.0 $\implies$ $\epsilon-\delta$ reasoning
$\forall V, f(x_0) \in V$. Thus, $x_0 \in f^{-1}(V)$.
Because $f^{-1}(V)$ is open set, $\exists \delta > 0 ~~s.t.~~ B(x_0,\delta) \subset f^{-1}(V)$.
Therefore, $\forall x \in X ,~~x \in B(x_0,\delta) \implies x \in f^{-1}(V)$.
$x \in B(x_0,\delta) \iff d(x,x_0) < \delta$
$x \in f^{-1}(V) \iff f(x) \in V$.
here, let radius of V is $\frac{\epsilon}{2}$.
Then, because $f(x) \in V$, $~~d(f(x),f(x_0)) < \epsilon$.

At result, because V is arbitrary,
$$\forall \epsilon ~~\exists \delta ~~s.t.~~ d(x,x_0) < \delta \implies d(f(x),f(x_0)) < \epsilon$$

- $\epsilon-\delta$ reasoning. $\implies$ Define 1.0
$\forall x \in f^{-1}(V)$, because $f(x) \in V$ and $f(x_0) \in V$, $d(f(x),f(x_0)) < \epsilon$. here, let $\frac{\epsilon}{2}$ is redius of V.
by $\epsilon-\delta$  reasoning, $\exists \delta > 0 ~~s.t.~~ d(x,x_0) < \delta$.

Let $B(x,\delta):= \{y \in X|d(x,y) < \delta \}$.
$\forall y \in B(x,\delta),$ because $d(x,y) < \delta$, $d(f(x),f(y)) < \epsilon$. Thus, $f(y) \in B(f(x),\epsilon) \subset V$ because $V$ is open set.
Because $f(y) \in V$, $y \in f^{-1}(V)$.
Thus, $y \in B(x,\delta) \implies y \in f^{-1}(V)$.
At result, $B(x,\delta) \subset  f^{-1}(V)$.
Therefore $f^{-1}(V)$ is open set.

Q.E.D