Topological space: Difference between revisions

Latest revision as of 14:26, 12 April 2026

A topological space is a fundamental mathematical structure that generalizes the concept of geometrical spaces and continuity. A topological space is equipped with a collection of open sets, capturing the intuitive idea of "nearness" without necessarily defining a metric. Topological spaces are the objects of study in general topology.

Definition

An ordered pair $(X, τ)$ is a topological space on set $X$ , if $τ \subseteq 𝒫 (X)$ satisfies the following properties:

$X, \emptyset \in τ$ ,
if $𝒰 \subseteq τ$ , then $⋃_{U \in 𝒰} U \in τ$ ,
if $U_{1}, U_{2}, \dots, U_{n} \in τ$ , then $⋂_{i = 1}^{n} U_{i} \in τ$ .

Elements of $τ$ are called open sets.

Examples

Standard topology

The real line $ℝ$ equipped with the standard topology $τ_{ℝ}$ is a topological space.

The standard topology on $ℝ$ is defined by taking all open intervals as a basis. A set $U \subseteq ℝ$ is open, if for all point $x \in U$ , there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq U$ .

Proof

$\emptyset \in τ_{ℝ}$ vacuously; every point $x \in ℝ$ belongs to some open interval, like $(x - 1, x + 1)$ , which is open in $ℝ$ . Therefore by the definition, $ℝ \in τ_{ℝ}$ .
Let ${U_{i}}_{i \in I}$ be open sets and $U = ⋃_{i \in I} U_{i}$ . Take $x \in U$ , then $x \in U_{i}$ for some $i \in I$ . Because $U_{i}$ is open, there exists $(a, b)$ such that $x \in (a, b) \subset U_{i} \subset U$ . Therefore $U$ is open.
The finite intersection property can be proved by induction. Let $U, V$ be open and $x \in U \cap V$ . By definition of openness, there exists $(a_{1}, b_{1}) \subset U$ and $(a_{2}, b_{2}) \subset V$ such that $(a_{1}, b_{1}) ∋ x \in (a_{2}, b_{2})$ . Set $a = \max {a_{1}, a_{2}}$ and $b = \min {b_{1}, b_{2}}$ . Then $x \in (a, b) \subset (a_{1}, b_{1}) \cap (a_{2}, b_{2}) \subset U \cap V$ . Thus $U \cap V$ is open. By induction,

□

Basis

for every $x \in X$ , there exists $B \in ℬ$ with $x \in B$ ,
if $x \in B_{1} \cap B_{2}$ with $B_{1}, B_{2} \in ℬ$ , then there exists $B_{3} \in ℬ$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$ .

The topology generated by $ℬ$ consists of all unions of elements of $ℬ$ .

Topological properties

Some key properties of topological spaces include:

@@ Line 2: / Line 2: @@
 == Definition ==
-An [[ordered pair]] <math>(X, \tau)</math> is a topological space on set <math>X</math>, if <math>\tau\subseteq \mathcal{P}(X)</math> is a '''topology''', satisfying the following properties:
+An [[ordered pair]] <math>(X, \tau)</math> is a topological space on set <math>X</math>, if <math>\tau\subseteq \mathcal{P}(X)</math> satisfies the following properties:
 * <math>X, \varnothing \in \tau</math>,
@@ Line 13: / Line 13: @@
 === Standard topology ===
-The real line <math>\mathbb{R}</math> equipped with the '''standard topology''' <math>\tau_{\mathbb{R}}</math> is a topological space.
+The real line <math>\mathbb{R}</math> equipped with the standard topology <math>\tau_{\mathbb{R}}</math> is a topological space.
 The standard topology on <math>\mathbb{R}</math> is defined by taking all open intervals as a [[basis]]. A set <math>U\subseteq \mathbb{R}</math> is open, if for all point <math>x\in U</math>, there exists an open interval <math>(a,b)</math> such that <math>x\in (a,b)\subseteq U</math>.
@@ Line 19: / Line 19: @@
 * <math>\varnothing\in\tau_{\mathbb{R}}</math> vacuously; every point <math>x\in\mathbb{R}</math> belongs to some open interval, like <math>(x-1,x+1)</math>, which is open in <math>\mathbb{R}</math>. Therefore by the definition, <math>\mathbb{R}\in\tau_{\mathbb{R}}</math>.
 * Let <math>\{U_i\}_{i\in I}</math> be open sets and <math>U=\bigcup_{i\in I}U_i</math>. Take <math>x\in U</math>, then <math>x\in U_i</math> for some <math>i\in I</math>. Because <math>U_i</math> is open, there exists <math>(a,b)</math> such that <math>x\in (a,b)\subset U_i\subset U</math>. Therefore <math>U</math> is open.
-* Let <math>U, V</math> be open and <math>x\in U\cap V</math>. By definition of openness, there exists <math>(a_1,b_1)\subset U</math> and <math>(a_2,b_2)\subset V</math> such that <math>(a_1,b_1)\ni x \in (a_2,b_2)</math>. Set <math>a=\max\{a_1,a_2\}</math> and <math>b=\min\{b_1,b_2\}</math>. Then <math>x\in(a,b)\subset (a_1,b_1)\cap(a_2,b_2)\subset U\cap V</math>. Thus <math>U\cap V</math> is open. By induction, the finite intersection property holds.
+* The finite intersection property can be proved by induction. Let <math>U, V</math> be open and <math>x\in U\cap V</math>. By definition of openness, there exists <math>(a_1,b_1)\subset U</math> and <math>(a_2,b_2)\subset V</math> such that <math>(a_1,b_1)\ni x \in (a_2,b_2)</math>. Set <math>a=\max\{a_1,a_2\}</math> and <math>b=\min\{b_1,b_2\}</math>. Then <math>x\in(a,b)\subset (a_1,b_1)\cap(a_2,b_2)\subset U\cap V</math>. Thus <math>U\cap V</math> is open. By induction, <math></math>
-Therefore, <math>\tau_\mathbb{R}</math> is indeed a topology on <math>\mathbb{R}</math>.
 }}
-=== Discrete topology ===
-Let <math>X</math> be an arbitary set and define the '''discrete topology''' of <math>X</math> by <math>\tau=\mathcal{P}(X)</math>. Every subset of <math>X</math> is open in <math>\tau</math>.
-{{Proof|proof=* <math>\emptyset, X\subseteq X</math>, thus <math>\emptyset, X\in \tau</math>.
-* Let <math>X\supseteq A,B\in \tau</math>, since <math>A</math> and <math>B</math> are subsets of <math>X</math>, their union <math>A\cup B</math> and intersection <math>A\cap B</math> are also a subsets of <math>X</math>, which are in the topology <math>\tau</math>.
-Therefore the discrete topology of <math>X</math> is a topology.}}
-=== Indiscrete topology ===
-Let <math>X</math> be an arbitary set, the '''indiscrete topology''' of <math>X</math> is defined by <math>\tau=\{\emptyset, X\}</math>.
-{{Proof|proof=* By definition, <math>\emptyset, X\in \tau</math>.
-* <math>\emptyset \cup X=X\in\tau</math>.
-* <math>\emptyset \cap X=\emptyset\in\tau</math>.
-Therefore the indiscrete topology of <math>X</math> is a topology.}}
 ==Basis==