Table of Contents
Definition
Let (X, τ) be a topological space, then the subset ‘A’ of X is said to be a closed set if Ac ∈τ (Ac is open in X).
OR
Let (X, τ) be a topological space and A⊆ X and if A is an open set then Ac is called a closed set in topological space.
Remarks
- In (X, τdiscrete) every subset of X is a closed set.
- Φ and set X are always closed.
Example
Let X= {1,2,3} and τ= {φ, {1}, {2, 3}, X} is a topology on set X. if τ is a topology on X then elements in τ are called open sets.
Φc =X−φ= X ∈τ
{1}c=X−{1}= {2, 3} ∈τ
{2, 3}c=X− {2, 3} = {1} ∈τ
Xc=X−X= φ ∈τ
So,
Φ, {1}, {2, 3} and X are closed subsets of X.
Theorem
In a discrete topological space, every subset of set X is closed.
Proof
Let (X, τ) be a discrete topological space, then τ=P(X) so every subset of X is a member of τ and therefore complement of every subset of set X is belongs to τ. This shows that every subset of set X is a closed subset.
Example
In a discrete topological space, every subset of set X is closed.
Solution
Let X= {a, b}
In a discrete topology τ=P(X) (contains all subsets of set X)
τ= {φ, {a}, {b}, x}
- As Φ ∈ τ, X ∈ τ
⇒ Φ, X ∈ τ
The first condition for topology is satisfied.
- φ∩{a}=φ∈ τ, φ∩ {a} = φ∈ τ, φ∩{b}=φ∈ τ, φ∩ X=φ∈ τ
{a}∩ {b} =φ∈ τ, {a}∩X={a} ∈ τ
{b} ∩ X= {b} ∈ τ
φ∩ {a}∩ {b} ∩ X=φ∈ τ
As the intersection of a finite number of members of τ belongs to τ.
- φ∪{a}={a}∈ τ, φ∪ {b} = {b,} ∈ τ, φ∪ X=X∈ τ
{a}∪ {b} = X ∈ τ, {a}∪X=X∈ τ
{b} ∪ X= X ∈ τ
φ∪ {a}∪ {b} ∪ X=X∈ τ
As the union of any number of members of τ belongs to τ.
So τ is a discrete topology on x and every subset of X is closed.
Because the compliment of every subset of X belongs to τ so every subset of X is a closed set.
Φc =X−φ= X ∈τ
{1}c=X− {1}= {2} ∈τ
{2 }c=X− {2} = {1} ∈τ
Xc=X−X= φ ∈τ
So φ, {a}, {b} and x are both open and closed subsets of X.
Theorem
Let (X, τ) be a topological space, Then
- φ and X are closed set.
- The intersection of any number of closed subsets of X is closed in topological space.
- The union of a finite number of closed subsets of X is closed.
proof
- The complement of φ is φc=X Which is open so φ is closed similarly, the complement of X is Xc=φ which is open
Therefore, φ is a closed set and X is a closed set.
\begin{array}{l}2.~~~Let\text{ }(X,\tau )\text{ }be\text{ }a\text{ }topological\text{ }space,\text{ }\{{{A}_{i}}:\text{ }i\in I\}\text{ }be\text{ }an\text{ }arbitrary\text{ }\\collection\text{ }of\text{ }closed\text{ }subsets\text{ }of\text{ }X.\text{ }since\text{ }each\text{ }{{A}_{i}}is\text{ }a\text{ }closed\text{ }set,\text{ }\\so\text{ }each\text{ }{{A}_{i}}is\text{ }an\text{ }open\text{ }set.\\\Rightarrow \bigcup\limits_{{i\in I}}{{{{A}_{i}}^{c}}}\in \tau \\\Rightarrow \bigcup\limits_{{i\in I}}{{{{A}_{i}}^{c}}}\in \tau \text{ is open }\\\Rightarrow {{\left( {\bigcap\limits_{{i\in I}}{{{{A}_{i}}}}} \right)}^{c}}is\text{ open (By de-Morgan }\!\!'\!\!\text{ s law) }{{(A\cup B)}^{c}}={{A}^{c}}\cap {{B}^{c}}^{{~~~}}\\~~~~~~~~~we\text{ }write\text{ }complement\text{ }as\\\Rightarrow X-\bigcap\limits_{{i\in I}}{{{{A}_{i}}}}\text{ is open}\\If\text{ }the\text{ }complement\text{ }of\text{ }the\text{ }intersection\text{ }of\text{ }collection\text{ }{{A}_{i}}is\text{ }open,\text{ }so\text{ }by\text{ }definition\\\text{ }of\text{ }the\text{ }closed\text{ }set\text{ }we\text{ }can\text{ }say\text{ }that\text{ }intersection\text{ }of\text{ }{{A}_{i}}is\text{ }closed\\\text{ }because\text{ }its\text{ }complement\text{ }is\text{ }open.\\\Rightarrow X-\bigcap\limits_{{i\in I}}{{{{A}_{i}}}}\text{ is open}\\\Rightarrow \bigcap\limits_{{i\in I}}{{{{A}_{i}}}}\text{ is closed }\end{array} \begin{array}{l}3.~~~Let\text{ }(X,\tau )\text{ }be\text{ }a\text{ }topological\text{ }space,\text{ }Let\text{ }{{F}_{1}}{{,}_{{}}}{{F}_{2}},\text{ }{{F}_{3}},\text{ }\ldots ,\text{ }{{F}_{n}}be\text{ }the\\\text{ }collection\text{ }of\text{ }finite\text{ }closed\text{ }subsets\text{ }of\text{ }X\text{ }\\\left( {also\text{ }we\text{ }write\text{ }{{F}_{i}}collection\text{ }and\text{ }i=1,\text{ }2,\text{ }3,\text{ }\ldots ,\text{ }n} \right).\\{{F}_{1}}^{c},{{F}_{2}}^{c},{{F}_{3}}^{c},...,{{F}_{n}}^{c}\text{ is open}\\{{F}_{1}}^{c},{{F}_{2}}^{c},{{F}_{3}}^{c},...,{{F}_{n}}^{c}\in \tau \\\text{By definition of topology}\\{{F}_{1}}^{c}\bigcap {{F}_{2}}^{c}\bigcap {{F}_{3}}^{c}\bigcap ...\bigcap {{F}_{n}}^{c}\in \tau \\{{F}_{1}}^{c}\bigcap {{F}_{2}}^{c}\bigcap {{F}_{3}}^{c}\bigcap ...\bigcap {{F}_{n}}^{c}\text{ is open}\\\text{By De-morgan }\!\!'\!\!\text{ s law }{{(A\cap B)}^{c}}={{A}^{c}}\cup {{B}^{c}}\\{{({{F}_{1}}\bigcup {{F}_{2}}\bigcup {{F}_{3}}\bigcup ...\bigcup {{F}_{n}}\text{)}}^{c}}\text{ is open}\\\text{we also write complement as}\\X-{{F}_{1}}\bigcup {{F}_{2}}\bigcup {{F}_{3}}\bigcup ...\bigcup {{F}_{n}}\text{ open}\\\Rightarrow {{F}_{1}}\bigcup {{F}_{2}}\bigcup {{F}_{3}}\bigcup ...\bigcup F\text{ closed}\end{array}
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