## Table of Contents

#### Theorem statement

Let X be a non-empty set.

τ= {φ, X, A⊆X such that A^{C }= X−A is finite} prove that τ is a topology on X.

##### Proof

- φ ∈ τ

X^{C}=X−X =φ finite

⇒X ∈ τ

- Let A
_{1}, A_{2}, A_{3}, —, A_{n}∈τ

We want to show that A_{1}∩ A_{2}∩ A_{3}∩ —∩ A_{n} ∈τ

A_{1}, A_{2}, A_{3}, —, A_{n} ∈τ_{ }

A_{1}^{c}, A_{2}^{c}, A_{3}^{c}, —, A_{n}^{c} are finite

As the union of a finite number of a finite set is finite.

A_{1}^{c}∪ A_{2}^{c}∪ A_{3}^{c}∪ —∪ A_{n}^{c} are finite

By De Morgan’s law (A∩B)^{C}=A^{C}∪B^{C}

(A_{1}∩ A_{2}∩ A_{3}∩ —∩ A_{n})^{ c} is finite

⇒ A_{1}∩ A_{2}∩ A_{3}∩ —∩ A_{n} ∈τ

The second condition for topology is satisfied

- Let A
_{1}, A_{2}, A_{3}, — ∈τ_{ }

⇒ A_{1}^{c}, A_{2}^{c}, A_{3}^{c}, — are finite

As the intersection of a finite set is finite.

A_{1}^{c}∩ A_{2}^{c}∩ A_{3}^{c}∩ — are finite

By De Morgan’s law (A∪B)^{C}=A^{C}∩B^{C}

(A_{1}∪ A_{2}∪ A_{3}∪ —)^{ c} is finite

⇒ A_{1}∪ A_{2}∪ A_{3}∪ — ∈τ

The third condition for topology is satisfied

Hence τ is a topology on X.

**Ex**ample

Let X be a non-empty set.

τ= {φ, X, A⊆X such that A^{C }= X−A is finite} prove that τ is a topology on X.

##### solution

- X, φ ∈τ

This shows that X and φ belong to τ.

So the first condition for topology is satisfied

∪ | φ | A | A^{c} | X |

φ | φ | A | A^{c} | X |

A | A | A | X | X |

A^{c} | A^{c} | X | A^{c} | X |

X | X | X | X | X |

This shows that the union of any number of numbers of τ belongs to τ.

The second condition for topology is satisfied.

∩ | φ | A | A^{c} | X |

φ | φ | Φ | Φ | Φ |

A | Φ | A | Φ | A |

A^{c} | Φ | Φ | A^{c} | A^{c} |

X | Φ | A | A^{c} | X |

This shows that the intersection of any number of numbers of τ belongs to τ.

The third condition for topology is satisfied.

This shows that τ is the topology on X.

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