Question

Prove that if X is an infinite set and τ is the set consisting of ∅ and every subset of X whose complement is finite, then (X,τ) is a topological space that is not a Hausdorff space. @ganeshie8 @myininaya @UnkleRhaukus @RadEn @robtobey @chmvijay @Luigi0210 @eliassaab @genius12 @Vincent-Lyon.Fr @Preetha @Kainui @ehuman @wolfe8 @lucaz @INeedHelpPlease? @khadeeja @thadyoung @A_clan @tester97 @Andras @Confusionist @linh412986 @BlackLabel @leozap1 @JMark @Microrobot @zacharyf @Rubio101 @NaomiBell1997 @MeganEdward @theballer225 @link,zoro @Goldenmoon @CrayolaCrayon_

Answer 1

It is easy to prove that is it is a topological space. Use the fact that the Complement of A intersection B is the complement of A union the complement of B.
\[
A\in \tau, \quad B\in \tau \implies A\cap B\in \tau

\]
Also
\[
A_i \in \tau \implies \cup_1^\infty A_i \in \tau
\]
Use the fact that
\[
(\cup_1^\infty A_i )^c = \cap_1^\infty A_i^c
\]

Answer 2

Suppose that
\[
x \ne y
\]
If O is an open set containing x, the complement of O is finite and cannot contain an open set. That shows that it is not Hausdorff (Why?)

Answer 3

@amistre64 Medal Elias please :D