Show that the collection of subsets of $\Bbb R$
$\mathscr T =\{(a,+\infty)|a ∈ \Bbb R\} ∪ \{\Bbb R\} ∪ \{\emptyset\}$
is a topology i.e. a runs over all real numbers, i.e. for every a we have $(a,+\infty) \in \mathscr T $.
Show that $( \Bbb R, \mathscr T )$ is not a Hausdorff topological space
So I think I should be basing my answer for first part like part c from Prove that the following collection of subsets of $\Bbb R$ is a topology? but i'm not really sure, can anyone give any hints to help?
Best Answer
Let $\mathscr T =\{(a,+∞)|a ∈ \Bbb R \} ∪ \{\Bbb R\}∪\{\emptyset\}$
$\mathscr T$ is not $T_2$: let $x,y\in \Bbb R$ s.t. $x < y$. Let $A \in \mathscr T$ an open not empty set s.t. $x\in A$. If $A= \Bbb R$ then $y \in A$ if $A=(a, +\infty)$ then $y\in A$. So you can't find $A,B$ disjoint open neighborhoods of $x$ and $y$