Let $$X=(-\infty,0) \cup \{5,8\}$$ $$\mathcal{B}=\{(a,b)\mid a < b < 0\} \cup \{(a,0) \cup \{5\} \mid a < 0\} \cup \{(a,0) \cup \{8\} \mid a < 0 \} $$ Show that $\mathcal{B}$ is a base for a topology on $X$ and it is not Hausdorff
To show it is a base I show that for all $$x \in X$$ there exist $$W \in \mathcal{B}$$ s.t $$x \in W \subseteq X$$ I take the cases $$x \in (-\infty,0)$$ or $$ x \in \{5,8\} $$ and for the second condition that for $B_1 , B_2$ in $\mathcal{B}$ s.t $$ x \in B_1 \cap B_2 $$ there exists $$B_3 \in \mathcal{B} \text{ s.t. } x \in B_3 \subseteq B_1 \cap B_2 $$ I take the 6 cases but to show that it is not Hausdorff can I take $x=5$ , $y=8$ and thus there exist no open basis disjoint s,t $$ x \in U \text{ and } y \in V $$
is it true ??
Yes, it’s true: $5$ and $8$ do not have disjoint open neighborhoods. This is easy to see and not hard to prove. If $U$ is an open nbhd of $5$, and $V$ is an open nbhd of $8$, then there are negative real numbers $a$ and $b$ such that $(a,0)\cup\{5\}\subseteq U$ and $(b,0)\cup\{8\}\subseteq V$. Let $c=\max\{a,b\}$: then $U\cap V\supseteq(c,0)\ne\varnothing$.