Exercise On topology not separated but satisfy the Borel-Lebesgue property I have a set $E\neq \emptyset $ with the cofinite topology.< The question is if $E$ is infinite prove that is not separated but it satify the Borel-Lebesgue property. We say that $E$ is separated if $\forall x,y\in E, \exists U\in \mathcal{V}_x, V\in\mathcal{V}_y, U\cap V=\emptyset $ < How to find $x,y\in E$ such that all the neighborhoods are not disjoint, i.e how to prove that $E$ is not separated? Thank you.

Any two points will do to show that $E$ is not Hausdorff (or separated). Take any $x \
eq y$ in $E$. Let $U_x$ be any open neighbourhood of $x$ in $E$, and $U_y$ be any open neighbourhood of $y$. Then $E \setminus U_x$ is finite and $E \setminus U_y$ is finite, so their union is finite as well. The set $E \setminus ((E \setminus U_x) \cup (E \setminus U_y))$, is infinite as $E$ is infinite. But that set is just $U_x \cap U_y$ by de Morgan, so the intersection of $U_x$ and $U_y$ is non-empty.

So any two non-empty open sets of $E$ intersect (in an infinite set). This shows that $E$ is not Hausdorff in a strong way, but also that $E$ is connected, e.g.