Suppose that $X$ is a topological space (not supposed to be Hausdorff). Denote $X^*$ its Alexandroff extension.
It's well known that $X^*$ is Hausdorff if and only if $X$ is Hausdorff and locally compact.
My question is about weaker separation axiom (in particular $T_1$).
Do we have : $X$ is $T_1$ if and only if $X^*$ is $T_1$ ?
Yes, this is pretty much immediate from the definitions. If $X^*$ is $T_1$, then $X$ is $T_1$ since it is a subspace. Conversely, if $X$ is $T_1$, any singleton $\{x\}\subseteq X$ is a closed compact subset of $X$ and hence is still closed in $X^*$, and $\{\infty\}$ is closed in $X^*$ since its complement $X$ is open in $X^*$.