Why do we need $X$ to be $T_1$ to be able to construct its Wallman compactification?

Question

I am struggling to understand the importance of $X$ to be $T_1$ to construct its Wallman compactification ($wX$).

Can a map $w:X\rightarrow wX$ be an embedding if $X$ is not $T_1$? Can $X$ be a dense subspace of $wX$ if $X$ is not $T_1$?

Note that $wX$ in our case is not Hausdorff.

Please help.

Answer 1

You need $T_1$ to associate every point with a closed ultrafilter: the embedding maps $x$ to $\mathcal{F}_x=\{F:x\in F\}$. Without $T_1$ this map would be ill-defined: if $\{x\}$ is not closed and $y$ is in the closure of $\{x\}$ then $\mathcal{F}_x\subseteq\mathcal{F}_y$ and if the two are not equal then $\mathcal{F}_x$ is not an ultrafilter and if they are equal then the map is not injective.