Abstract:
If $X$ is a regular hereditary Souslin space and $x\in X$ then either there exists a sequence $\{x_n:n=1,2,\dots\}\subset X\{x\}$ such that $x\in[{x_n:n=1,2,\dots}]$, or the pseudocharacter of $x$ in $X$ is no greater than countable. In other words, if $X$ is a hereditary Souslin bicompactum which is a $\chi$-space, then $X$ is a Frechet–Urysohn space.