A question on right-separated space 
Q1, How to take the subspace $Y$ from $X$ which is right-separated by the indexation? 
Q2, Why cannot $Y$ be dually metalindelof?
Thanks for your help.
 A: It’s well-known that $hl(X)$ is the supremum of the cardinalities of right-separated sequences in $X$. Here $hl(X)\ge\kappa^+$, so $X$ must contain a right-separated sequence of length $\kappa^+$. To see that $Y$ is not dually metaLindelöf, for $\xi<\kappa^+$ let $U(x_\xi)=Y_{\xi+1}$; $\mathscr{U}=\{U(x_\xi):\xi<\kappa^+\}$ is a nbhd assignment for $Y$. Suppose that $Z\subseteq Y$ is metaLindelöf, and $\mathscr{U}(Z)=Y$; clearly $|Z|=\kappa^+$. Recall that $hd(Y)=\kappa$, so $Z$ has a dense subset $D$ of cardinality $\kappa$. Let $\mathscr{V}=\{U(x_\xi)\cap Z:x_\xi\in Z\}$; this is an open cover of $Z$, and $|V|\le\kappa$ for each $V\in\mathscr{V}$. Suppose that $\mathscr{R}$ is an open refinement of $\mathscr{V}$ covering $Z$. $D$ is dense in $Z$, so $R\cap D\ne\varnothing$ for each $R\in\mathscr{R}$, and therefore $Z=\operatorname{St}(D,\mathscr{R})$. $|D|\le\kappa$, and $|Z|>\kappa$, so there must be a point $z\in Z$ such that $|\{R\in\mathscr{R}:z\in R\}|>\kappa\ge\omega$. In particular, $\mathscr{R}$ is not point-countable, and therefore $Z$ is not metaLindelöf.
