A question from Arhangel’skii-Buzyakova

The question is also posted here, however there is no answer.

Recently, I am reading the paper: On linearly Lindelöf and strongly discretely Lindelöf spaces by Arhangel'skii and Buzyakova. Here is the Lemma 2.2 in paper. (Sorry for the picture is not clear.)

The fifth line from last. How could I see that for any $a\in H$ and $z\in Z\setminus H$, there exists an element $V$ of $\mathcal{U}$ such that $a\in V$ and $z\notin V$? Thanks very much.

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1 Answer

The assertion seems wrong, but the final conclusion is correct:

Fix $x \in X \setminus H$. For each $a \in H$ there exists an element $V$ of $\mathcal{U}$ such that $a \in V$ and $x \notin V$ (if $a \in Y$ then $V \in \xi_a$ and if $a \in H \setminus Y$ then $V \in \gamma_\alpha$ for an $\alpha$ for which $a \in K_\alpha$). So by compactness of $H$ there is a neighborhood $W$ of $H$ in $Z$ for which $x \notin W$, such that $W$ is the union of a finite subcollection of $\mathcal{U}$.

Therefore the family $\mu$ actually satisfies $$X \cap \bigcap \mu = X \cap H$$ even though you might not get $\bigcap \mu = H$. But this is enough because the left-hand side is a $G_{2^\omega}$-set in $X$ and the right-hand side is $Y$.

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 Thanks for the answer. So it is wrong at $x\in Z\setminus H$, where it should be $x\in X\setminus H$. Am I right? – Paul Mar 7 at 0:07 Yes that´s right, but then also the claims that ⋂μ=H and that $H$ is a $G_{2^\omega}$-set in $Z$ are not justified. – Ramiro de la Vega Mar 7 at 11:39