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.)

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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.


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 $Y$.

  • $\begingroup$ 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? $\endgroup$ – Paul Mar 7 '13 at 0:07
  • $\begingroup$ 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. $\endgroup$ – Ramiro de la Vega Mar 7 '13 at 11:39

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