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In the Picture blew:enter image description here The paper can be downloaded here. Why $z \in \overline{A}$? Thanks.

A point $x$ of a space $X$ is called $G_\omega$-separated from a subset $Y$ of $X$ if there is a closed $G_\omega$ -set $P$ in $X$ such that $x \in P$ and the sets $Y$ and $P$ are disjoint.

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    $\begingroup$ Downvoted, although I haven't voted to close (yet). $\endgroup$
    – Yemon Choi
    Feb 25, 2013 at 16:28
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    $\begingroup$ Given the existing votes to close, I suggest that discussion (about closing or keeping open this question) takes place at tea.mathoverflow.net/discussion/1539/… $\endgroup$
    – Yemon Choi
    Feb 25, 2013 at 16:40
  • $\begingroup$ @Yemon Choi: You are not very welcome. It seems that you only can do this, what other could you do? $\endgroup$
    – Paul
    Feb 26, 2013 at 3:45
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    $\begingroup$ @Luke: perhaps you would get a better reception if you gave some explanation of why you have posted so many questions in general topology, often in complicated areas but with the questions usually seeming to have quick answers $\endgroup$
    – Yemon Choi
    Feb 26, 2013 at 6:56
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    $\begingroup$ Speaking for myself: I'm afraid I don't react well to someone who uses to different names on different sites, and in neither case is upfront about their motivation for questions or previous experience $\endgroup$
    – Yemon Choi
    Feb 26, 2013 at 6:58

1 Answer 1

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I may be missing something because this is quite elementary, but:

Since $Z$ is regular, for any open neighborhood $U$ of $z$ there is an open $V$ such that $z\in V$ and $\overline{V}\subseteq U$ (apply regularity to $z$ and the complement of $U$). Do this countably many times and take intersection, and you'll find that for any open neighborhood $U$ of $z$ there is a closed $G_\omega$-set $P$ such that $z\in P\subseteq U$.

If follows easily that if $z$ is not $G_\omega$-separated from $A$ (that is, if (b) fails), then $z$ must be in the closure of $A$.

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  • $\begingroup$ You may find tea.mathoverflow.net/discussion/1539/… provides some background $\endgroup$
    – Yemon Choi
    Feb 25, 2013 at 16:25
  • $\begingroup$ @Todd Eisworth: How could you see that $\cap V_i$ is closed as a closed $G_\omega$-set $P$ of $z$? $\endgroup$
    – Paul
    Feb 26, 2013 at 3:50
  • $\begingroup$ The nesting guarantees that the intersection of the V_i is the same as the intersection of their closures, hence closed. $\endgroup$
    – Todd Eisworth
    Feb 26, 2013 at 3:56
  • $\begingroup$ That is, you make sure $V_{i+1}\subseteq\overline{V}_{i+1}\subseteq V_i$, and then $\cap V_i = \cap \overline{V}_i$ is closed and a $G_\omega$ set. $\endgroup$
    – Todd Eisworth
    Feb 26, 2013 at 4:01
  • $\begingroup$ @Todd Eisworth: Yes. I got it. Thanks! $\endgroup$
    – Paul
    Feb 26, 2013 at 4:11

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