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I came across a question of a Buzyakova's 2005 paper. This is a paragraphy of his paper, which is the last example in his paper. (I know it is a little complex to ask the complete question without the paper. ) enter image description here

I want to know in the last line, why $N-\cup_{\beta <\alpha}N_\beta$ is not empty.

Thanks for any help and everyone who comes across this question, please vote me:)

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up vote 3 down vote accepted

Let $N$ be an infinite subset of $Q$ that is closed in $\bigcup_{\beta<\alpha}X_\beta$; then every point of $\left(\bigcup_{\beta<\alpha}X_\beta\right)\setminus N$ has a nbhd disjoint from $N$. In particular, for each $\beta<\alpha$ there is a $B_\beta\in\mathcal{B}_{x_\beta}$ such that $N\cap B_\beta=\varnothing$. If you go on to read the construction of the local base $\mathcal{B}_{x_\alpha}$, you’ll see that each $\mathcal{B}_{x_\beta}$ is countable and nested, so that if $N\cap B_\beta=\varnothing$, then $N$ intersects at most finitely many members of $\mathcal{B}_{x_\beta}$. By definition, therefore, $N\notin\mathcal{N}_\beta$ for any $\beta<\alpha$, i.e., $N\in\mathcal{N}\setminus\bigcup_{\beta<\alpha}\mathcal{N}_\beta$.

(By the way, Raushan Buzyakova is a woman.)

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Haha:) I'm sorry for I think she is a man. What's meaning of "nested"? – Paul Jul 12 '12 at 6:09
@John: In this case it means that each local base has the form $\{B_n:n\in\omega\}$, where $B_0\supseteq B_1\supseteq B_2\supseteq\ldots$. – Brian M. Scott Jul 12 '12 at 19:16
Oh, I see. Thanks, Brain. – Paul Jul 13 '12 at 0:26

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