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Let A=$\{a_n : n\in \omega \}\subset 2^{\omega\times\omega}$ be nonempty countable without isolated points (i.e. homeomorphic to $\mathbb{Q}$), and satisfy $ \forall n\in \omega \exists^\infty m|\{k:a_n(m,k)=1\}|=\omega $. Does there exist $ a\in cl(A)\setminus A$ satisfying $\exists^\infty m|\{k:a(m,k)=1\}|=\omega $?

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what's the meaning of "$\exists^\infty$"? – Pietro Majer Oct 13 '12 at 15:57
"There are infinitely many" – Stefan Geschke Oct 13 '12 at 16:15
up vote 3 down vote accepted

I think the answer is no, as shown by the following set $A$ (whose elements I will describe as subsets rather than binary functions).

For any map $\sigma:\omega\to\omega+1:=\omega\cup\{\omega\}$ define its hypograph $$\mathrm{hypo}(\sigma):=\{(m,k)\in\omega\times\omega: k < \sigma(m)\}\subset\omega\times\omega\\ .$$

An increasing map $\sigma:\omega\to\omega+1$ takes the value $\omega$ if and only if its hypograph contains a subset of the form $[n,\omega)\times \omega$, thus, if and only if it has the property (P) stated in the question (there exist infinitely many $m$ such that there exist infinitely many $k$ such that $(m,k)\in \mathrm{hypo}(\sigma)$; which is indeed verified quite in a strong way).

Let $A$ be the set of all hypographs of all increasing maps that take the value $\omega$. Clearly, the set $A$ is countable, with no isolated points; and all its elements enjoy property (P). A point in $\mathrm{cl}(A)\setminus A$ is exactly the hypograph of an increasing $\omega$-valued map, which never satisfies property (P).

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Nice example Pietro. – Ramiro de la Vega Oct 14 '12 at 3:28
Tanks for your answer,you are right. – Jialiang He Oct 14 '12 at 6:44

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