most recent 30 from http://mathoverflow.net 2013-05-19T20:07:43Z http://mathoverflow.net/feeds/question/109534 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109534/a-question-about-q Jialiang He 2012-10-13T14:28:59Z 2012-10-14T00:18:32Z

<p>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 $? </p>

http://mathoverflow.net/questions/109534/a-question-about-q/109556#109556 Pietro Majer 2012-10-13T21:12:27Z 2012-10-14T00:18:32Z

<p>I think the answer is no, as shown by the following set $A$ (whose elements I will describe as subsets rather than binary functions).</p> <p>For any map $\sigma:\omega\to\omega+1:=\omega\cup\{\omega\}$ define its hypograph $$\mathrm{hypo}(\sigma):=\{(m,k)\in\omega\times\omega: k &lt; \sigma(m)\}\subset\omega\times\omega\ .$$</p> <p>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 <strong>(P)</strong> 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). </p> <p>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 <strong>(P)</strong>. A point in $\mathrm{cl}(A)\setminus A$ is exactly the hypograph of an increasing $\omega$-valued map, which never satisfies property <strong>(P)</strong>. </p>