A compactness property for Borel sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T04:42:04Z http://mathoverflow.net/feeds/question/64095 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64095/a-compactness-property-for-borel-sets A compactness property for Borel sets Alex Simpson 2011-05-06T08:15:53Z 2011-05-07T04:26:55Z <p>Is the following generalised compactness property of Borel sets in a Polish space consistent with ZFC?</p> <p>(*) Let $\mathcal{B}$ be a family of $\aleph_1$-many Borel sets. If $\bigcap \mathcal{B} = \emptyset$ then for some countable $\mathcal{C} \subseteq \mathcal{B}$ it holds that $\bigcap \mathcal{C} = \emptyset$.</p> <p>In the special case that $\mathcal{B}$ is a family of open sets, it is not hard to show that property (*) is a consequence of $\neg\mathrm{CH} + \mathrm{MA}$ (where $\mathrm{CH}$ is the Continuum Hypothesis and $\mathrm{MA}$ is Martin's Axiom) - more specifically, a consequence of $\mathrm{MA}_{\aleph_1}$. Also, trivially, this special case of (*) itself implies $\neg \mathrm{CH}$.</p> <p>However, I am stuck as to the consistency of (*) for general Borel sets. Actually, I have been unable to show consistency even when $\mathcal{B}$ is restricted to families of $F_\sigma$ sets.</p> http://mathoverflow.net/questions/64095/a-compactness-property-for-borel-sets/64102#64102 Answer by Clinton Conley for A compactness property for Borel sets Clinton Conley 2011-05-06T09:54:17Z 2011-05-06T10:08:59Z <p>This compactness property is never true, even for collections of $F_\sigma$ subsets of an uncountable Polish space. One way to see this is to fix your favorite example of an $F_\sigma$ graph $G$ with clique number $\aleph_1$ and a maximal $G$-clique $K$. Then let your family be $\{G_x : x \in K\}$, where $G_x$ is the set of neighbors of $x$ in $G$.</p> <p>For a simple example of such a graph, see, e.g., <a href="http://www.math.cas.cz/preprint/pre-207.pdf" rel="nofollow">http://www.math.cas.cz/preprint/pre-207.pdf</a></p> http://mathoverflow.net/questions/64095/a-compactness-property-for-borel-sets/64115#64115 Answer by Juris Steprans for A compactness property for Borel sets Juris Steprans 2011-05-06T11:52:38Z 2011-05-06T11:52:38Z <p>It is worth noting that a construction provided by Hausdorff more than a hundred years before the result of Kubis and Vejnar also provides a counterexample. A Hausdorff gap is a family of subsets of the natural numbers $A_\xi, B_\xi$ for $\xi\in\omega_1$ such that $A_\xi \subseteq^* A_\eta \subseteq^* B_\eta \subseteq^* B_\xi$ for $\xi&lt; \eta$ but such that there is no subset of $\mathbb N$ such that $A_\xi \subseteq^* X \subseteq^* B_\xi$ for all $\xi$. (Here $\subseteq^*$ means inclusion except for a finite set.) Letting $S_\xi$ be the Borel set of all $X\subseteq \mathbb N$ such that $A_\xi \subseteq^* X \subseteq^* B_\xi$ yields the counterexample.</p> http://mathoverflow.net/questions/64095/a-compactness-property-for-borel-sets/64170#64170 Answer by Liang Yu for A compactness property for Borel sets Liang Yu 2011-05-07T03:51:49Z 2011-05-07T04:26:55Z <p>Here is a even simpler example.</p> <p>Let $\{x_{\alpha} \}_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}=\{y\mid y\geq_T x_{\alpha}\}$. Each $B_{\alpha}$ is a boldface $\Sigma^0_3$ set.</p> <p>Then for any countable ordinal $\beta$, $\bigcap_{\alpha&lt;\beta}B_{\alpha}$ is not empty but $\bigcap_{\alpha&lt;\omega_1}B_{\alpha}=\emptyset$.</p>