most recent 30 from http://mathoverflow.net 2013-05-23T08:58:51Z http://mathoverflow.net/feeds/question/55178 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55178/how-much-choice-do-we-need-for-regularity-of-product-of-regular-spaces Silver 2011-02-12T03:10:52Z 2011-02-12T11:51:58Z

<p>It is usually stated that the (possibly uncountable) product of regular topological spaces is regular. However the only proof that I know of this fact seems to use the full axiom of choice :</p> <p>See <a href="http://topospaces.subwiki.org/wiki/Regularity_is_product-closed" rel="nofollow">here</a> (proof based on Mukres' Topology p.197). </p> <p>Do we really need AC (does this imply AC) ? Would it hold in ZF, or in ZF + a weaker form of choice ?</p>

http://mathoverflow.net/questions/55178/how-much-choice-do-we-need-for-regularity-of-product-of-regular-spaces/55195#55195 François G. Dorais 2011-02-12T11:51:58Z 2011-02-12T11:51:58Z

<p>The standard proof that the product of regular spaces is regular does not use the Axiom of Choice.</p> <p>Suppose $X_i$ ($i \in I$) are regular spaces and let $X = \prod_{i \in I} X_i$. Given a point $x \in X$ and a basic neighborhood $U = \prod_{i \in I} U_i$ of that point, we can always find a basic subneighborhood $V = \prod_{i \in I} V_i$ of $x$ such that $\overline{V} \subseteq U$. Indeed, when $U_i = X_i$ let $V_i = X_i$ too. For the finitely many $i \in I$ such that $U_i \neq X_i$, pick a neighborhood $V_i$ of $x_i$ such that $\overline{V}_i \subseteq U_i$. This only involves making finitely many choices, so the Axiom of Choice is not necessary.</p>