most recent 30 from http://mathoverflow.net 2013-05-19T04:07:48Z http://mathoverflow.net/feeds/question/67962 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67962/cardinality-of-connected-manifolds Adam Epstein 2011-06-16T15:47:24Z 2011-06-16T23:05:47Z

<p>Consider the assertion:</p> <p>Every connected, but not necessarily paracompact, n-manifold is of cardinality $2^{\aleph_0}$ (at least assuming the axiom of choice). </p> <p>For n=1 this may be proved via enumeration of the short list of examples. The essential point is that while there is a Long Line, there is no Extra Long Line.</p> <p>What is the situation for n>1?</p>

http://mathoverflow.net/questions/67962/cardinality-of-connected-manifolds/67985#67985 Stephen S 2011-06-16T18:47:55Z 2011-06-16T18:47:55Z

<p>A connected Hausdorff manifold with more than one point has cardinality $2^{\aleph_0}$.</p> <p>Here's a proof sketch.</p> <p>For each point $x$ of the manifold, let $U_x$ be an open Euclidean neighbourhood of $x$. Define a transfinite sequence of subsets $V_\alpha$ of the manifold as follows. Choose some point $y$ of the manifold, and put $V_0=U_y$. For each ordinal $\alpha$, let $V_{\alpha+1}$ be the union of $U_x$ over all $x$ such that $x$ is a limit of a sequence in $V_\alpha$. Take unions at limit ordinals.</p> <p>Each $V_\alpha$ is open, and $V_{\omega_1}$ is clearly sequentially closed, and therefore closed (as manifolds are first countable), and is therefore the whole space (by connectedness). As we are assuming that the manifold is Hausdorff, sequential limits are unique, so it follows easily by transfinite induction that $V_{\omega_1}$ has cardinality $2^{\aleph_0}$.</p>

http://mathoverflow.net/questions/67962/cardinality-of-connected-manifolds/68015#68015 Juris Steprans 2011-06-16T22:54:33Z 2011-06-16T22:54:33Z

<p>Stephen's argument can also be phrased in the language of model theory. Given a connected manifold $X$, consider an elementary submodel of a large fragment of set theory $\mathfrak{M}$ that (1) contains all the reals, (2) is closed under countable subsets, (3) contains $X$ as an element and (4) is of size $2^{\aleph_0}$. It suffices to show that $X\subseteq \mathfrak{M}$. But $X\cap \mathfrak{M}$ is open since each point of $X$ has a neighbourhood of size $2^{\aleph_0}$ and, by elementarity there is a bijection from the reals to this neighbourhood and, since the $\mathfrak M$ contains the reals it must also contain the image of this bijection, namely the neighbourhood. But $X\cap \mathfrak{M}$ is also closed since $\mathfrak M$ contains all sequences from $X\cap \mathfrak{M}$ and hence their unique (by Hausdorffness) limits. By connectedness $X\cap \mathfrak{M} = X$.</p>