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Consider the assertion:

Every connected, but not necessarily paracompact, n-manifold is of cardinality $2^{\aleph_0}$ (at least assuming the axiom of choice).

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.

What is the situation for n>1?

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  • $\begingroup$ Your assertion is about the size of manifolds, but your comment is about the number of such manifolds. What is your exact question? $\endgroup$ Jun 16, 2011 at 16:06
  • $\begingroup$ As posed, about cardinality. One could pose the weaker question, of whether there is any upper bound on cardinality. That in turn is equivalent to the motivating question, of whether there are only 'set-many' homeomorphism types of connected manifolds. Assuming paracompactness, there is a trivial 'yes' answer. $\endgroup$ Jun 16, 2011 at 16:25
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    $\begingroup$ You should add that you want Haudsdorff, since without this there is a classical counterexample. $\endgroup$ Jun 16, 2011 at 17:15
  • $\begingroup$ True enough. Meanhwile, which of the above statements would be violated? $\endgroup$ Jun 16, 2011 at 17:37
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    $\begingroup$ @Adam: A non-Hausdorff manifold of arbitrary cardinal larger than the continuum can be constructed by considering $\mathbb R$ with many, many origins, generalizing the usual construction of the line with two. $\endgroup$ Jun 16, 2011 at 21:12

2 Answers 2

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A connected Hausdorff manifold with more than one point has cardinality $2^{\aleph_0}$.

Here's a proof sketch.

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.

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 $2^{\aleph_0}$.

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  • $\begingroup$ That's a really nice argument. $\endgroup$
    – Todd Trimble
    Jun 16, 2011 at 23:39
  • $\begingroup$ Indeed! It is the natural thing to do, yet it is not quite obvious how to carry it out. $\endgroup$ Jun 17, 2011 at 0:04
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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 $X\cap \mathfrak{M} = X$.

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  • $\begingroup$ It is obvious that such an $\mathfrak M$ exists? $\endgroup$ Jun 17, 2011 at 0:06
  • $\begingroup$ Yes, this follows from the Lowenheim-Skolem Theorem. $\endgroup$ Jun 17, 2011 at 0:34
  • $\begingroup$ @Juris: Nice solution! Technically speaking, Lowenheim-Skolem cannot be invoked unless we are working in Kelley-Morse theory of classes (where there is a truth-definition for $(V,\in))$; but what "saves the day" and makes your proof implementable in $ZF$, is the $ZF$-Reflection Theorem. $\endgroup$
    – Ali Enayat
    Jun 17, 2011 at 16:46
  • $\begingroup$ @Ali: The phrase "elementary submodel of a large fragment of set theory" does sweep a bit under the rug. I had in mind simply, for any manifold taking a submodel of $V_\kappa$ where $\kappa$ is larger than he cardinality of the manifold. Of course, this also the strategy for proving reflection. $\endgroup$ Jun 18, 2011 at 20:09
  • $\begingroup$ @Juris: right, I see your point. By the way, is there a similar proof [using elementary submodels] of Arhangel'ski's theorem about the maximum cardinality of first countable Lindelof Haussdorf spaces? $\endgroup$
    – Ali Enayat
    Jun 21, 2011 at 20:09

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