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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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Your assertion is about the size of manifolds, but your comment is about the number of such manifolds. What is your exact question? – Juris Steprans Jun 16 '11 at 16:06
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. – Adam Epstein Jun 16 '11 at 16:25
You should add that you want Haudsdorff, since without this there is a classical counterexample. – Juris Steprans Jun 16 '11 at 17:15
True enough. Meanhwile, which of the above statements would be violated? – Adam Epstein Jun 16 '11 at 17:37
@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. – Mariano Suárez-Alvarez Jun 16 '11 at 21:12

2 Answers 2

up vote 14 down vote accepted

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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That's a really nice argument. – Todd Trimble Jun 16 '11 at 23:39
Indeed! It is the natural thing to do, yet it is not quite obvious how to carry it out. – Mariano Suárez-Alvarez Jun 17 '11 at 0:04

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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It is obvious that such an $\mathfrak M$ exists? – Mariano Suárez-Alvarez Jun 17 '11 at 0:06
Yes, this follows from the Lowenheim-Skolem Theorem. – Juris Steprans Jun 17 '11 at 0:34
@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. – Ali Enayat Jun 17 '11 at 16:46
@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. – Juris Steprans Jun 18 '11 at 20:09
@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? – Ali Enayat Jun 21 '11 at 20:09

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