MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A manifold is usually defined as a second-countable hausdorff topological space which is locally homeomorphic to Rn. My understanding is that the reason "second-countable" is part of the definition is to make sure that the space is paracompact, which you want so that you get locally finite partitions of unity. Once you have locally finite partitions of unity, basically anything you can multiply by a function can be constructed locally (any presheaf that is a module over the sheaf of functions is automatically a sheaf), a property you want manifolds to have.

But do we unnecessarily throw out some paracompact topological spaces which "should" be manifolds by requiring second-countability. A boring example is an uncountable disjoint union of manifolds, but there are other more interesting spaces that kind of look like they should be manifolds.

In particular, is the long line paracompact? Should I consider it a manifold?

share|cite|improve this question
up vote 9 down vote accepted

Here's another proof, which shows that any connected paracompact locally Euclidean space X is second-countable. Cover X by Euclidean charts and take a locally finite refinement. Say an open set is good if it only intersects finitely many of the charts. Now take any point x and take a good neighborhood of it. The charts that intersect that good neighborhood can then themselves be covered by countably many good open sets. There are then only countably many charts intersecting those good open sets, and those charts can be covered by countably many good sets. Iterating this countably many times, you get an open set U associated to x which is covered by countably many charts such that if a chart intersects U, it is contained in U. It follows that the complement of U is also a union of charts, so by connectedness U is all of X. Thus X can be covered by countably many charts and is second-countable.

share|cite|improve this answer

For connected locally Euclidean spaces, paracompactness is equivalent to second countable. Therefore if we truly insisted that our manifolds be second countable then the only thing that we would lose would be arbitrary coproducts. However, we don't insist that. Although it's often stated in the definition early on in a differential topology course or book, in my experience that's just because it's easier to explain than paracompactness (I tend to pick metrisable, myself). Of course, that early in a course we're probably not too concerned about arbitrary coproducts either.

Moreover, because manifolds split into a coproduct of their connected components, we tend to deal with connected manifolds unless we really can't avoid it. And even when they aren't connected, they most often have countably many components. So in practise the distinction doesn't arise.

I don't have it in front of me here, but I believe that an appendix to the first volume of Spivak's "Introduction to differential geometry" contains a proof of four equivalent conditions for locally Euclidean spaces (perhaps requiring Hausdorff). If I remember aright, the conditions are: paracompact, second countable, metrisable, and σ-compact.

There was a paper on the arxiv on Monday, 0910.0885, which lists 107 conditions for a connected locally Euclidean Hausdorff space equivalent to that it be metrisable. Amongst them are paracompactness and second countable.

share|cite|improve this answer

The Handbook of Set-Theoretic Topology is the bible for this kind of question. It has a chapter on the long-line and a chapter on questions of paracompactness of manifolds.

share|cite|improve this answer

The wikipedia article states that the long line is not paracompact. Here is a proof that the long ray is not paracompact (so neither is the long line):

Start with the open covering [0, α) for every ordinal α < ω1. Let X be some refinement of that covering. Let S be the limit ordinals below ω1. S is a stationary subset of ω1. For each β in S, pick a Yβ in X so that β is in Yβ. Consider the function f from S to ω1 which sends β to the least ordinal in Yβ. For any β in S, f(β) < β since β is a limit ordinal. So by Fodor's Lemma, There is a stationary subset of S, call it S', so that f is constant on S'. Let γ be the value of f on S'. Then γ is in Yβ for all β in S'. So the refinement is not finite (or even countable) at γ.

share|cite|improve this answer
I guess the problem with linking to so many Wikipedia pages in my question is that I didn't read them all. Thanks for the proof. – Anton Geraschenko Oct 9 '09 at 19:23

Let me give another proof that the long line is not paracompact. A theorem of Tamano states that a completely regular space $X$ is paracompact if and only if $X\times\beta X$ is normal where $\beta X$ denotes the Stone-Cech compactification of $X$.

Suppose that $L$ is the long line. To make our lives easier, let $X=L\cup\{0\}$ be the totally ordered set with least element $0$. Then $X$ is a manifold with boundary $0$. Then $\beta X=X\cup\{\infty\}$ where $\infty\not\in X$ and $X\cup\{\infty\}$ is given the one-point compactification topology.

Then $X\times\beta X=X\times(X\cup\{\infty\})$. Let $C=\{(x,x)|x\in X\}$ and let $D=\{(x,\infty)|x\in X\}$. Then by an easy application of Fodor's lemma, it can be shown that if $U,V$ are open sets with $C\subseteq U,D\subseteq V$, then $U\cap V\neq\emptyset$. Therefore $X\times\beta X$ is not normal, so $X$ is not paracompact.

Before I go, I should mention that the result and its proof stating that every paracompact connected manifold is second countable follows from the more general fact that a locally compact space is paracompact if and only if it is the disjoint union of open $\sigma$-compact sets. In particular, every locally compact connected space is paracompact if and only if it is $\sigma$-compact.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.