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A problem in Lee's Introduction to Topological Manifolds got me thinking about this question. I can easily construct a locally euclidean space that is not second countable, by taking a disjoint union of $\mathbb R^n$ over an uncountable index set, but this is not connected of course. Similarly one can easily find connected first countable spaces that are not second countable, I presume an infinite dimensional Hilbert Space would suffice (no?).

The difficulty that I find in constructing such a space, is a lack of "tools". Essentially I can either take a disjoint union or a connected sum. It's appealing then to attempt to take an uncountable connected sum of n-manifolds, but it's not apparent to me if this is well-defined. For instance we can't enumerate an uncountable index, so we can't just attach one after another. Then it seems we would have to attach all of them at the same time, but because an $n$-manifold is second countable there can't be an enough open balls to do this.

Is there a clever way to construct such a space with connected sums? The only thing I can think of is to assume the negation of the axiom of choice and take the limit of an infinite chain of connected sums of $n$-manifolds, then this doesn't have to be countable.

If it isn't true how do you prove it? I can't see any clear reason why it would be false.

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Just wondering, why can't you enumerate uncountable cardinals? You are assuming the axiom of choice, aren't you? :-) – Asaf Karagila Mar 8 '11 at 22:53
I've never taken a formal course on set theory past the basics and wasn't aware that uncountable cardinals could be enumerated assuming the axiom of choice. – JSchlather Mar 8 '11 at 23:11
up vote 7 down vote accepted

The long line.

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Okay as a final question can you define a covering projection from the open long line to the real line? – JSchlather Mar 8 '11 at 23:15
@Jacob: no, the real line is simply-connected, hence has no nontrivial connected covers. – Chris Eagle Mar 9 '11 at 2:20

Problem 3-8 in Lee constructs the line with two origins (maybe this is the problem that got you thinking about this question).

Well, what about a line with uncountably many origins?

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It's actually a later question in chapter 10, in one of the questions he asks to prove that if X is an n-manifold then it's cover space is an n-manifold. As far as I can tell though there's no reason why you can't make a cover space of the circle out of the long line. Thank you though, that's a much more simpler construction than the long line. – JSchlather Mar 9 '11 at 1:56
@Jacob: You can't put a Riemannian metric on the long line, so it is not a covering space if the circle. – George Lowther Mar 9 '11 at 2:24

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