# Question about definition of topological manifold

The following definition of topological manifold is given in Lee's Introduction to topological manifolds (2000) on page 33:

A topological manifold is a second countable Hausdorff space that is locally Euclidean.

I omitted the dimension since it's not relevant to my question so it's not a faithful quote.

My question is: Doesn't Hausdorffness follow from being locally Euclidean? Pick $x,y$ in $M$. Then for each point there exists an open set homeomorphic to an open subset of $\mathbb R^n$. Let's call them $U_x, U_y \subset M$. Then either they are already disjoint or if they intersect, by Euclidenaity, we can pick smaller open sets inside them so that the smaller sets don't intersect. So we have a Hausdorff space.

Question 1: What am I missing? Why does this not work?

Question 2: I guess we have to require second countability because otherwise the disjoint union $\bigsqcup_{r \in \mathbb R \setminus \mathbb Q} (0,1)$ would be a $1$-manifold (that doesn't admit a countable basis) . Is that a correct counter example?

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On the first question: DrKW has already given you a counterexample. The error in your proof is that if $U_x, U_y$ intersect, it's not guaranteed that the smaller open sets you want to take will actually contain your original points $x$ and $y$.

On the second question: An uncountable disjoint union of intervals is indeed locally Euclidean but not second countable, if you take the disjoint union topology on them. I'm not sure exactly what you mean by your notation, but I'm worried that you're trying to take some subspace of $\Bbb{R}^2$, which would be guaranteed second countable as second-countability is a hereditary property. (An uncountable union of disjoint intervals in $\Bbb{R}^2%$ would also fail to be locally Euclidean — even if the intervals are disjoint, neighborhoods of points on one interval will be forced to include points on other intervals.)

You can also find connected spaces which are locally Euclidean but not second-countable; the standard example is the long line. In some ways this is a more important counterexample than the disjoint union one, as its non-second-countability also leads it to be nasty in other ways (e.g., it doesn't admit a partition of unity).

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Thank you! I was thinking of an uncountable number of $(0,1)$ lying around in some space and I wasn't sure how to write that. I was not thinking of them as embedded into $\mathbb R$ or $\mathbb R^2$. Now that you mentioned it I am not sure how line segments in $\mathbb R^2$ can fail to be locally Euclidean! I think of $(0,1)$ as line segments scattered across the plane, in this case. But I don't see that a nbhood of a point will be forced to contain points on other intervals. (Perhaps you are assuming I attach each line to an irrational?) – Rudy the Reindeer Aug 29 '12 at 6:00
@Matt: The fact that you were labeling your index set by $\Bbb{R} \backslash Q$ (which I thought might be a typo for $\Bbb{R} \backslash \Bbb{Q}$) meant I was wondering if you were attaching lines to irrationals, yes. But it actually doesn't matter; if you're taking a subspace of $\Bbb{R}^n$ consisting of uncountably many intervals, they can't possibly form a topological disjoint union (meaning the subspace won't be locally Euclidean). This is exactly because any such subspace is forced to be 2nd countable, while the disjoint union wouldn't be; ... – Micah Aug 29 '12 at 6:11
@Matt: ... i.e., $\Bbb{R}^n$ "isn't big enough" to hold uncountably many intervals without their open covers bumping into each other. Fortunately for the purposes of your counterexample, you can build topological disjoint unions abstractly, so this is a side issue. – Micah Aug 29 '12 at 6:12
Yes, this $Q$ is indeed a typo. But I merely wanted an uncountable index set. I guess I could have written "let $I$ be any uncountable set" and then used $I$ instead. But I did indeed mean to build an abstract disjoint union. OTOH, I'm glad that you pointed out that it cannot work in $\mathbb R^2$ since I hadn't thought about that. Thank you! – Rudy the Reindeer Aug 29 '12 at 6:17
@Matt: Exactly! The reason to require manifolds to be second countable is hinted at in my last sentence -- it just makes our lives so much easier. Intuitively, we want to be able to construct global objects on our manifold by piecing them together from some kind of local counterpart; partitions of unity are an extraordinarily useful tool for doing that, and you need some condition like second-countability to guarantee their existence. – Micah Oct 20 '12 at 1:37

To reply to your first question. Consider the real line with two origins. That is, take disjoint copies of the real line and identify $x$ in the first copy with $x$ in the second copy except for the point $x=0$. This space is locally 1-Euclidean and second countable, but not Hausdorff. Any two open sets containing either origin will have points in common.