# Meaning of conditions on definition of topological manifold

This is my first time to study on the manifold.

I've studied that the topological manifold is:

• Hausdorff,
• locally homeomorphic to Euclidean space, and
• second countable

topological space.

With these 3 conditions, manifold is one of generalization of Euclidean space since we don't need a full Euclidean, or full flat space, but only local features can also give a bunch of good properties as well as in Euclidean space. Thus, I believe we can deal with differentiation and integration on manifold.

But my question is more fundamental. Why those three conditions are required to define a manifold well?

I think only second condition (locally Euclidean) is needed to satisfy our demands of generalization.

Why are Hausdorff and second countable included?

I admit I can answer this question when if I study futher about manifold. Probably some good properties of manifold will be come out from those two conditions.

But before studying deeply, I want to understand intuitively. I am very unconfortable and cannot accept naturally about this definition.

Someone please give a good explanation. Thanks.

• The point of second countability is to get rid of examples like the long line. If your space isn't second countable, you don't have partitions of unity, and you absolutely need those if you're going to do any smooth manifold theory on your topological manifolds; probably there are constructions you want to do on a topological manifold where you want partitions of unity too. The point of Hausdorffness is to get rid of examples like the line with two origins. I have no idea how far the theory can be taken with these. If you don't assume this you can't always eg embed your manifold in $\Bbb R^n$. – user98602 May 9 '15 at 1:46
• Also another reason for second countability is to get rid of uncountable discrete spaces. Any point set which is uncountable and given the discrete topology is automatically Hausdorff (any space with the discrete topology is) and locally Euclidean (because each point is a neighborhood of itself homeomorphic to $\mathbb{R}^0$). – Chill2Macht Mar 7 '17 at 15:01