# Why do we need Hausdorff-ness in definition of topological manifold?

Suppose $M^n$ is a topological manifold, then $M^n$ locally looks like $\mathbb{R}^n$. $M^n$ is locally Hausdorff, since $\mathbb{R}^n$ is Hausdorff and Hausdorff-ness is a topological invariant.

All I want to understand is:

1) Does locally Hausdorff-ness imply Hausdorff-ness? (I can not imagine a locally Hausdorff topological space that is not globally Hausdorff)

2) Why do we need Hausdorff-ness in definition of the topological manifold? Is locally Hausdorff-ness not sufficient? If not, why?

Can anyone say anything that might be helpful?

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There is the Line with two origins which is even locally euclidean and second-countable, but only locally Hausdorff. –  Stefan Hamcke Aug 21 '13 at 18:15

There are non-Hausdorff spaces that are locally Euclidean; some people include them in the class of manifolds, and some prefer to exclude them by requiring a manifold to be Hausdorff. The line with two origins is a simple example of a non-Hausdorff manifold: it is locally Hausdorff, since it has a base of open sets homeomorphic to $(0,1)$, but the two origins cannot be separated by disjoint open sets.

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