Why do we generally require that a differentiable manifold be Hausdorff and second countable? Is this universally accepted in the definition? My Professor only required the Hausdorff condition for example, but most book I have read require both. Thanks

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    $\begingroup$ I don't have a good answer, but have you heard about the long line? It is the canonical example of a “manifold” which is not second countable. If you look at the wikipedia article you will see some odd behaviour when you try to put a differentiable structure on it. $\endgroup$ – Harald Hanche-Olsen Mar 25 '13 at 21:40
  • $\begingroup$ Very interesting, thank you for the response. $\endgroup$ – mike Mar 25 '13 at 21:44
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    $\begingroup$ No, it’s not universal: some people study non-metrizable manifolds. This question and its answers, and the accepted answer to this question should go a long way towards answering your question. $\endgroup$ – Brian M. Scott Mar 25 '13 at 21:45
  • $\begingroup$ It is fairly common to require paracompactness instead of second-countability (it can be shown that a manifold is paracompact iff each connected component is second-countable). $\endgroup$ – Eric Wofsey Jan 14 '16 at 7:09

Those conditions provide the metrizability of the manifold. That is we want a metric space as a result. Actually we want to produce a Riemannian manifold. Some texts require paracompactness, which provides partitions of unity and helps to define a Riemannian metric. Look at Urysohn metrizatiın theorem and other metrization theorems, you will see that these conditions are in their hypothesis.


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