# Riemannian metrics on non-paracompact manifolds

After proving the existence of Riemannian metrics on manifolds, one of the students asked if the "paracompactness" is necessary. Of course the standard proof with the partition of unity uses this assumption, and for the only non-paracompact manifold I know, namely the long line, a similar argument seems to show the existence of a metric. So, just out of curiosity, does there exist a "manifold" (of course, non-paracomapact) which does not admit any Riemannian metric?

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@Greg Kuperberg: Thank you for clarification. I had carelessly assumed that it is possible to cover the long line with an uncountable number of open intervals in a locally finite way, similar to ${\mathbb R}$. But this cannot be done all the way to the end. –  Keivan Karai Feb 8 '11 at 14:18