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On page 330 of John G. Ratcliffe's text on hyperbolic manifolds, he defines a manifold $M$ to be a locally Euclidean Hausdorff space. By locally Euclidean, he means as usual that for each $x \in M$ there is an open neighbour $U$ of $x$ which is homeomorphic to an open subset of $\mathbb{R}^n$.

But he omits the usual paracompactness assumption. I would guess that $M$ being locally Euclidean and Hausdorff is not sufficient to imply that it is paracompact, given this problem from Lee's book.

Does anyone know of such a counterexample?

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There's the long line. $\qquad$

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  • $\begingroup$ Thanks! Looks like that does the trick. $\endgroup$ – ಠ_ಠ Jun 10 '17 at 12:51
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Note that every metrizable space is paracompact and $\mathbb R^n$ with usual topology is metrizable.

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    $\begingroup$ Also, a paracompact Hausdorff locally metrizable space is metrizable. So for manifolds, paracompactness and metrizable are equivalent. $\endgroup$ – Henno Brandsma Jun 10 '17 at 14:43

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