Question: a Hausdorff differentiable manifold (locally Euclidean space): $$ \text{is metrizable} \iff \text{is paracompact} \iff \text{admits a Riemannian metric} \,?$$

Does one also have for a locally Euclidean Hausdorff space (not necessarily differentiable): $$\text{second countable} \iff \text{metrizable with countably many connected components}? $$

Thus second countability is strictly stronger for such spaces than metrizability/paracompactness/existence of a Riemannian metric?

(According to this question, it seems they are all equivalent when connectedness is assumed, and according to this question even equivalent to the existence of a universal cover when connected.)

Attempt: By the Smirnov metrization theorem, a locally metrizable space (e.g. locally Euclidean spaces) are metrizable if and only if they are Hausdorff and paracompact.

According to Wikipedia, Riemannian manifolds are metrizable.

(Is this only true for connected Riemannian manifolds, or can we use the trick where we make the metric less than 1 on each connected component and 1 for distances between points in different components? Or perhaps only when there are at most countably many components?)

Finally, according to Wikipedia, any differentiable paracompact manifold admits a Riemannian metric (I'm not sure if the differentiable hypothesis is necessary -- this question seems related).

Thus (at least for connected, differentiable Hausdorff manifolds) "admits Riemannian metric" $\implies$ "metrizable" $\implies$ paracompact $\implies$ "admits Riemannian metric".

Context: This question is motivated by how the definition of manifold used by Spivak in Comprehensive Introduction to Differential Geometry (see here for a related question) is different from the one used by Lee in Introduction to Smooth Manifolds (where second countability is required). In particular, I had thought that the definition my professor was using (Hausdorff, metrizable, locally Euclidean) was equivalent to Lee's, until Spivak started mentioning a bunch of counterexamples which I recognized as not being second countable. Although I don't remember if my professor specified countably many components, which would rule out most of Spivak's counterexamples consisting of uncountable spaces with the discrete topology and metric.

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    $\begingroup$ Seems correct to me, if it helps you (and you don't know it already) Uryshon's Metrization theorem, insures that any manifold is metrizable (where by manifold I mean a second-countable, Hausdorff with a smooth atlas). $\endgroup$
    – user321268
    Commented Jan 27, 2017 at 21:42
  • $\begingroup$ @mayer_vietoris I don't think I knew the application of Urysohn's metrization theorem here, so this definitely helps me -- I appreciate it. $\endgroup$ Commented Jan 28, 2017 at 10:57

1 Answer 1


It seems like the answer to many of these questions can be found on p.459, Appendix A, of Spivak's Comprehensive Introduction to Differential Geometry, 3rd edition 1999.

Specifically, the Theorem appaears to address my confusion regarding how the cardinality of the number of components of a manifold plays into these properties/definitions.

Note that this seems to hold even for topological manifolds, i.e. not necessarily smooth ones.

Theorem The following properties are equivalent for any manifold $M$:

(a) Each component of $M$ is $\sigma$-compact.
(b) Each component of $M$ is second countable (has a countable base for the topology).
(c) $M$ is metrizable.
(d) $M$ is paracompact.

(In particular, a compact manifold is metrizable.)


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