This question stems from Jeff Rubin's earlier MO question and a follow-up that I posted.
The former recalls the following result proved by both Serge Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag):
Theorem: A connected Hausdorff Banach manifold with a Riemannian metric is a metric space.
That said, consider 27.6 (pdf pp. 262-263) in The convenient setting of global analysis (AMS, 1997), and in particular the example given at the end of it, which concludes with: "Then the same results are valid, but $X$ is now even second countable."
My question: Is this second countable $X$ a counterexample to the above theorem?
I'm hoping someone can shed some light on this matter, either by explaining why it fails as a counterexample (offhand, I'd deem this the more likely scenario) or by proving/sketching why it might actually suffice.
Edit 1: Here's a sketch of why one might even consider this example:
If indeed the proposed space $X$ described in 27.6 of the link above is second-countable, then at least one source I have found claims that $X$ would, as a result, admit a Riemannian metric. [NB: It has been pointed out that this source states its claim strictly in the context of finite-dimensional manifolds.] Furthermore, $X$ is described as a modification (where "the same results are valid") of a space that is a connected Hausdorff Banach manifold that is separable and not regular.
To summarize, we might have $X$ as a connected Hausdorff Banach manifold with a Riemannian metric, which is separable and not regular (hence non-metrizable by Urysohn's Theorem), in which case, $X$ would be a counterexample to the above-stated theorem.
Sub-question 1: can anyone find other sources (preferably with proof) that a second-countable connected Hausdorff manifold necessarily admits a Riemannian metric? Alternatively, can anyone find a counterexample to this? [NB: Particularly in the context of infinite dimensional manifolds.]
Sub-question 2: can anyone prove (or sketch a proof of) the connectedness of $X$? Alternatively, can anyone show that $X$ is not connected? [NB: This has been answered: $X$ is connected.]
I'd appreciate even a partial answer to my original question or either of my sub-questions. Also, if you should know (of) anyone who is doing work in this area of mathematics, perhaps you could direct them to my query.
Thanks!
Edit 2: My second sub-question has been answered in the affirmative by Wolfgang Loehr: $X$ is indeed a connected space.
I see numerous mentions of the result mentioned in my first sub-question (that second-countability alone implies a connected Hausdorff manifold admits a Riemannian metric) but I'm wondering whether this is in fact only a theorem for finite dimensional manifolds.
Nonetheless, my initial question still stands: is the space $X$ described in the AMS book on Global Analysis a counterexample to the theorem stated above?
Edit 3: As time winds down on the question's bounty, I wonder whether anyone has helpful thoughts with regard to non-regular manifolds that admit Riemannian metrics. More precisely, how could one prove that $X$ does or does not admit a Riemannian metric?
Post-bounty Edit: I awarded the bounty since my sub-question 2 was answered entirely. There is still no conclusion as to whether or not the space referenced above is a counterexample to the aforementioned theorem, but it is increasingly clear that there is a fair bit of confusion surrounding when theorems about Banach manifolds do or do not extend from the finite dimensional case to the infinite dimensional one.
