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I am working on some infinite dimensional differential geometry. I have tried proving a somewhat weaker statement than the above by replacing locally Banach with locally metrizable. But after some research this is obviously wrong as the K-topology is a contradiction to such a statement, see A locally metrizable, Lindelöf Hausdorff space that is not metrizable.

My goal is to use this to prove that second countable locally Banach, Hausdorff spaces are metrizable, which is equivalent to locally metrizable, Hausdorff and paracompactness. That is to prove that we need not worry about convention used to define Banach manifold. Does anyone have a reference for an article or such that treats these questions? I have looked into some books on Banach manifolds but without any luck.

If it is not the case that every locally Banach, Hausdorff space is regular, then does anyone have a counterexample at hand? (If this is the case the article I am trying to use would be incorrect on this point).

Martin Sleziak
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ZuperPosition
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1 Answers1

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A standard metrisation theorem (by Smirnov (1951)) states that a locally metrisable paracompact Hausdorff space is metrisable.

But if we weaken paracompactness or omit it altogether, while strengthening locally metrisable to locally Banach, I'm not so sure what happens: in the finite-dimensional (locally Euclidean) case, things are easy: we are locally compact and so Hausdorffness implies Tychonoffness (stronger than regular). But the infinite dimensional case will be nowhere locally compact. I don't see a handy theorem to guarantee regularity. (If $X$ is even second countable, the regularity gives paracompactness and then Smirnov applies)

There could well be a locally infinite dimensional Banach space that is Hausdorff and not regular. I don't see why not.

Martin Sleziak
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Henno Brandsma
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  • I agree with all your statements, thank you for your input! I think it was proven by Eells and Elworthy in 1970 that infinite dimensional Banach manifolds are subspaces of the modelling space itself, thus such Banach manifolds are obviously regular. I don't want to use this, as I want to look into a somewhat weaker notion of differentiability. Therefore, I want to make the argument purely topological if possible. – ZuperPosition Oct 17 '18 at 12:39
  • @ZuperPosition How did Eels and co prove this? With some extra structure theory only available on infinite-dimensional Banach spaces? Did they use some stricter definition? I'm no expert on Banach spaces.. – Henno Brandsma Oct 17 '18 at 22:48