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After reading this MO post, I am wondering:

Is every (connected) Hausdorff Banach manifold a regular space?

Though unjustified, page 53 of this paper nonchalantly states: "Note that a Hausdorff Banach manifold X is a regular space."

But does anyone know of a proof of this statement (or a counterexample)?

Of course, the real difficulty arises in proving the statement for the infinite-dimensional version, since such a Banach manifold will not be locally compact.

Follow-up: Now that Theo Buehler has kindly pointed to a counterexample (i.e. a connected Hausdorff Banach manifold which is not regular) perhaps it will give someone an idea about how to tackle the question that provided the inspiration for this one.

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2 Answers 2

up vote 12 down vote accepted

Apparently the answer is no, not every connected Hausdorff Banach manifold is regular, not even when it is modeled on a separable Hilbert space.

I quote (verbatim) from J. Margalef-Roig, E. Outerelo-Dominguez, Differential Topology, North Holland Mathematics Studies 173, 1992, page 44f.

It is well known the result of General Topology that every Hausdorff locally compact topological space satisfies the Tychonoff axiom [M-O-P, V.2, pg. 231]. By this and the Riesz's theorem every Hausdorff locally finite dimensional differentiable manifold satisfies the Tychonoff axiom. This last affirmation is not true for arbitrary Hausdorff differentiable manifolds. In [M.O.1] there is an example of a Hausdorff connected differentiable manifold $X$ of class $\infty$, such that $\partial X = \emptyset$, $X$ is not regular and $X$ admits an atlas whose charts are modelled over an infinite dimensional separable real Hilbert space.

They continue to add the regularity hypothesis in their results whenever it is needed.

The cited references are:

  • [M.O.P.] MARGALEF, J.-OUTERELO, E.-PINILLA, J.L.: Topologia, I-V. Alhambra, Madrid 1975, 79, 79, 80 and 1982.

  • [M.O.1] MARGALEF, J.-OUTERELO, E.: Una variedad diferenciable de dimension infinita, separada y no regular. Rev. Mat. Hisp.-Am, IV, V.42, 1982, 51-55. (QuickView link).

Edit: As was pointed out by Benjamin Dickman in the comments, the example also appears in English in A. Kriegl, P.W. Michor, The convenient setting of global analysis, AMS (1997), 27.6 Non-regular manifold, page 266. The book is available as a pdf file on Kriegl's homepage.

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A Google search for the title of the relevant paper gives various hits, so the paper appears to be online, but, unfortunately, none of the links seems to work right now. –  Theo Buehler Aug 27 '12 at 2:42
Thanks for these additional pointers, I incorporated them into the answer. –  Theo Buehler Aug 27 '12 at 20:41

I'm confused as to why (failure of) local compactness is an issue. I think one can just prove this directly as follows. It's rather trivial, though, so maybe I'm missing something.

Let $x\in X$ be a point and $K\subseteq X$ a closed subset with $x\notin K$. Find some coordinate chart near $x$, that is, a map $f:E\to X$ which is a homeomorphism onto some open set containing $x$, where $E$ is a Banach space and $f(0)=x$. Now $f^{-1}(K)$ is a closed subset of $E$ which does not contain $0$. Thus there exists $\epsilon>0$ such that $\|k\|\geq\epsilon$ for all $k\in f^{-1}(K)$. Let $B(0,a)$ denote the open ball in $E$ centered at $0$ with radius $a$. Then we simply observe that the following two open sets are disjoint and "separate" $x$ and $K$: $$x\in f(B(0,\epsilon/3))$$ $$K\subseteq X\setminus f(\overline{B(0,2\epsilon/3)})$$ I'm using Hausdorffness to conclude that $f(\overline{B(0,2\epsilon/3)})$ is closed in $X$.

EDIT: As the comments point out, the point here is exactly to show that $f(\overline{B(0,2\epsilon/3)})$ is closed, which in finite dimensions (i.e. locally compact) follows easily from Hausdorfness. So, as of now this argument is incomplete.

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How do you get closedness of $f(\overline{B(0,2\epsilon/3)})$ from Hausdorffness? –  Wolfgang Loehr Aug 20 '12 at 9:35
Is that where local compactness is an issue? It's a simple fact that a continuous map from a compact space to a Hausdorff space is closed, so the argument should work if you had local compactness. –  Paul Reynolds Aug 20 '12 at 13:18
@Benjamin, I wasn't disputing that fact. I was just pointing out where I think the above argument might not work. However, I am certainly no expert! –  Paul Reynolds Aug 20 '12 at 13:51
Having said that, $f$ is a homeomorphism so is closed anyway. –  Paul Reynolds Aug 20 '12 at 15:15
$f$ is a homeomorphism onto an open set in $X$, so the image is a priori relatively closed in that open set. Why does it have to be closed in $x$? –  Wolfgang Loehr Aug 20 '12 at 15:46

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