187 reputation
9
bio website cmcc.ufabc.edu.br/…
location Cotia, Brazil
age 35
visits member for 2 years
seen Dec 16 at 2:02

I'm a professor at the Center of Mathematics, Computation and Cognition of the Federal University of the ABC, Santo André, Brazil.


Dec
15
awarded  Caucus
Nov
16
answered Gathering books on Lorentzian Geometry
Feb
22
suggested rejected edit on Are $L_p$ spaces of functions with separable support separable?
Jun
6
comment Open neighborhoods of a $G_\delta$ set
@Cameron: nice indeed, but see my last comment on Asaf's answer...
Jun
6
comment Open neighborhoods of a $G_\delta$ set
It seems that disconnectedness of some of the sets involved plays a common role in all these counter-examples... My last move: what if $X$ is complete and connected, $A$ is connected and $A_n$ is connected for all $n\in\mathbb{N}$?
Jun
6
awarded  Commentator
Jun
6
comment Open neighborhoods of a $G_\delta$ set
Never mind, Asaf provided a suitable counter-example...
Jun
6
comment Open neighborhoods of a $G_\delta$ set
I've deleted a pair of messed-up comments before I could see your reply to one of them... You're right, $X$ here is totally disconnected.
Jun
6
awarded  Scholar
Jun
6
awarded  Supporter
Jun
6
accepted Open neighborhoods of a $G_\delta$ set
Jun
6
comment Open neighborhoods of a $G_\delta$ set
What if $X$ is assumed complete and connected as in your counter-example, but also that $A\neq\varnothing$?
Jun
6
comment Open neighborhoods of a $G_\delta$ set
Yeah, I got it... Can the possibility that $A$ is open and a proper subset of $A_n$ for each $n\in\mathbb{N}$ be evaded by assuming completeness and (say) connectedness of $X$? The latter property seems to fail in both counter-examples...
Jun
6
comment Open neighborhoods of a $G_\delta$ set
What if one assumes in addition that $X$ is complete (which is clearly false in your counter-example)?
Jun
6
awarded  Student
Jun
6
asked Open neighborhoods of a $G_\delta$ set
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
That's OK. I've edited the answer to account for the above discussion - now it goes in the direction of suggesting a possible counterexample, if I understand correctly the situation now. Thanks to both you and Asaf Karaglia.
Dec
10
revised Does the open mapping theorem imply the Baire category theorem?
added 194 characters in body
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
So, could it be something like "OMT holds for all strict, countable inductive limits of Baire spaces", which includes non-Baire spaces such as the ones I've discussed (this fact uses only ZF and hence avoids circularity)? If so, BCT fails with the axiom of choice (AC) for sure. I don't know what happens if AC is abandoned. The proof of the closed graph theorem in the above class of spaces is a bit involved (Theorem 5.4.1, pp. 92-93 of Jarchow's book) - I'm not sure enough of its details to be able to say that AC (or even DC, to avoid circularity) is not needed besides Baire's "black box".
Dec
10
comment Does the open mapping theorem imply the Baire category theorem?
OK, I accept your apologies and I've got your point. A negative answer to your question could be in a form of a counterexample only if it uses nothing else but ZF. More precisely, one needs a pair of topological vector spaces for which the open mapping theorem holds (and this should be proven using only ZF - here's the catch) and Baire's theorem doesn't hold (also to be demonstrated with only ZF). Indeed, in the situation I described above, I need the fact that Baire's theorem holds for the elements in the sequences defining the inductive limits in order to prove the open mapping theorem.