@AlessandroCodenotti Thoughts on earlier comments? right idea / wrong idea?

@Ted Okay, I've just looked back at the last thing you said. Are you suggesting that I project the (top sheet of the) hyperboloid $z^2 = x^2 + y^2 + a$ onto the plane $z = \log a$?

I don't think I was suggesting that.

Hmm. I may have misread your intent then.

Don't worry about log and exp (that's just to turn $\Bbb R^+$ into $\Bbb R$).

I was suggesting filling up the inside of the cone with hyperboloids rather than with cones.

I see. I think the way I did it, I ended up filling the cone with sections of spheres.

But I can see how the hyperboloid angle (heh) would be easier.

I mean, one wants to use level sets that fill up (foliate) the set.

Cones are obvious, but not smooth.

That's why I made up the problem years ago.

It's a fun one. I like that it confirms some basic intuitions one might have about open sets in $\Bbb R^k$ being diffeomorphic to $\Bbb R^k$ under certain conditions.

In general, any convex open set is diffeo to all of Euclidean space, but I actually don't know a proof. We've discussed that here before and someone found a slick reference.

Huh. That seems obvious, but I have no idea how you'd prove it.

If it's bounded, you could try to use distance to the boundary, but the boundary might be very non-smooth.

In fact, any star-shaped region works. Here is a surprisingly nontrivial reference.

Huh, the Riemann mapping theorem sounds a bit like my first thought.

Well, that's very $\Bbb C$-specific. And far more general.

Is there a case where a simply connected open set in $\mathbb{R}^k$ is not diffeo to $\mathbb{R}^k$?

The Whitney Extension Theorem is an amazing analysis result ... I learned that first year of grad school and was dumbfounded.

Take a tubular neighborhood of the 2-sphere? I think that'll work

It needn't even be homeomorphic. Alexander horned sphere stuff.

@Daminark Oh yeah. Good point.

Yeah, Demonark wins.

@Fargle you should start with contractible before anything else

then you should ask homeomorphic before diffeomorphic. in 4D there are many exotic $\Bbb R^4$s (homeo to standard, but not diffeo) that live as open subsets of $\Bbb R^4$

(in other dimensions, homeo => diffeo)

@MikeMiller Ah yeah, I've seen that notion before but couldn't put two and two together there.

Math Enthusiast Gets BTFO by Immediate Counterexamples

For homeomorphic you should learn about the Whitehead manifold, a contractible 3-manifold which is not homeomorphic to $\Bbb R^3$, and you should learn about the fundamental group at infinity, the invariant that distinguishes them

Then you should learn about the (Siebenmann's?) theorem that in dims $n \geq 5$, an open $n$-manifold which is contractible and simply connected at infinity is in fact homeomorphic to $\Bbb R^n$

Stalling's maybe

Poor Stallings is like poor Stokes. He gets misaprostrophed.

Ask a silly question, get a surprisingly in-depth and thorough answer.

Thanks @Mike, @Ted, @Dami

Topology is crazy subtle with non-standard structures, both topological and smooth.

That's why I stuck to complex varieties :) Of course, lots of holomorphic structures there, but that's a more standard thing. :)

I mean, there's a reason one of the most popular topology books is a compendium of counterexamples, right?

Um, NO.

D:

Most mathematicians don't hold up weird point-set topology as the banner of mathematics :P

Some do.

But I actually loved teaching point-set.

@Fargle: Did you see my $x^4-10x^2+1$ challenge earlier?

I did not.

Show that that polynomial (which you may have encountered) is irreducible/$\Bbb Q$ but reducible mod every $p$. That should fit your number theory course nicely.

Well, it would if we were past the Euclidean algorithm. :|

I'll give a crack at it though.

Well, the rational roots theorem basically immediately says that it's irreducible over $\Bbb Q$, so that's not too bad.

OOh, no, that doesn't suffice.

Lol I only actually had to deal with point-set considerations that weren't obvious once. Functional analysis problem, show that if a Banach space has separable dual or is reflexive, then there's a sequence on the unit ball which converges weakly to 0

@TedShifrin It doesn't?

@Fargle: Not if $\deg\ge 4$.

Oh right right right duh.

Err, unit sphere

Irreducible $\neq$ no linear factors

Duh duh duh

That was one of my huge deals teaching the first semester of algebra, @Fargle. Don't mess this up if you expect to pass :P

I am legally allowed 3 brain farts a day, thank you. :P

Let me see that law.

It's, uh, common law precedent. Since time immemorial.

I'll have to google "Fargle + legal brain farts"

Well, it's for everyone. They didn't make it for me.

Anyway, I've hit my quota for the day in the past hour, so I'll need to tighten up

The interesting part of the question is the other part, regardless. :)

I'm sure.

Okay, I got irreducibility in $\Bbb Q$. Now for the hard part.

I will let you play w/o hints for a while. But there is a good hint I'll pass along at some point.