Mathematics - 2017-03-21 (page 4 of 6)

6:00 PM

Particularly young women.

@ShaVuklia think of that sum as starting from k=1 then take away upto k=m-1 what you should have left over is your original sum

Ted Shifrin

They had done just fine in our Spivak course (not taught by me — I might have been a bit tougher).

Ali Caglayan

which should be easier to work with

Ted Shifrin

@Ali: I'm trying to suggest an alternative technique, though.

Liad

@TedShifrin give me the iso. :P

Ted Shifrin

6:01 PM

If I know how to do $1+1/2+1/4+1/8+\dots$ and instead I see $1/16+1/32+1/64+\dots$, what can I do?

Ali Caglayan

@Sha ignore me in that case, Ted's technique is probably better :P

Eric

Ah, that's so unfortunate.. I swear when I am a professor I will work hard to not be so exclusive, that stuff just bums me out.

Daminark

Is Marianna confusing in the Soug way?

Ted Shifrin

@Liad: In your head, you keep trying to fit everything into $[0,1)$, which is impossible (at least for me).

Sha Vuklia

@Ali I'll consider your technique too later!

@Ted So we have $\sum_{k=m}^\infty\frac{1}{2^{k-1}}=\frac{1}{2^{m-1}}+\frac{1}{2^m}+\frac{1}{2^{‌m+1}}+\dots$

Ted Shifrin

6:02 PM

@Eric: Part of good teaching if you teach hard courses is caring, being welcoming, and being a cheerleader.

OK, @Sha, so think factoring.

Liad

@TedShifrin $(n,x) \to nx $ does not work

Ted Shifrin

@Liad: Do you really not see the picture I keep suggesting? Draw it on your paper.

Ali Caglayan

Yeah @Ted s technique is miles better than mine

Ted Shifrin

You said it right. You want $\{k\}\times [0,1)$ to end up at $[k-1,k)$ (it would be easier to start $\Bbb N$ with $0$, the way the French do, admittedly.

Eric

lol @Daminark Marianna is just like... insanely smart. And I think she has a really hard time judging what would be difficult or easy for us considering that at our age she was solving open problems and stuff.

Daminark

6:03 PM

And yeah our class size had dropped hard at the beginning. Soug did give a lot of pep talks which were appreciated, but the whole 60 problems/week thing was way too much for many people to bare

Eric

@Ted I will definitely keep this in mind.

Ted Shifrin

Demonark, I hope you weren't bare.

Eric

Wait @Daminark idk if I ever asked you this but did Souganidis ever give you tell you that doing math is like playing basketball

Sha Vuklia

wait

sorry

Ted Shifrin

LOL

Sha Vuklia

6:04 PM

:P time pressure

Daminark

Yeah, he definitely said that

Ted Shifrin

Do the concrete case I wrote down above, @Sha. It's good to start with concrete things first and then generalize.

@Eric: We all have our pet phrases. I personally liked "proof by intimidation" and a few others that some people keep noticing in the videos.

Liad

@TedShifrin $(k,x) \to k + x$

Ted Shifrin

Bingo, @Liad, except we probably need $k-1$ on the right.

Eric

lmao I've heard him use this analogy so many times.

@Ted that's a classic one

Ali Caglayan

6:06 PM

Nov 1 '16 at 0:30, by Ted Shifrin

@Ali: I never regretted my lectures ... I regretted Hippa.

Liad

Why? i will start $N$ with $0 $:P

Eric

proof by intimidation is definitely a good descriptor sometimes lol

Ted Shifrin

LOL, @Ali. He was obnoxious for a bit, but we're still friends. :)

Oh, fine, then, @Liad. Then you have it.

Liad

ok, so we have $N \times [0,1)$ iso. to $[0,\infty)$ and we can say $X_{\alpha} \times [0,1)$ iso. to $[0,1)$ correct?

Ted Shifrin

Once you show why $[0,\infty) \cong [0,1)$, @Liad, yes.

Daminark

6:08 PM

One time in office hours he was like "Look, math is like basketball, not a spectator sport, if you want to get good you need to spend a lot of time. Some people prefer giving 10 tricky problems but I don't think that's how you learn, it's best that you just see every type of problem"

Ted Shifrin

Lots of practice, working your way from routine to the challenging, yes. I still think I probably made my students do a lot more concrete computations, though, Demonark.

Liad

@TedShifrin could you take a look at this http://math.stackexchange.com/questions/2187586/prove-that-for-every-countable-ordinal-alpha-alpha-times0-1-is-order-is
i think he is right that there is a need for induction, do you?

Sha Vuklia

I see it now @Ted It's a matter of factoring out $\frac{1}{2^{m-1}}$, and then realising that the geometric series for the simple case we worked towards is equal to 2 (which is $\frac{1}{2^{-1}}$)

Daminark

I might be mixing different convos but yeah

Ted Shifrin

There you go, @ShaVuklia.

Of course, you could always do a proof by induction, by why work that hard? :)

Ali Caglayan

6:10 PM

@TedShifrin

Eric

lol @Daminark im pretty sure ive heard what you just said word for word like 4 or 5 times

Ted Shifrin

@Ali: Where'd you get that!?

arctic tern

what in the world

Ali Caglayan

In the chat archives

Sha Vuklia

Haha true! Thanks @Ted :)

Ali Caglayan

6:11 PM

Hippa from 2014

Daminark

Yeah that was in defense of the pset size, which was somewhat unprompted but yeah

Ted Shifrin

I never saw that one, @Ali. I just saw his somewhat hurtful memes. :P

Ali Caglayan

@TedShifrin here is 'that' conversion chat.stackexchange.com/transcript/message/16152286#16152286

The image seems to have been deleted

Daminark

I mean it's pretty fair, I've def learned a lot from that class

Ali Caglayan

But I faintly recall keeping a copy of it on an old hard drive, I will try to fish it out this weekend

Daminark

6:13 PM

But it is kinda sad that we tanked from 40 to 25 so quickly, I kinda wish people were able to stick around.

Eric

I mean I get why he assigns so many problems, but when I was in my first quarter of college, seeing this gigantic analysis pset was definitely very intimidating @daminark, you at least had a year to be mentally prepared for that kind of thing.

Ted Shifrin

Demonark: In my younger years, when my multivariable course was first starting, I worked hard to "keep" people in the class who I knew would have done fine but who just didn't really want to be there. In the latter years, I let them go without a fight.

@Liad: I don't have the patience to look at that whole thing. Sorry.

Daminark

Yeah, we talked to you guys and we kinda were expecting things for the most part. First year group got hit hardest with Soug, and then 5 second years dropped with Schlag which balanced it out somewhat

At least Soug eased us in somewhat

Ali Caglayan

Wow 2014 was such a simpler time

Ted Shifrin

I suspect my courses at UGA were perceived much the same way, but I contend that my courses were much more supportive and sensible. Just my biased viewpoint.

Ali Caglayan

6:16 PM

1914 must have been trivial

Ted Shifrin

@Ali: Don't worry. In a few months, Trump et al. will have the world back to 1914.

Eric

I guess my crop of first years was also insanely prepared. @Daminark, I don't think many of us dropped.

Daminark

We lost basically half of the first years by the midterm. Most of our group had multi, and I think maybe one had looked at Rudin and thus knew topology

Ted Shifrin

@Ali: In my own defense, what Hippa didn't acknowledge in his mean memes was that I was teasing my students and it was all quite friendly. They weren't hurt by it.

Ali Caglayan

@TedShifrin I know it was a lovely conversation

Ted Shifrin

6:19 PM

Anyhow, Hippa and I are past that, and I hope I get to meet him.

Daminark

Second years started in better shape because we all kinda knew the basics of topology from the 160s, and most of us did the REU and thus knew linalg

Ali Caglayan

Talking of blasts from the past

Does anybody remember Chris'sis?

Ted Shifrin

Anyhow, Demonark, I would rather we encouraged more people to work hard and excel at math, rather than discouraging so many.

Ali Caglayan

I know you were a #1 fan @Ted

Ted Shifrin

Although in today's America, we know learning and science are bad.

Eric

6:20 PM

^^^

applauds

Ted Shifrin

Ha, yes. @Ali. Her tone was very smug and defensive. It got old.

Daminark

In the second half, though, it was multi, which meant that the first years who were at the first barely holding on because Soug was going fast through unfamiliar material were in better shape, and it all balanced out

I mean it wasn't much multi

Ted Shifrin

@Eric: I'll do my best to make snide remarks to you as you learn differential geometry :D

Daminark

We only did that for a total of 3-4 weeks

Ted Shifrin

Demonark: I still contend that you know $\varepsilon$ of the multi I think you should.

Daminark

6:21 PM

But it at least helped a bit

Eric

LOL got it @Ted

Ted Shifrin

Well, maybe $\sqrt\varepsilon$ :)

Eric

Chicago's math honestly doesn't make much sense

Ali Caglayan

Last time I spoke with him, he was supposedly collaborating with someone in the US, I wonder how it worked out. I haven't heard from him in quite a while.

Daminark

Haha, yeah we still didn't do any integration

Eric

6:22 PM

we don't teach anywhere near enough of the fundamentals before people jump into using fancy words

Ted Shifrin

@Ali, "he"? I was pretty sure Chris'ssis was a woman.

Eric

we didn't even have a linear algebra class in the department until this year

Ali Caglayan

@TedShifrin I am fairly sure, I remember seeing his name.

Ted Shifrin

Demonark: I had physics majors in my course. I wanted them more competent, not less, when they got to the E&M course in physics.

Ali Caglayan

Although I won't dig further in the interests of privacy

Ted Shifrin

6:24 PM

Well, interesting. I was partly swayed by the name, partly by his/her fanaticism about Simona Halep (tennis), but admittedly that's a weak link.

Daminark

I'd be using your book @Ted but it's been checked out from the library, so I've gotten ahold of Fleming

Ted Shifrin

Demonark: If you email me, maybe we can do something. Ssshhhh.

Danu

Hi @Ted

Daminark

Oh yeah before linalg was like, a bit in 163/159, a bit in 204/7, and mostly in 255/8

Ted Shifrin

Hi @Danu

Danu

6:25 PM

I got somewhere, sort of, I think, with the question Pontryagin class question

Ted Shifrin

I haven't thought further, @Danu. I'm still swamped with Europe plans/purchases.

Danu

I was looking for $p(T\pi)$ where $T\pi$ are the tangent vectors to the (2-dimensional, $\Bbb CP^1$) fiber

So this is rank 2, hence $p(T\pi)=e^2(T\pi)$.

Using naturality of the Euler class, we see that $\langle \iota^*e(T\pi),[\Bbb CP^1]\rangle=2$

Eric

@Daminark yeah this makes no sense to me lol, students should probably know linear algebra pretty solidly before they get taught about modules imo.

Danu

So now I just have to find out what $\iota_*[\Bbb CP^1]$ is

Daminark

Also I heard next year they're fixing functional analysis which will be nice for sure

Ali Caglayan

6:27 PM

We had a talk today about how to glue things together and produce G2 manifolds

Ted Shifrin

So you need the intersection structure on homology, or product structure on cohomology, of the bundle, @Danu.

Wow, cool, @Ali.

Eric

the version in the analysis sequence or the standalone class or the grad class @Daminark

?

Danu

@AliCaglayan Oh, tell me about that. How does one make $G_2$ manifolds

Daminark

Like, non-honors analysis has 2 sections, one of which does Lebesgue stuff, one which does not

Danu

I generally want to understand special holonomy better

Daminark

6:28 PM

Standalone undergrad class

Ted Shifrin

BBIAB ... going to check my snail mail.

Danu

or rather, understand anything about it at all

Eric

wait... what??

I did not know that

Danu

see you soon, Ted

Daminark

Math 272, functional analysis

Eric

6:29 PM

Glad I just took grad functional then...

Daminark

It's been having a problem because not everyone knows about Lebesgue theory before going in to the class

Kenig taught this quarter and basically made it into measure theory

Ali Caglayan

So there is some structure called a building block which is a pair $(Z, \Sigma)$ such that $f:Z\to \Bbb P^1$ is some map and $\Sigma=f^{-1}(\infty)$ + some additional topological conditions

$\Sigma$ is a smooth K3 surface

SoumyoB

"I once shot an elephant in my pajamas. . . . How he got into my pajamas, I'll never know."

Daminark

Since half the class didn't know it

Ali Caglayan

Then some twisted connected sum of these things makes a G2 manifold

SoumyoB

6:31 PM

@AkivaWeinberger I don't know what drugs you're on, but whatever it is, I want it

Ali Caglayan

I'll be honest, after that talk, I don't think I learnt much

Danu

@AliCaglayan What's $Z$?

Ali Caglayan

but I wrote down some of the important papers

Ted Shifrin

@SoumyoB: It's Groucho Marx.

Ali Caglayan

@Danu some nice space apparently

Liad

6:32 PM

@TedShifrin do we have lex. order on $N \times [0,1)$ when we show the iso. ?

Ali Caglayan

Let me find the literature

Ted Shifrin

Yes, @Liad.

Eric

@Daminark I thought we taught it in the regular analysis sequence

Daminark

I think they're now adding a class on Lebesgue stuff to bridge the gap

SoumyoB

wow I didn't know that actor had a drug patented on his name!

oh wait he was quoting the actor ._.

Daminark

6:33 PM

Only the Rudin one

Ali Caglayan

@Danu arxiv.org/pdf/1510.03836.pdf

and diving into the references should supplement that

Daminark

The Sally class doesn't do it at all, third quarter is Riemann integration on $\mathbb{R}^n$, then vector calc

Eric

Wait the regular sequence doesn't use Rudin @Daminark?? TIL

Ohhhh

Liad

one problem i see is that we showed iso. $N \times [0,1)$ to $[0,\infty)$ and $[0,1) $ with $[0,\infty)$ but this does not imply we have a homeomophism between $N\times [0,1)$ with $[0,1) $ @TedShifrin

Learning user

Hi

Ted Shifrin

6:35 PM

From the movie, @SoumyoB. Check out this book.

Learning user

shall i ask one question

A sells a horse to B for Rs. 9720, thereby losing 19 per cent, B sells it to C at a price which would have
given A 17 per cent profit. Find B’s gain.

Ted Shifrin

@Liad: Compose functions. $\Bbb N\times [0,1) \to [0,\infty) \to [0,1)$.

Learning user

what is the answer

Danu

@AliCaglayan There was a paper about "$G_2$ instantons" a few days ago on arxiv

Learning user

guide me the steps also

Liad

6:36 PM

@TedShifrin yeah i see that, but is it a homeo' ?

SoumyoB

@TedShifrin is that the actor's biography?

Ali Caglayan

@Danu oh really, you got a link?

Whats it in dg?

Ted Shifrin

No, @SoumyoB. History of all the brothers and the movies. Lots of pictures, quotes from the movies.

Semiclassical

Hi chat

Learning user

Hi @Semiclassical

Ted Shifrin

6:36 PM

@SoumyoB: Google Marx Brothers movies.

Hi Semiclassic.

Eric

I'm gonna go get lunch; bye chat folk

Learning user

How are you? @Semiclassical

Danu

arxiv.org/abs/1703.06329

Ted Shifrin

@Liad: Yes, composition of homeos.

Bye, @Eric. Me too soon.

Danu

I know a bit about SW monopoles (just a little bit) so it sounded interesting enough to me

Mike Miller

6:37 PM

@Danu Take a torus $\Bbb R^4/\Bbb Z^4$. Mod out by the action of $\pm 1$. You get $2^4$ singular points. Delete a neighborhood of these. Now for the hard part: solve the Calabi conjecture for asymptotically locally Euclidean manifolds. This means you can find Ricci-flat metrics on things that look like Euclidean space at infinity (a noncompact generalization of Yau's theorem).

@Danu this one?

G'night @MikeM.

@Eric See you!

A sells a horse to B for Rs. 9720, thereby losing 19 per cent, B sells it to C at a price which would have
given A 17 per cent profit. Find B’s gain. @Semiclassical

Liad

@TedShifrin the iso. between $[0,1) $ with $[0,\infty)$ is $0\to 0 $ and $x \to 1/x$ for $x \in (0,1)$ ?

Learning user

6:37 PM

Guide me please

Liad

wait no.

Danu

@MikeMiller ...OK?

Ted Shifrin

No, @Liad. You want it order-preserving. You have to do better than that.

Ali Caglayan

oh yes @Danu

Semiclassical

My favorite Marx bros line of late: youtu.be/xsOf0TZPPWY

SoumyoB

6:38 PM

wow they're so old @TedShifrin

Mike Miller

Now I'm trying to remember how this works. I think you apply this to a punctured $\Bbb{CP}^2$, and then glue those in at the singularities. Your solution to the Calabi conjecture ensures that the final thing will be smooth and Ricci-flat.

You've just constructed a K3 surface with its Ricci-flat metric.

Semiclassical

@Learninguser not available for that today

Ted Shifrin

Dead a long time, yes, @SoumyoB. :P Like me soon. :P

SoumyoB

like everyone

Danu

@MikeMiller wat^wat

Learning user

6:39 PM

@Semiclassical i am not understand

Mike Miller

Joyce's construction of $G_2$ manifolds (though not as general as Sa Earp's there via twisted connected sums) is the analagous thing in 7-dimensions. Solve a $G_2$-Calabi conjecture on asymptotically locally Euclidean things, and then take $\Bbb R^7/\Bbb Z^7$, quotient by pm 1, delete the singularities, and glue in good model spaces.

Learning user

I think you are busy now @Semiclassical

Mike Miller

OK, nevermind.

Semiclassical

yeah. Just stopping by momentarily

Danu

@MikeMiller Was the thing you outlined supposed to give... the Kaehler analog or something?

Mike Miller

6:40 PM

Hyperkahler. But yes.

Ted Shifrin

@MikeM: Is someone doing a $G_2$ Calabi conjecture?

Danu

@MikeMiller What's hyperkaehler supposed to be again?

Mike Miller

@TedShifrin No, but that was the easiest way to phrase Joyce's result I could think of.

holonomy SU(2)

aka K3

oh no you're right, not hyperkahler, just calabi-yau

in this case that's hyperkahler but that's just good luck

SoumyoB

by the way I squatted 165 pounds today

Semiclassical

Here's a math question while I'm here, though. Is there an obvious / well-known representation of $e^{i x}$ which makes the periodicity explicit?

SoumyoB

6:42 PM

almost got crushed

Ted Shifrin

LOL, @SoumyoB: You can crush lots of us.

Semiclassical

Obviously there's Euler's formula but that pushes it onto the trig functions

Ted Shifrin

@Semiclassic: I assume you do not acknowledge periodicity of sin and cos?

Ali Caglayan

I can spend 165 pounds

Ted Shifrin

LOL @Ali.

Mike Miller

6:43 PM

@TedShifrin In particular he finds asymptotically Euclidean metrics on resolutions of $\Bbb R^7/G$, $G$ a finite group.

SoumyoB

@TedShifrin I'm probably too short to do that

Mike Miller

With that he can G2-resolve-singularities.

Ted Shifrin

I see, @MikeM ...

Semiclassical

Not for these purposes, no

Danu

How come you know about so many diverse areas, Mike?

Ted Shifrin

6:43 PM

@SoumyoB: I'm quite short.

@Danu: Working on this kind of stuff you have to.

SoumyoB

shorter than 5'6"?

Ted Shifrin

Precisely 5'6" :P @SoumyoB.

Mike Miller

At some point I got interested in the G2 instanton story so I read Joyce's book on special holonomy. But then I never got around to reading eg Walpuski or Sa Earp's work.

I just know what Walpuski's told me.

Ted Shifrin

@Liad: Did you figure it out? Can you think of a precalculus-setting function that will map $[0,\infty)$ homeomorphically to $[0,1)$? Think of a graph.

SoumyoB

@TedShifrin wow that makes the two of us :D

Liad

6:46 PM

@TedShifrin something like $arctan$ ?

Ted Shifrin

Perfect, scaled appropriately.

You can do it with a rational function, too.

Danu

Oh, I have a question about characteristic classes: For a rank 2 bundle, how do I get $p_1(S^2H)$?

Mike Miller

What is $S^2$?

Liad

$x/(1+x ^ 2)$ ?@TedShifrin

Mike Miller

Symmetric matrices?

Danu

6:47 PM

Can I use something like $S^2H\oplus \Lambda^2 H=H\otimes H$ (waits to be slapped)

yeah

Don't disturb

I haven't seen lately the awesome user @ShaVuklia

Mike Miller

Seems pretty reasonable.

Ted Shifrin

Right direction, but graph that one, @Liad.

Mike Miller

In fact, $p(\Lambda^2 H) = 1$, so...

Ted Shifrin

It's certainly not one-to-one with the right range.

Danu

6:48 PM

@MikeMiller Right. The case I'm interested in is actually complex rank 2, I just realized

Mike Miller

Oh, there's a tensor product on the right side. Good fucking luck.

9

Ted Shifrin

LOL.

Danu

??

Mike Miller

I guess you could work out the details with the Chern character if you're willing to work over Q

Ted Shifrin

such language ...

Danu

6:49 PM

@MikeMiller Why is that bad?

Don't disturb

@Danu Prove that $$3\zeta(2)\zeta(5)+\frac{3}{4}\zeta(3)\zeta(4)-6 \zeta(7)>0$$

Mike Miller

How do you calculate the classes of a tensor product?

Danu

@Don'tdisturb No.

@MikeMiller I guess the Chern characters, yeah..

Don't disturb

@Semiclassical are you done with the inequality above?

Mike Miller

Only thing I know that's well-behaved like that. Try the Chern character and work over Q.

Danu

6:50 PM

Can't you just do it brute force by the splitting principle?

Ted Shifrin

When it's not tensoring with a line bundle, you get a horrific mess.

Danu

Yeah, but not impossible, right?

Ted Shifrin

I haven't thought about this in centuries.

Danu

Just like... 24 terms or something that you have to simplify

Liad

@TedShifrin x/1+x will work ?

Ted Shifrin

6:52 PM

LOL.

Yup, @Liad.

Liad

why did i add ^2 in the first place

:P

Ted Shifrin

Trying to get a 0 limit at infinity, which you didn't want? :)

Liad

right :)

Give me a break

@Danu OK. Being inspired by your answer I felt the need to change my username.

Liad

you sure $x/1+x$ works ?@TedShifrin doesnt it tend to 1/2 near 1 ?

Give me a break

6:56 PM

I don't see robjohn around these days.

Ted Shifrin

Graph it. Use a bit of algebra or calculus ...

You're going from $[0,\infty)$ to $[0,1)$, remember?

Liad

huh right right

sorry

thanks :P

Danu

@Givemeabreak I'm happy to have inspired you.

3

Give me a break

@Danu I had to star you.

Liad

@TedShifrin what is $f \ ^ {-1}$ ? i want to show $f$ is homeo'

Mike Miller

6:58 PM

@Danu I'm not going to do it, at least.

Ted Shifrin

You know how to find an inverse function, @Liad. Do it. :)

Liad

Do i ?

Give me a break

OK, no time here for jokes. I'm out for some more research.

Ted Shifrin

You did in high school algebra, yes.

Danu

@MikeMiller ^^ I'm pretty sure we actually did this as an exercise in the course on SW theory.

Mike Miller

7:00 PM

Feel free.

Liad

@TedShifrin found it :P

Danu

@MikeMiller Meaning the answer is already written out for me :)

$c_2(V\otimes W)=2(c_2(V)+c_2(W))+c_1^2(V)+c_1^2(W)+3c_1(V)c_1(W)$ :) (V,W of complex rank 2)

Mike Miller

Oh, right, I remember.

Ted Shifrin

I think I needed such things at some point in my research life, but ... hell if I remember.

Mike Miller

Well there you go then.

Danu

7:02 PM

Indeed the splitting principle with 24 terms :P

Mike Miller

You probably mean $c_1(V)c_1(W)$.

Danu

yeah

Ted Shifrin

Pfeh. I like Chern character better.

Danu

Me too!!!

Mike Miller

@TedShifrin Same argument, I think.

Danu

7:02 PM

It's a bit more beautiful or something, at least in my mind

Both are quite tedious/mechanical though

Mike Miller

only over Q, feh

Danu

I think I should look for some simple software to just do these computations for me

Ted Shifrin

Yup. My days and days of checking formulas with my coauthor back in the 80s would have been very simple with Maple or Mathematica.

You just have to figure out how to get UsingRules or whatever to work.

OK, lunchtime. Ciao.

Daminark

See you @Ted!

Liad

@TedShifrin i found a problem

@TedShifrin if $\alpha$ is countable ordinal then we showed $\alpha $ iso. to $\alpha +1$

and this can't be

Alessandro Codenotti

7:21 PM

That's false

If iso means order isomorphic

Liad

@AlessandroCodenotti this is the question: math.stackexchange.com/questions/2196882/…

i showed (with the help of ted ) that $N \times [0,1)$ with lex. order is iso. to $[0,1)$ as ordered set

and now i conclude that if $\alpha$ is countable then it is iso to $N $ and i have the iso to $[0,1)$

@AlessandroCodenotti what do you think?

arctic tern

what is $N$?

Danu

@TedShifrin mathematik.uni-kl.de/~hannes/s/sing_2232.htm That'll do it!

Liad

@arctictern natural numbers

Sha Vuklia

@Givemeabreak Haha, I'm pretty often here! Especially since my testweek is around the corner:p

He guys, I have one question:

(*) Let $(S,d)$ be a complete metric space. Then the space $S$ is not a union of a sequence of nowhere dense subsets of $S$.

Category 1 consists of sets that are unions of sequences of nowhere dense subsets of $S$.

Category 2 consists of the other subset of $S$.

Now the “Baire Category Theorem” states:

A complete metric space $(S,d)$ is of the second category in itself.

However, how is this different form (*)? In my eyes, those two states state the same, so I’m assuming I’m interpreting something wrong here.

arctic tern

7:28 PM

@Liad how do you conclude that $\alpha$ is iso to $\Bbb N$?

Liad

@arctictern i did not learned yet set theory so i thought the fact that it is countable is enough

user193319

Hello. I am working on this problem: math.stackexchange.com/questions/528230/…

And I have a question: Why is it necessary to go all the work that Brian Scott went through? If p denotes the restriction of the metric d onto A, then (A,p) forms a metric topology space whose basis is given by B_p(a,e). Now, as a subspace of X, the subspace topology on this is given by A \cap B_d(x,e). It seems pretty straightforward that B_p(x,e) = A \cap B_d(x,e), implying that the basis elements are identical. Wouldn't that be the end of the proof?

arctic tern

@Liad well, to be isomorphic as sets it's enough that they're both countable and infinite, but to be order isomorphic it's more than just having the same number of elements.

Liad

@arctictern huh, so that's makes everything i did useless :/

arctic tern

I don't know anything you did

Liad

7:31 PM

i showed $N \times [0,1)$ iso. to $[0,1)$ and concluded $\alpha \times [0,1)$ also.

arctic tern

okay

Liad

this is not true as you said because $\alpha$ is not iso. to $N$

arctic tern

just because $\alpha$ and $\Bbb N$ are not order isomorphic doesn't mean $\Bbb N\times[0,1)$ and $\alpha\times[0,1)$ aren't

Liad

please elaborate :P

arctic tern

what's there to elaborate on?

"$\alpha$ and $\Bbb N$ are not isomorphic" simply does not imply "$\Bbb N\times[0,1)$ and $\alpha\times[0,1)$ are not isomorphic"

Liad

7:37 PM

@arctictern how to show $\Bbb N \times [0,1)$ iso. to $\alpha \times [0,1)$ ?

arctic tern

show they're both iso to $[0,1)$

since $\alpha$ is an arbitrary countable ordinal, it suffices to show $\alpha\times[0,1)$ is iso to $[0,1)$

Liad

the question is to show $\alpha \times [0,1)$ iso to $[0,1) $ :-) @arctictern

Ted Shifrin

@Danu: Aluffi is a big Chern class computer.

Heya mr tern

Daminark

@Sha Those two are the same statement

Sha Vuklia

@Daminark Strange! They're listed as two separate theorems in Ross :P But whatever, I guess

Liad

7:42 PM

@TedShifrin we couldn't conclude $\alpha$ iso. to $\Bbb N$

Ted Shifrin

What does it mean to say $\alpha$ is a countable ordinal?

Liad

that there is a bijection to $\Bbb N$

not that it preserves the order

Danu

@TedShifrin It seems so!

Ted Shifrin

@Danu: McCrory and I were overlapping Aluffi a few times on things.

Daminark

@Sha How does it prove each one?

Sha Vuklia

7:46 PM

@Daminark the second one isn't proven, but merely added as "Baire Category Theorem"

the first one is proven by using that the union of a sequence of nowhere dense subsets of S has a dense complement.

Ted Shifrin

@Liad: I told you I was no good at set theory. With the axiom of choice, can't we well-order and assume the bijection is order-preserving?

@Alessandro: What about that?

arctic tern

@TedShifrin you can't prove $\alpha\times[0,1)$ is order iso to $[0,1)$ by changing the lex order on $\alpha\times[0,1)$

Ted Shifrin

Ah, fair enough.

Alessandro Codenotti

I have a movie starting in 10 minutes at the cinema, I'd rather not start a discussion I can't finish

Ted Shifrin

LOL, OK, @Alessandro.

Night :)

OK, so I at least got @Liad seeing/understanding what's going on for the case of $\Bbb N$. I guess I don't see the general argument.

Daminark

7:50 PM

@Sha My guess is that they're just taking that statement which they proved and giving it a name

Sha Vuklia

@Daminark yea I think so too

Let $S$ be a complete metric space. Is it true that the countable union, say $X$, of nowhere dense subsets of $S$ is again nowhere dense? We know that its complement $(S\setminus X)$ is dense. However, I think that I would need so show that $(S\setminus X^-)$ is dense.

The reason I would like to show this, is because my book says that the countable union of sets of the first category (which consists of countable unions of nowhere dense subsets) is again of the first category. So my first guess was to try to show that a countable union of nowhere dense subsets is nowhere dense.

Liad

@TedShifrin i thought of your suggestion with axiom of choise too :P i did not took set theory course yet so i dont know if what you say is legal :P

Ted Shifrin

@Liad: But what mr tern said is right. We can't go that route.

Perhaps mr tern can give a better hint, but I think you now have a better intuition for things.

Liad

i am , thanks to you :)

Daminark

@Sha This is not true, the rationals are a dense set, but a countable union of points

Sha Vuklia

7:57 PM

Are the rationals a complete metric space tho?

Mike Miller

@AkivaWeinberger I looked through that answer I posted again and yeah, it is true that every orientation-preserving homeomorphism of $S^n$ is isotopic to the identity

Sha Vuklia

or what were you referring to actually, my statement or the statement in the book?

Liad

@arctictern could you give a hint? a full solution would work fine too , im tired of this question :P

Daminark

No, they're incomplete. I'm referring to your statement, since you wanted to prove that a countable union of nowhere dense sets is nowhere dense. The counterexample comes from how $\mathbb{Q}$ is not nowhere dense, but it is a countable union of points, which are nowhere dense

But try this

arctic tern

@Liad do you have any results about countable ordinals and embedding in reals?

Daminark

7:59 PM

If you take a countable union of countable sets, the result is still countable, yeah?

Liad

@arctictern i just got familiar with ordinals in this question, so no..

Sha Vuklia

yea, I guess

Ted Shifrin

mr tern, can we do (order-preserving) induction with an arbitrary countable cardinal/ordinal?