I wonder how this will go.

Let's see. I'm abstaining myself for the moment.

By the way, the answer given in the paper linked to in the dual modules question is quite nice, but I really don't understand how passing to the dual category makes this argument any more transparent.

How low do fruits grow? =)

@Srivatsan yeah, looks like you dug up a carrot :)

@tb Oh, digging is work.

:)

@Srivatsan plucked out of the soil, then?

@tb My proof of convergence is a real horror.

Ok, see you tb.

@Srivatsan I'm still there :) Let me have a closer look at the convergence.

@Srivatsan I don't think it's that bad. Why is it a horror? See you later, then.

Didn't look so bad to me either.

oh, my...

Wow.

Do you ever upvote something just because it looks like a lot of work

I've done that, yes, and I'm pretty sure I received quite a few of such votes, too. But in this case I really fear that eng will suffer a heart attack.

@tb Ok. =)

So whatever Pacciu did, does that make sense?

I didn't have the patience to read that too closely, but what I did read makes sense.

Wow! Brian was in a funky mood

Guess who said this: "Really? And somebody is forcing you to accept answers (1) which you do not understand, (2) 19 minutes after you submitted the question, and (3) while your brain is not working? Wow..." =)

Robert hasn't responded to my query in the comment. Should I post it as a question?

Didier was in the mood of enumerating

Hi guys

@Srivatsan Well, it would certainly make a good question. Yes, Robert rarely responds to comments.

Hi Benjamin

Hi, Ben

@t.b. @Srivatsan Did I interrupt something important?

No, not as far as I'm concerned.

@BenjaminLim No, nothing important.

ah ok

I am reading through a section of Munkres, and am confused about something involving bases and topologies

okay

I don't get the difference between the topology generated by a basis and the elements in the basis being a topology itself

for example we have the product topology

e.g. X x Y where X and Y are themselves topological spaces

look at a more familiar example first.

R^2?

Let's even stick to R

ok

Before I start: Do you have the MathJaX Bookmark installed now?

I don't know how to do it

What browser do you use?

google chrom

chrome

I don't even understand the jargon

"When you are trying to find a solution, you can pretty much do whatver you like, including dancing a bit in honor of Tlaloc, the God of Rain." =)

@Ben First, click Options and check "Always show bookmarks bar" option

yeah that's already checked

Ok. Copy the code javascript:blabla from the answer. Right click on the bar, click add page.

@tb That may not work in chrome. It didn't when we (me, Ilya..) tried.

okay, removed

$\mathbb{R}$

ah ok

$\mathcal{T}$

it's rendering

@BenjaminLim cool, =)

Every time you load chat, you should click that button.

@t.b. what were you going to say?

@srivatsan yeah thanks

Okay, I'm back. Let's look at $\mathbb{R}$. What is your favorite basis for the usual topology of $\mathbb{R}$?

the set of all open intervals

all open intervals. Okay. So $(-1,0) \cup (2,3)$ isn't in the basis, but it's open, right?

yeaah

A basis for a topology only satisfies 1) that for each point $p$ in an open set $U$ there's a basis element $B$ such that $p \in B$ and $B \subset U$.

yes

and 2) that the intersection of two basis elements again contains a basis element.

There's nothing about unions here.

I see what you mean

it is not closed under taking unions

Exactly. However, the main point is: every open set is a union of basis elements.

Can you prove that?

so similarly in the product topology if we take the union of two rectangles say

it's not necessarily a rectangle

right.

Will this suffice? To prove that the topology $\mathcal{T}$ generated by the basis $\mathcal{B}$ is equal to the union of all elements in $\mathcal{B}$?

I mean if you take the basis consisting of all open intervals in $\mathbb{R}$

(wait, I'll get to that)

then a product of basis elements is a rectangle.

So you get a basis for the product topology by taking products of open intervals, i.e., rectangles.

@BenjaminLim Every open set is a union of basis elements, so the topology is the collection of all possible unions of basis elements.

I get that yeah

So we need to understand first why every open set is a union of basis elements.

yes

Which brings me back to my question: can you prove that?

well

if $U$ is open in $X$ a topological space

then by definition for every $u \in U$ there is a $B \in \mathcal{B}$ such that $u \in B \subset U$

so if we do $\bigcup_{u \in U} B_u$

$B_u \in \mathcal{B}$

one way we have that because every point of $U$ is in the cover we have $U \subset \bigcup_{u \in U} B_u$

however on the other hand because $B_u \subset U$ by definition, $\bigcup_{u \in U} B_u \subset U$

so that $U$ is a union of basis elements

precisely. Very good

thanks

Now the second question I have: Given a collection $\mathcal{B}$ with the following property: If $B_1, B_2 \in \mathcal{B}$ then for every $p \in B_1 \cap B_2$ there exists $B_p$ such that $p \in B_p \subset B_1 \cap B_2$ then $\mathcal{B}$ is the basis of some topology $\mathcal{T}$. Can you describe $\mathcal{T}$?

I could be wrong

Is it just me or LaTex not showing up in chat?

is it the topology generated by unions of finite intersections of elements of $\mathcal{B}$?

@ZeeshanMahmud It works fine for us.

@ZeeshanMahmud If you have the bookmark installed, you need to click it each time you enter chat. Otherwise you need to install it

@BenjaminLim You're not wrong, but it suffices to take $\mathcal{T}$ to be all possible unions of elements of $\mathcal{B}$. No need to take intersections.

do you mean "all possible unions of elements of $\mathcal{B}$?

Yes, sorry.

Thanks.

Ok, you posed the second question because...

... I wanted to make the point that in this description it is convenient to have $\mathcal{B}$ "small".

So, I prefer the basis of $\mathbb{R}$ consisting of all intervals with rational endpoints.

Because it is countable.

Do you see why that's a basis?

well there is a lemma I know of

that is if $X$ is a topological space equipped with a topology $\mathcal{\tau}$ on it

and $\mathscr{C}$ a collection of open sets

such that for every open set $U$ in $X$ and each $x$ in $U$ there is a $C \in \mathcal{C}$ such that $x \in C \subset U$

then $\mathscr{C}$ is a basis for the topology $\mathcal{\tau}$

right.

But instead of using this lemma, I'd do it by hands.

so in $\mathbb{R}$ it suffices to show that for every open interval $(a,b)$ and a point $x \in (a,b)$ there is an interval $(c,d)$ where $c,d \in \mathbb{Q}$ such that $x \in (c,d) \subset (a,b)$

okay. and how do you find such an interval?

but then this follows immediately from the density of $\mathbb{Q}$ in $\mathbb{R}$

yes, that's what I wanted to hear.

more or less...

completeness axiom -> archimedean property -> density of rationals in R

okay, okay, that's fine!

man you are asking me questions KGB style that's why!!

do you have a copy of munkres with you now?

no worries, but if someone knocks on your door after you got a question wrong, I'd be worried if I were you :)

@BenjaminLim I guess I should point out one subtlety here. The lemma you stated required you to check for every open set $U$ in $\mathbb R$. But you started off with "it suffices to check for every open interval $(a,b)$ and..." Do you see why it suffices?

because every open set is just the union of these guys?

@tb what do you mean?

Ya. In fact, the open intervals form a basis for the euclidean topology, that's why. This is one advantage in keeping the basis as small as possible. You need to check smaller number of things.

In other words: for every $p \in U$ there's some inerval $(a,b)$ such that $p \in (a,b) \subset U$ and now you can take $(c,d)$ with rational endpoints with $p \in (c,d) \subset (a,b)$.

@BenjaminLim oh, that was just a silly allusion to my agents down under.

@tb Do you have many "agents" in universities here?

@BenjaminLim We might have one in your university as well. Where did you say you study? =)

I laughed so hard.

@Srivatsan you have 5 seconds to view the next comment, then I will delete it.

okay, I've fetched my copy of Munkres.

@BenjaminLim Oh boy. I was just kidding; I didn't mean that really...

[obviously, I hope :)]

Kinda reminds me of that scene in MI 2

Tom Cruise had 10 seconds before the sun glasses explode was it?

@tb RIght. There was this proof on the section on product topologies

I think it's theorem 15.2, if we're using the same edition (year 2000)

This one?

which is...

yeah

how did you do that?

black magic

the proof seems to me he is showing that the topologies are equal

Let me check the definitions he gives, just a sec.

Okay, I'm ready.

Yes, he has the product topology $\mathscr{T}$ and the topology $\mathscr{T}'$ generated by $\mathscr{S}$

He wants to show that $\mathscr{S}$ is a subbasis for the topology $\mathscr{T}$, right?

$\mathscr{S}$

yeah

So, what is your question?

In the proof it seems he is showing that the topology $\mathscr{\tau}$ is equal to the topology generated by $\mathscr{\tau}'$

I don't get how that is the same as showing that $\mathcal{S}$ is a sub-basis for the product topology on $X \times Y$

In fact this brings me to an even more important question

I don't know how the elements in $\mathcal{S}$ even relate to the original topology on $X \times Y$

First, you need to understand that the sets in $\mathcal{S}$ belong to the product topology.

wait

Ok, I understand that

So, a set in $S \in \mathcal{S}$ is either of the form $U \times Y$ for some open $U \subset X$ or of the form $X \times V$ for some open $V \subset Y$

yeah

Hence $S$ is certainly open in the product topology.

Which is good, because that tells us that $\mathcal{S} \subset \mathcal{T}$.

ok

Hence $\mathcal{T}' \subset \mathcal{T}$

one thing

okay

why is it that because $\mathcal{S} \subset \mathcal{T}$, that $\mathcal{T}' \subset \mathcal{T}$?

Well, $\mathcal{T}'$ is the smallest topology containing $\mathcal{S}$.

= =

?

What does that mean?

I think it means blur face

I think it's a face.

I see. What's not clear?

I don't understand what you mean that $\mathcal{T}'$ is the smallest topology containing $\mathcal{S}$

I see. So: What do you mean by saying that $\mathcal{T}'$ is the topology generated by $\mathcal{S}$?

that every element in $\mathcal{T}'$ is the union of stuff in $\mathcal{S}$?

Topologies on a set $X$ are, among other things, subsets of the power set $\mathcal P(X)$ of $X$.

@tb There are two meanings, which -- I think -- is causing some confusion to Ben: (a) It is the intersection of all topologies that contain $\mathcal S$. (b) it is the union of finite intersections of members of $\mathcal S$.

Yeah, I wanted Ben to say that... :)

@Srivatsan actually you know what, (a) was actually a problem in munkres

@BenjaminLim I am sure =) Either a problem or a lemma or claim or something...

By the way: the comments here are becoming postmodern :)

So (a) is saying that $\mathcal{T}'$ is the smallest topology containing $\mathcal{S}$, no?

(assuming that we know that an intersection of topologies is again a topology)

that is trivial to show

sure.

Can we assume (a) or do we need to think about it?

Problem is: How does (b) mean the same as (a)

give me some time

BTW, my (b) was a bit off. It should be: $\mathcal T^{\; \prime}$ is the collection of all unions of finite intersections of members of $\mathcal S$.

Okay. Let $\mathcal{T}_a$ be the topology as described in (a) and let $\mathcal{T}_b$ be the topology as described in (b).

We want to know that $\mathcal{T}_a = \mathcal{T}_b$, right?

yes

Which inclusion is causing trouble?

1. $\subset$, 2. $\supset$, 3. both, 4. none?

I don't know how to tell you. Now I am a bit confused

So, my guess is 3. :)

I guess the confusing bit is that there a lot of concepts, all look similar

No worries, it is confusing.

I feel dumb for asking this but what formula is used to find this

@BenjaminLim So, we have our collection of subset $\mathcal{S}$.

We have our intersection $\mathcal{T}_a$ of all topologies containing $\mathcal{S}$

Ok, for now how about this perhaps simpler problem, if $\mathcal{B}$ is a basis for a topology $\mathcal{T}$ on X, then the topology generated by $\mathcal{B}$ is equal to the intersection of all topologies on $X$ that contain $\mathcal{B}$

@ZeeshanMahmud If it's just a random set of values, then I don't think there'll be any formulas.

Do the numbers have any pattern?

@BenjaminLim Let's stick to $\mathcal{S}$, the basis thing doesn't make it (much) easier.

What crap...?

no, it's just an expression meaning "my suggestion does not help"

@Srivatsan No.. '2' is missing..so is 16..18

@ZeeshanMahmud I don't get it. What is the pattern again?

Okay. Again: $\mathcal{T}_a$ is the intersection of all topologies containing $\mathcal{S}$.

I wonder how many good mathematicians are Mensa members.

yeah

I would suspect very few if any, but I don't know.

@Srivatsan I don't see any pattern

@DylanMoreland There are many self-infatuated mathematicians.

@ZeeshanMahmud Then hard luck, I believe.

$\mathcal{T}_b$ is all unions of finite intersections of elements of $\mathcal{S}$

@Srivatsan Trial and error?

So $\mathcal{T}_b$ is at topology contontaining $\mathcal{S}$.

@ZeeshanMahmud Ya, that would work. May be one could be clever in special cases.

Trial and error never fails. :)

[just kidding]

@tb which $\mathcal{T}$?

let's move here

ok

Personally I am not a fan of IQ unlike a teacher i know...but once in a while like their puzzles.

@tb Exactly. It always surprises me but it's true.

@DylanMoreland Do you know any mensa members?

Many are inexplicably defensive about it. I feel like it has eaten at Shimura for most of his life, for example, and he's a professor at Princeton. How can you still feel small?

@Srivatsan No one who will admit it, at least.

@DylanMoreland Wait, people are mensa members but they are ashamed about it?

@DylanMoreland what's this Shimura bit?

What makes you think it has eaten at him?

Maybe I should stop here, but if someone told me that they were a member I would probably start laughing, assuming that were safe to do.

He has an autobiography. There's other evidence as well.

Ok time to retire in my...

@tb Sorry about that. Got carried away. The other two should share a third of the blame too =) Anyway, ignore that.

@Srivatsan No problem... Don't worry about it.

I planned to make the move as soon as others want to talk.

@tb Ok.

Ok, I have to leave now. See you, @tb.

See you @Srivatsan

Why, if a holomorphic function has constant absolute value on both boundary rings of an annulus, is it constant on that annulus?

math.stackexchange.com/a/92406/16980 Is this a correct answer??

@Potato There's some Cauchy integral formula on the annulus, right?

Everything is determined by boundary values.

See 4.2.5 here.

Ah, ok, I see now

Thank you.

What exactly has eaten Shimura for most of his life, Dylan?

Bread.

?

Also, I see how knowing the values on the boundary would allow us to compute everything inside, but we only know it has constant modulus.

Ohh sorry. I misread you.

Is that even true?

@Potato What about like $f(z) = z$? It's constant on rings around zero.

Er, the modulus is, rather.

And yet it isn't constant.

anywhere

Not of the top to morrow you, what.

Hi Asaf.

No silliness now, Asaf! You gotta earn some points... :)

Good morning, btw.

@tb It's not about earning points. It's about finding the answer for myself. I like these sort of questions. I think about them often.

@AsafKaragila oh, I was thinking of the hit-squad math.SE will have when you cap today

But good to know, of course

Wow, I could only find one thing: Ramsey's theorem holds in that model.

I think.

Cohen's second model?

Oh wait no, I misread. It does not hold there either.

We are witnessing history.

LOL! I came to post EXACT same comment

about witnessing a miracle

Ok I need to delete the Lol comment

Well, time to go to the university. First, a class. Then a long day of researching into this answer! :-)

How great it is that we're on a strike...

hi all

know any interesting study problems in symplectic and contact geometry?

exercises?

btw, i just recently read about orientation in Hatcher

and the way it's defined using homology is beautiful 8)

@AsafKaragila The morrow, of what top and not to you.

@JM :-D

And now we leave! So long, for now.

...I must step out, as well. TTFN!

@JM: later

More snow!!!!

Good morning everyone.

Bye JM.

@Matt we had rain earlier tonight. No snow here for over 10 years.

@robjohn Do you like it there?

@Matt I do, though I liked Princeton when I lived there, and Cupertino when I lived there.

The winters in Princeton were probably easier to deal with as a grad student, when I had no house or car to maintain.

We don't have either : )

@Matt house or car?

@robjohn Yes, no house and no car. If we need a car there is a very cool car sharing system.

@Matt There was a car rental place in Princeton that I used once in a while.

@Asaf: please consider pinning this comment by JM

@robjohn Princeton is quite elite. : ) Did you do your undergrad studies there?

@Matt no, I did undergrad at UCLA.

@Alexei I agree, it's very nice.

@robjohn I think if I could choose freely I'd do part 3 in Cambridge and then try to get a PhD position there. I really like England and Cambridge is lovely.

@Matt I've never been to England, but my wife says that I'd like it.

@Skullpatrol: You have returned.

@robjohn ;-)

@robjohn I worked there for a while for a games company as a Junior SWE. The housing is not so nice but I found English people, at least the ones at work, nice.

@Skullpatrol yet your profile is still absent.

@Matt I'm sure there are good and bad everywhere.

Chain homotopy. Fun. :-/

@robjohn Of course.

@AsafKaragila I thought you weren't around! You know, I'll be doing forcing today : )

I am in class. I will help later if you need me.

@AsafKaragila is that something that you put on your tires when it snows?

Since we have a strike today, I won't teach so I'll have more time to loaf around the chat.

Strike??

@AsafKaragila An air strike?

Not Quite.

: )

Bobby J.: what snow? :-)

@AsafKaragila ?

Rats! MathJax doesn't support \sout{...}

@Matt !

@AsafKaragila chains go on tires when it snows.

When does it snow???

@robjohn What did he mean by that? Can you explain him to me?

@Matt what did who mean by what?

@robjohn Asaf saying "Bobby J.: ..."

@Matt Bobby is a children's nickname for Robert and J is the first initial of my last name

@Matt Although no one called me Bobby even as a child

@robjohn Oh. facepalm

And I am an eternal child...

Also internal child. Not external though.

Oh, this is sad.

It was over a week ago, but I just noticed it.

Here is an interesting tool.

Hah! Awesome!

I have that on my iPhone too.

@AsafKaragila I will have to try it on my iPhone

$\unicode{x2603}$

@Matt very seasonal

$\Huge\unicode{x2603}$

Even better : )

You need to adjust the ChatJax to iPhone too.

$\Huge\unicode{x263a}$

@AsafKaragila ChatJax?

The bookmark?

I wish there were a way to have the iPad keyboard show $ and \ when I'm typing on this site.

Paging to get to those two characters gets old.

good morning guys

@robjohn Quite.

Nice. It's snowing again.

Where? It's hot in here.

@Matt: You said that you study set theory from Combinatorial Set Theory: With a Gentle Introduction to Forcing, right?

@AsafKaragila Only partially but yes. I have Kunen here, too. They recommended Kunen for the course but I think Jech might have been a better book.

Jech is a good book as a reference. Personally, of course, I much prefer the Boolean-valued models approach to forcing, but in practice you don't really work with that. So the approach in Kunen might be useful for modern forcing techniques.

In Kunen, $p \leq q$ reads as $p$ is stronger than $q$ and in the lecture it reads as $p$ is weaker than $q$. I wonder why they do it one way or the other. It seems to make no difference so why not stick to the book.

It's the Jerusalem tradition vs. the rest of the world.

I may be in troubles if I'll do my Ph.D. with Magidor. He's in Jerusalem, so the Jerusalem notation is appropriate, but he himself switched to the worldly notation... so I should follow the place or the advisor??

Make up your own mind and use what you prefer?

I also wonder why all the letters have to be either curly, have a tilde or have a dot somewhere. I don't think it looks pretty. Why not just use regular letters like e.g. $U$ instead of $\mathcal{U}$ to denote an ultrafilter?

Put another way: can you stand up to Magidor and use the notation you're comfy with?

@Matt I don't care, as long as the introduction section says "We shall denote $p\le q$ as $p$ being stronger than $q$", or replace "stronger" by "extends".

@JM Of course. I have too much backbone. I even stand up to people that I don't have to stand up to, or that I know from the beginning will rule me over and make me change things.

This is one of the reason I do my best to drink daily. Alcohol is known as "courage juice"!

Also @JM: If you can put the MathJax link (as starred by Bobby J.) in a titled form (like I did with the ChatJax link) I'll pin it to the top.

Speaking of drinking, I am out of beer $\stackrel{\circ\cdot\circ}{\frown}$

The fridge is like... two whole meters from me!

In the meantime, I'll go get some coffee.

LaTeX commands supported by MathJax

@AsafKaragila Done.

@Asaf: there was already a 2 star comment by JM. I was thinking you could simply pin that existing starred comment :-)

Nah. If I want to pin something for "formal" purposes then I prefer it to be nicely written.

Okay...

I'm going out for lunch. See you later alligators.

Bon Apétit

After some whiles, crocodiles.

@AsafKaragila I think that might require talking to the people who wrote the mobile math.SE site.