Mathematics - 2012-08-12 (page 1 of 3)

BenjaLim

12:04 AM

@MarianoSuárez-Alvarez There is a problem with my answer here: math.stackexchange.com/questions/181277/…

David Wheeler

@PeterTamaroff k+1

David Wheeler

12:17 AM

@PeterTamaroff but yes, the maximal difference is 1/(3^k)...what is the problem?

Peter Tamaroff

@DavidWheeler Because earlier you said $k-1$ ¬¬

And I was :confus:

David Wheeler

no, i did not, i said you needed to go "one more digit"

Peter Tamaroff

@DavidWheeler Sorry, you said $k+1$, true.

Jonas Teuwen

12:34 AM

Good night guys!

user19161

12:54 AM

@JonasTeuwen I am going to sleep now, see you in my dreams!

Srle

1:14 AM

If matrix A is invertible, then is A^m * A^n=A^(m+n) for m,n∈Z?

David Wheeler

@Srle what do YOU think?

Srle

i think it is correct

David Wheeler

can you see why without loss of generality, you may assume m ≥ n?

Srle

no :D

David Wheeler

well let's suppose than m ≥ n

Peter Tamaroff

1:18 AM

@Srle Do you know what is meant by "without loss of generality"?

David Wheeler

we can do the other case later

the general idea is you want to prove it for n ≥ 0, then for n < 0.

Peter Tamaroff

@DavidWheeler Dude. So we agree that if $x$ and $a$ are equal up to $k$, then $|f(x)-f(a)|\leq 3^{-k}$. Now, note that since $f$ is one one, $P_k=\bigcup_{i=1}^k p_i^{-1}(a(i))\subset f^{-1}B(f(a);\epsilon)$ is equivalent to $f(P_k)\subset B$. Thus what we do is choose $k$ large enought so that $3^{-k}<\epsilon$, then every $x\in P_k$ will satisfy $|f(x)-f(a)|<\epsilon$, from where $f(P_k)\subset B$ and $f$ is continuous. Correct?

David Wheeler

$3^{-(k+1)}$

Peter Tamaroff

@DavidWheeler $\eqalign{ \sum\limits_{n = k + 1}^{ + \infty } {\frac{2}{{{3^n}}}} = \sum\limits_{n = 0}^{ + \infty } {\frac{2}{{{3^n}}}} - \sum\limits_{n = 0}^k {\frac{2}{{{3^n}}}} &\cr = 2\frac{1}{{1 - \frac{1}{3}}} - 2\frac{{{{\left( {\frac{1}{3}} \right)}^{k + 1}} - 1}}{{\frac{1}{3} - 1}} &\cr = 3 + 3\left( {{{\left( {\frac{1}{3}} \right)}^{k + 1}} - 1} \right) = \frac{1}{{{3^k}}} &\cr}$

¿?¿?

David Wheeler

yes, that is true.

Peter Tamaroff

1:26 AM

That is the "extreme" case, right?

The difference can only be smaller.

David Wheeler

yes, the "worst case scenario", is that one number is "all 2's" afterwards, and the other is "all 0's".

Peter Tamaroff

Why was it again you need $k+1$, then?

You're allowed to deep sigh

David Wheeler

if they are the same for k digits, their difference is at most $1/3^{k+1}$

Peter Tamaroff

Dear Lord. I have just worked out the difference is at most $3^{-k}$, didn't I?

David Wheeler

oh shoot...now i am getting confused.

Peter Tamaroff

1:31 AM

@DavidWheeler Hahaha, please look above there!

It is $x(n)-x(a)=2-0$ for each $n\geq k+1$.

It is the "worst case scenario" from $k$ onwards.

David Wheeler

ok, first digit different (0 digits the same)...possible difference of 1. first digit the same (second digit different), possible difference of 1/3

so if k digits are the same, yes, the max difference is 1/3^k

Peter Tamaroff

@DavidWheeler ;) Good.

So now we agree taking $k$ such that $3^{-k}<\epsilon$ should do the work?

I think azarael just took that, halved. Because he said choose $N$ so that $$\frac{1}{2}\sum\limits_{n =N}^{ + \infty } {\frac{2}{{{3^n}}}} = \sum\limits_{n = N}^{ + \infty } {\frac{1}{{{3^n}}}} < \varepsilon $$

@DavidWheeler Sorry that this talk got so long.... but there was some confusion around.

David Wheeler

we can pick N even bigger, if we like.

Peter Tamaroff

@DavidWheeler Yes, yes.

David Wheeler

all we need to do is find one that works. it's really not that important which one that works it is.

Peter Tamaroff

1:37 AM

@DavidWheeler OK. Thanks.

David Wheeler

if you imagine the situation "geometrically", we just need to trap f(x) on the same "level" as f(a), because we know the "pieces" get smaller and smaller on each level.

Peter Tamaroff

@DavidWheeler Yeah, ;)

David Wheeler

can you see why we need the product topology and not the full discrete topology?

Peter Tamaroff

@DavidWheeler If we had the full discrete topology, $f$ would be trivially continuous, wouldn't it?

David Wheeler

yes, it would

now what is wrong with that?

Peter Tamaroff

1:45 AM

@DavidWheeler From what point of view?

David Wheeler

it's not that f would be "bad", but consider f:X→f(X).

Peter Tamaroff

@DavidWheeler OK.

David Wheeler

that is a bijection, right?

Peter Tamaroff

We're still talking about $f$ up there, right?

David Wheeler

same f, and X is the product of countably many copies of {0,2}

Peter Tamaroff

1:47 AM

I would write $f^*:X\to f(X)\;/\;f^*(x)=f(x)$ =P

Just to avoid circularity =)

David Wheeler

f is bijective from X to f(X), since its injective.

Peter Tamaroff

Yes, yes.

Forgot to answer.

David Wheeler

now, is the inverse continuous?

Peter Tamaroff

@DavidWheeler I think by the argument we just made, it is.

I mean

David Wheeler

ok, what does an "open set" look like in f(X)?

Peter Tamaroff

1:50 AM

@DavidWheeler Oh, wait. We changed out topological space, right?

Now it is $f(X)$ as a subspace of $[0,1]$?

Or $\Bbb R$, better.

David Wheeler

$f^{-1}$ is continuous if, given any open set in X, f(X) is open, right?

Peter Tamaroff

@DavidWheeler Yes.

@DavidWheeler Wait wait.

David Wheeler

well, technically $(f^{-1})^{-1}(X)$

Peter Tamaroff

$f^{-1}$ is continuous iff whenever $O\subset \mathcal S$ it follows $f(O)\subset \Pi$ - where $ \mathcal S$ is the subspace topology and $\Pi$ is the product topology.

David Wheeler

i'm saying, suppose we had the discrete topology on X, can you see what that does to $f^{-1}$?

robjohn

1:54 AM

@BillDubuque: I made it back in, but I didn't do anything other than post to SO and wait.

Peter Tamaroff

@DavidWheeler But then the open sets would be everywhere! What is your point?

David Wheeler

for example, we have the "open set" 0.022222222222......

Peter Tamaroff

@DavidWheeler Yes.

David Wheeler

which f maps to the single point 1/3.

Peter Tamaroff

@DavidWheeler Oh. Right.

David Wheeler

1:56 AM

is 1/3 an open set in the relative topology on the Cantor set?

(hint: every point is the Cantor set is a "cluster point")

aka "a point of accumulation"

which means every epsilon ball around 1/3 contains some other point of the Cantor set.

Peter Tamaroff

@DavidWheeler Right. It is open.

David Wheeler

which means that (1/3- ε,1/3+ε) intersect C can't possibly be {1/3}

so, no, the image of a single point in X, is NOT open.

the discrete topology has "too many open sets" to make $f^{-1}$ continuous.

so even though f is a continuous bijection, it's not a homeomorphism, if X has the discrete topology.

Mark Dominus

@robjohn I raised a flag several hours ago asking for this question of mine to be made CW. The flag has not been either accepted or declined. Do you know what the holdup might be?

David Wheeler

so the million dollar question is: if we take the product topology, is $f^{-1}$ continuous then?

robjohn

@MarkDominus I've been trying to get fully reconnected for a while. I just got back. I will look.

Mark Dominus

2:05 AM

Thanks very much.

Peter Tamaroff

@MarkDominus Could you proofread this, please?

@MarkDominus It is topology, so I guess it will be routine for you.

Mark Dominus

@PeterTamaroff Looking.

Peter Tamaroff

@robjohn Do you know how to set up tables in $\LaTeX$? I have a problem in one of my answers.

Mark Dominus

@PeterTamaroff Ha.

Peter Tamaroff

@MarkDominus bites nails

robjohn

2:09 AM

@MarkDominus Done

@PeterTamaroff where?

Peter Tamaroff

It is crooked

Mark Dominus

@robjohn Many thanks.

Henry T. Horton

@PeterTamaroff Post a screenshot, it looks fine to me

Other than a missing \hline at the top

Mark Dominus

@PeterTamaroff Correct me if I am crazy, but is it not the case that (a) a product of discrete spaces is discrete, and (b) every function on a discrete space is continuous?

robjohn

@PeterTamaroff what needs changing?

Peter Tamaroff

2:13 AM

@MarkDominus I don't have that as a theorem. GAWD.

Mark Dominus

I am not sure about (a).

Peter Tamaroff

Henry T. Horton

The lines go all the way across for me

David Wheeler

@MarkDominus i do not believe that the product topology on an infinite number of discrete factors is, in fact, discrete.

Mark Dominus

Okay, no, because otherwise the Cantor set would be a discrete space, which it isn't, as per exactly the example you are proving.

Peter Tamaroff

2:14 AM

@MarkDominus "Yes. In fact, a product of $A$ discrete spaces, with the product topology, each with more than one point is discrete if and only if $A$ is finite. – David Mitra"

Mark Dominus

Good!

Peter Tamaroff

"Yes. Finite products of discrete spaces are again discrete. Both X and Y have as a base the singleton sets, and the product topology on X×Y will have as a base, the product of singleton sets, meaning every point is open and closed, and hence the topology on X×Y is discrete.

An infinite product of discrete spaces need not be discrete however."

Henry T. Horton

Is &c supposed to be etc.

David Wheeler

which is exactly the point that this example brings into sharp relief

Peter Tamaroff

@HenryT.Horton Is etc. supposed to mean &c?

@DavidWheeler =)

robjohn

2:15 AM

Peter Tamaroff

@robjohn What do you use to get such neat screenshots?

robjohn

That's what I see. Perhaps it is a browser issue.

Mark Dominus

@HenryT.Horton "et" in "etc." is Latin for "and". "&" is a long-standing abbreviation for "and". The "c" stands for "cetera" = "the rest".

Peter Tamaroff

@HenryT.Horton You just got schooled by a guy as important as Jigglypuff.

robjohn

@PeterTamaroff I just take a screenshot (command-shift-3 on the Mac)

Henry T. Horton

2:17 AM

Yeah but who the hell uses &c for etc

David Wheeler

@PeterTamaroff the usual abberviation is "etc." short for "et cetera", Latin for "and so forth" (literally "and the other things").

Henry T. Horton

Just looks unprofessional. -1.

robjohn

@HenryT.Horton i see it all the time.

Peter Tamaroff

@HenryT.Horton Me. And when I drink, I ask for a Martinus. I ask for Martini when I really want to get wasted,

David Wheeler

have you ever said: "what a bunch of ani!"?

Peter Tamaroff

2:19 AM

@MarkDominus Do you think the proof is OK?

@DavidWheeler I talk in spanish most of the time, so no. The Martini thing is an exception.

David Wheeler

>.< 7 foot joke?

Mark Dominus

@PeterTamaroff In the next-to-last paragraph, do you really mean "for $n≥ k+1$"?

@HenryT.Horton In fact, the "&" symbol is itself a variation on the letters "et".

Peter Tamaroff

@MarkDominus Didn't I write that?

robjohn

I gotta log off for a bit. My wife needs to charge her phone and I only have one iPhone cable.

Mark Dominus

@PeterTamaroff It's okay. I just had to think about it.

I did not see any problems in the proof.

Peter Tamaroff

2:23 AM

@MarkDominus OK, cool.

I am reading about identification topologies now.

Mark Dominus

@PeterTamaroff Is that the same as quotient spaces?

Peter Tamaroff

@MarkDominus Yes. Actually "quotient topologies" =P

Mark Dominus

David Wheeler

"The phrase et cetera ("and so forth"), usually written as etc. can be abbreviated &c. representing the combination et + c(etera)" -Wikipedia

Mark Dominus

You can see the older "et"-like forms in the second and third ampersands in the middle row.

And a sort of intermediate form in the leftmost of the middle row.

Peter Tamaroff

2:25 AM

@MarkDominus OMG I will learn to write the "Hoefler Text Italic."

Mark Dominus

I know Hoefler. He was a year behind me in high school.

David Wheeler

from what i've read, the "Hoefler Text Italic" is actually a fairly "old" form

Peter Tamaroff

@MarkDominus Hehehe, look at that.

Mark Dominus

Yes, Hoefler is very witty.

Peter Tamaroff

@DavidWheeler Would you like to discuss a little about quotient spaces?

Mark Dominus

2:27 AM

We were not close friends, but we knew one another and were on good terms. I have been to his typography studio.

Peter Tamaroff

I'm trying to grokke the construction.

David Wheeler

when i scribble it, my own handwriting looks like the Palatino

Mark Dominus

I use a form a lot like the Palatino one myself.

David Wheeler

what's to grok? you have a quotient map. you want it to be continuous.

Peter Tamaroff

@DavidWheeler BRB.

David Wheeler

2:35 AM

the simplest example is the circle: start with the real line, and define the following equivalence: x~ y iff x-y is an integer.

Mark Dominus

I would have said that the simplest example is the circle: start with $[0,1]$ and define the smallest equivalence that contains 0~1.

David Wheeler

same thing really, i just have "larger equivalence classes", but there's an obvious 1-1 correspondence between the two

and it's the same quotient map $q: x \mapsto e^{2\pi i x}$

the counter-intuitive thing is that we don't want images of open sets, we want sets whose pre-images are open.

Peter Tamaroff

3:06 AM

@DavidWheeler I'm back.

Could you bear with me for a while?

It's not that I'm confused about something, don't worry! =)

@MarkDominus Are you there?

Mark Dominus

here

Peter Tamaroff

@MarkDominus Oh, good.

So. The author defines an identification $p:(E,\mathfrak I)\to (B,\mathfrak I')$ as a continuous function such that $p^{-1}(O)$ open in $E$ means $O$ open in $B$. Equivalently from Munkres "$U\subseteq \mathfrak I\iff p^{-1}(U) \in \mathfrak I'$.

Mark Dominus

Okay.

Peter Tamaroff

@MarkDominus He then proves a little lemma which says that if $p:E\to B$ is an identification and $G:E\to Y$ is continuous, and $p(x)=p(x')\Rightarrow G(x)=G(x')$, we can set up a continuous function $g:B\to Y$ by $g(b)=G(x)$, with $x\in p^{-1}(\{b\})$.

Mark Dominus

Sorry, I don't think I'm going to be able to help. It's too late at night here.

Peter Tamaroff

3:19 AM

@MarkDominus It's OK.

Laters.

Mark Dominus

Yes.

Peter Tamaroff

@HenryT.Horton

Henry T. Horton

I can't help you until Tuesday night

Peter Tamaroff

@HenryT.Horton You're working?

Henry T. Horton

I should be

Peter Tamaroff

3:22 AM

@HenryT.Horton OK. Best wishes.

Henry T. Horton

I'll need them :'(

Peter Tamaroff

@HenryT.Horton Why?

Henry T. Horton

I have a qualifying exam Tuesday

Peter Tamaroff

@HenryT.Horton Quialifying for what?

Henry T. Horton

Anal

I just did a practice one and answered a passing amount of questions in 25% of the time limit though, and I seem to know how to do all the questions almost right away

Peter Tamaroff

3:26 AM

@HenryT.Horton I mean, Grad studies?

Henry T. Horton

Yes a Ph.D qualifying exam

Peter Tamaroff

@HenryT.Horton Well, that is awesome, isn't it?

Henry T. Horton

Yes

mixedmath

I just noticed that Martin put the homework tag on the mysteriously-revived "should we have a homework tag" meta question. (meta.math.stackexchange.com/questions/621/… awesome.

Peter Tamaroff

@mixedmath LOL

@mixedmath Do you know about identification maps?

mixedmath

3:29 AM

identification maps?

Peter Tamaroff

@mixedmath Quotient maps, quotient topologies.

mixedmath

yes, I know some things about them, I guess. But I'm quick to say that topology has always been one of my weaker subjects, so if this is technical at all, I'll not be helpful.

BenjaLim

3:43 AM

@PeterTamaroff I can help you with quotient topologies.

@mixedmath Can I ask a quick/basic question on simplices?

Martin Sleziak

@mixedmath Some problem there? Since the thread was already bumped, I don't think that adding correct tags does any harm.

mixedmath

@MartinSleziak No - I thought it was excellent. It really pleased my dry humour side.

Martin Sleziak

Ok then. It sounded a bit sarcastic, so I asked.

BenjaLim

Let $K$ be a simplicial complex, and $\sigma, \tau$ distinct simplixes of $K$. Why is it if $x \in \sigma \cap \tau$, then $x \in \textrm{int} s \cap \textrm{int} t$ for some faces $t$ and $s$?

mixedmath

@MartinSleziak ah - I was composing my comment to Makoto at the time, and perhaps it rubbed a bit on me when I was typing in the chat, too. I'm sorry about that.

Martin Sleziak

3:47 AM

No problem. It was just my interpretation, obviously not a correct one.

BenjaLim

@PeterTamaroff the fact that you want your map to be constant on the fibers guarantees well-definiteness

@PeterTamaroff It's like the topological requirement that you quotient out by a normal subgroup/ideal

mixedmath

@BenjaLim As you tell me that on the one hand they are distinct, but on the other hand they have non-trivial intersection, I will assume that distinct means they have no common vertex and that they are defined by a set of vertices. So if a point is in both, but not a vertex, then it must be on faces.

BenjaLim

@mixedmath I copied out the words from elements of algebraic topology.

mixedmath

that's fine, my answer stands

I was just explaining the definition I was assuming

BenjaLim

hmmm now I realised I don't know what distinct means....

mixedmath

3:49 AM

if that's not the definition of a simplex to you, then we have a problem

aha - well, progress

one bit at a time

;p

BenjaLim

I have the definition of a simplex

mixedmath

I wouldn't be surprised if simplices came from a set of vertices. That's how Hatcher does it too.

BenjaLim

yes.

@mixedmath So in the case of 2-simplexes a.k.a. triangles

you say two triangles are distinct if

mixedmath

let's say that we are talking about the vertices [a,b,c,d], strictly speaking a 3-simplex (tetrahedron), and I wanted to talk about distinct 1-simplices (lines)

then, for example, [a,b] and [c,d] are distinct simplices

as they don't share a vertex

BenjaLim

ah ok.

now what about the 2-simplices (the triangles)?

I could not have [a,b,c] and [b,c,d] being distinct because they share the vertex $b$ yes?

mixedmath

3:54 AM

that's right

BenjaLim

thanks.

I am curious why munkres did not define what distinct simplices meant earlier.

mixedmath

Is his definition of a simplex a set of vertices, like I'm using?

BenjaLim

Definition:

Let $\{a_0,\ldots,a_n\}$ be a geometrically independent set in euclidean space. We define the $n$ - simplex $\sigma$

spanned by $a_0,\ldots, a_n$ to be the set of all points $x$ such that

$x = \sum_{i=0}^n t_i a_i$ where $\sum t_i = 1$

and $t_i \geq 0$ for all $i$.

mixedmath

close enough, I suppose

BenjaLim

Ok now that we have the definitions sorted out

why if a point is in both $\sigma $ and $\tau$

this forces it to be on a face?

mixedmath

3:58 AM

oh, my previous answer still stands.

BenjaLim

Yes.

But I don't understand why if it is in the intersection

and not a vertex

then it is on a face like you said.

Peter Tamaroff

Terrible electrical storm here. Power went off three times already.

BenjaLim

not a cloud in the sky :D

mixedmath

oh, I missed that you responded

I suspect that if you look up what a simplicial complex is, it will say that the intersection of any two simplices is a face of both simplices, too

Peter Tamaroff

@BenjaLim I did not see your pings. Thanks.

54 mins ago, by Peter Tamaroff

So. The author defines an identification $p:(E,\mathfrak I)\to (B,\mathfrak I')$ as a continuous function such that $p^{-1}(O)$ open in $E$ means $O$ open in $B$. Equivalently from Munkres "$U\subseteq \mathfrak I\iff p^{-1}(U) \in \mathfrak I'$.

53 mins ago, by Peter Tamaroff

@MarkDominus He then proves a little lemma which says that if $p:E\to B$ is an identification and $G:E\to Y$ is continuous, and $p(x)=p(x')\Rightarrow G(x)=G(x')$, we can set up a continuous function $g:B\to Y$ by $g(b)=G(x)$, with $x\in p^{-1}(\{b\})$.

@BenjaLim

mixedmath

4:15 AM

@BenjaLim In addition to seeing how simplicial complex is defined, I suddenly wonder also how you define a "face." Some authors say a face is a proper subset of the vertices, and others say any subset, so that a face might be the entire simplex

@BenjaLim I did not learn from Munkres, nor do I own it, but since I'm about to run away and you don't seem to be here, I'll say a few closing remarks. The advantage of having any subset of vertices be a simplex is that then every interior point of a simplex is in the interior of a unique face. And this is really useful in showing certain things about simplicial maps.

Anyhow - good night all

Peter Tamaroff

@mixedmath Night.

@BenjaLim Could you solve your problem?

BenjaLim

@mixedmath Sorry I was typing an answer on the main page

Peter Tamaroff

4:37 AM

@BenjaLim You there?

BenjaLim

4:47 AM

@PeterTamaroff Sorry I was busy with AT stuff.

Peter Tamaroff

@BenjaLim No problem.-

BenjaLim

Look at the main page :D

Peter Tamaroff

@BenjaLim I see your new question.

BenjaLim

@PeterTamaroff Simplicial complexes .....

Peter Tamaroff

@BenjaLim Yes?

@BenjaLim Is that what you're struggling with?=

BenjaLim

5:04 AM

yes

Peter Tamaroff

@BenjaLim Are you busy now?

@AlexBecker Hello.

Alex Becker

@PeterTamaroff Hey.

Peter Tamaroff

@Matt Hello

@BenjaLim Did I tell you I met Cedric Villani?

And Mariano, too.

BenjaLim

no

Peter Tamaroff

=)

Alex Becker

5:12 AM

@PeterTamaroff When did that happen?

Matt

Or why don't I call them Andre and Fred?

Hi Peter.

Peter Tamaroff

@AlexBecker On Friday.

Matt

This sucks so much. I missed t-bear last night.

Peter Tamaroff

@Matt I missed lunch with Cedric Villani!!!

Matt

@PeterTamaroff Never heard of, I'm sorry.

BenjaLim

5:14 AM

@PeterTamaroff I am very busy with AT.

Peter Tamaroff

@Matt Field Medallist. =P

BenjaLim

A lot of stuff is confusing as hell.

@PeterTamaroff Well I've met Paul Baum :D

@PeterTamaroff As in the paul baum in :en.wikipedia.org/wiki/Baum%E2%80%93Connes_conjecture

Peter Tamaroff

@BenjaLim OK. Good luck.

Alex Becker

@PeterTamaroff Pshaw. We've got two of those in the department :P

Matt

@PeterTamaroff I'd rather have lunch with t-bear than with a Field's medalist.

Peter Tamaroff

5:16 AM

@AlexBecker Hahaha well, I'm a newbie in the math world.

@Matt I'm sure you would.

Fortuon Paendrag

@PeterTamaroff And I met Endre Szemerédi the Abel prize winner of this year.

Matt

@PeterTamaroff : )

Peter Tamaroff

@Matt I also missed lunch with Mariano. =/

@FortuonPaendrag A recent recipient!

@AlexBecker Wanna talk a little about Quotient Topologies?

Alex Becker

@PeterTamaroff Sure!

BenjaLim

@PeterTamaroff I wish I could spend time talking to you. But I really have to finish this AT stuff.

Matt

5:19 AM

@PeterTamaroff Well well. : )

Alex Becker

I really should be working on my paper, but a distraction might be good too.

Peter Tamaroff

@AlexBecker Well, I read about them, I think yesterday,

I ended up with a really badass diagram.

But that's not the point.

The whole point of the chapter seemed to start with a function $f:X\to Y$ and define two functions $\pi_f:X\to X/\sim_f$ and $f^*:X/\sim_f\to Y$ such that $f=f^* \pi_f$ be continuous.

Here $x\sim_f x' \iff f(x)=f(x')$

Matt

@PeterTamaroff Can you put one vote on this question? I voted on all his questions already. It's for numerology play.

Peter Tamaroff

@Matt I happened to have that upvoted already ;)

Matt

@PeterTamaroff Oooh, : ) Too bad. What about any of his other questions?

He'll be 40, you know : )

Peter Tamaroff

5:23 AM

@Matt He'll turn 40?

Matt

@PeterTamaroff Yes, 40k : )

Peter Tamaroff

@Matt Dunnnnnn

Matt

What? Can you undo?

It's one too many now.

Peter Tamaroff

@Matt Oh, you want $40k$ exactly?

Matt

Yes! Of course : )

Peter Tamaroff

5:25 AM

Well, upvote him once. Downvote him 3 times =P

Matt

I don't do downvoting.

Peter Tamaroff

Accept one of his questions?

@AlexBecker

Matt

Why can you not just +1 one of the questions rather than two?

Peter Tamaroff

3 mins ago, by Peter Tamaroff

The whole point of the chapter seemed to start with a function $f:X\to Y$ and define two functions $\pi_f:X\to X/\sim_f$ and $f^*:X/\sim_f\to Y$ such that $f=f^* \pi_f$ be continuous.

@Matt Oh, duh!

Matt

With +5 it's 40k.

@PeterTamaroff W0000t!

You're awesome : )

Peter Tamaroff

5:27 AM

@Matt Hahaha, done.

Matt

Let's hope it'll last for a while.

Peter Tamaroff

@AlexBecker

Alex Becker

@PeterTamaroff What's your question?

Peter Tamaroff

@AlexBecker OK, sorry. Let me rollback.

Matt

3

Peter Tamaroff

5:28 AM

So, what we do is take a function $f:(X,\mathfrak I) \to (Y,\mathscr S)$. Any function.

Matt

Ok, my duty here is done. See you later! : )

Fortuon Paendrag

We can keep tweaking it all night, @Matt :).

Alex Becker

@Matt Bye

Fortuon Paendrag

Bye!

Peter Tamaroff

Then define $\sim_f\subset X^2$ by $x\sim_f x'\iff f(x)=f(x')$

Then we define $\pi_f:X\to X/\sim_f$ by $x\mapsto \hat x$, where $\hat x$ is the corresponding equivalence class.

Alex Becker

5:30 AM

Yes, this is fairly standard.

Peter Tamaroff

@AlexBecker Wait, wait, give me a while.

Alex Becker

Okay :)

Peter Tamaroff

Now, we give $X/\sim_f$ the identification topology determined by $\pi_f$.

Then $\pi_f$ is an identification, and $\pi_f(x)=\pi_f(x')\iff f(x)=f(x')$

J. M.

I'm leaning towards making this an faq. Opinions?

Peter Tamaroff

Now by a previous lemma, $f$ "induces" a continuous function $f^*:X/\sim_f \to Y$ such that $f=f^*\pi_f$

Alex Becker

5:33 AM

@J.M. I think it's a good idea.

Peter Tamaroff

The lemma is that if $p:E\to B$ is an identification and $G:E\to Y$ is continuous, and $p(x)=p(x')\Rightarrow G(x)=G(x')$, we can define a continuous $g:B\to Y$ by $g(b)=G(x)$ whenever $x\in p^{-1}(\{b\})$.

We say that $G$ induces the continuous function $g$.

Now, this is the last part.

Fortuon Paendrag

@J.M. Thirded !

J. M.

@AlexBecker I was trying to pick between that and this, but the other one is the only thing that has Faulhaber's formula written explicitly...

Peter Tamaroff

$f^*$ is one one (follows almost immediately by the more explicit defintion of $f^*$ which could be "$f^*(\hat x)=f(\alpha)$ for any $\alpha \in \hat x$."

Now:

@AlexBecker Hope I'm not boring you.

Alex Becker

@J.M. I think your choice is correct.

@PeterTamaroff You're fine. I'm working while you type.

Peter Tamaroff

5:38 AM

Since $f^*$ is continuous, we have that $f^*{}^{-1}(\mathscr S)\subset \Bbb I$, where $\mathscr S$ is the topology of $Y$ and $\mathbb I$ is the identification topology on $X/\sim_f$.

Or equivalently, since $f^*$ is one-one, we have $\mathscr S \subset f^*(\mathbb I)$

Then the author says: "Now, if $\mathscr S'$ were another topology on $Y$ such that $f$ were continuous, we would again have $\mathscr S'\subset f^*(\mathbb I)$ so that $\mathbb I$ is the weakest topology carried over to $Y$ by $f^*$ such that $f$ is continuous."

@AlexBecker

Alex Becker

Is this confusing you?

Peter Tamaroff

@AlexBecker I don't know where he wants to get to.

It seems the idea of all this is supplying with a technique to obtain from any function $f:X\to Y$ a continuous function $f:(X,\mathfrak I)\to (Y,\mathscr S)$.

Alex Becker

@PeterTamaroff I'm not sure what you mean.

@PeterTamaroff Ah. I think he's trying to motivate the precise definition of the quotient topology.

Peter Tamaroff

@AlexBecker Yes, something like that.

Alex Becker

What he's said there essentially boils down to saying the quotient topology is the weakest topology such that you can always obtain $f:(X,\mathfrak I)\to (Y,\mathscr S)$.

Essentially, the construction of the quotient space is categorically "correct".

Peter Tamaroff

5:47 AM

@AlexBecker He gives this construction and then he gives three example: attaching a space to another, shrinking a subset to a point, and covering the circle with the real line.

@AlexBecker What is the motivation behind the identification topology?

Alex Becker

Think of it in terms of the function $f$.

Let's say $f$ is equal to $0$ everywhere.

Peter Tamaroff

@AlexBecker OK

Alex Becker

This is a perfectly valid function on any space, but you don't need the whole space really.

Thinking of it as a function from $(\mathbb R^5\coprod S^{87})\times \mathbb C^\times$, for example, is retaining a lot of extra information you don't need.

Peter Tamaroff

@AlexBecker LOL; what? This is an Introductory Topology book.

Alex Becker

In order to minimize the amount of information you have to keep track of, you want to treat $f$ as a function from the "smallest/simplest" possible space.

Peter Tamaroff

5:52 AM

@AlexBecker OK.

Alex Becker

So in this case, we can treat $f$ as essentially a function defined on a single point. This is a whole lot simpler than $(\mathbb R^5\coprod S^{87})\times \mathbb C^\times$.

Peter Tamaroff

@AlexBecker But what is that thing!?!??

Alex Becker

All that stuff about defining $f^*$, etc. was to show that $f$ is essentially defined on the quotient space $X/\sim_f$.

Peter Tamaroff

@AlexBecker OK.

Alex Becker

@PeterTamaroff Something hopelessly convoluted. Don't worry about it. It was just an example of a space you don't want to work with.

Peter Tamaroff

5:55 AM

@AlexBecker Yes, I got that part. We identify all the $f(x)=f(x')$ with a single point. When I sa "identify" I mean it in the most mundane term, like "put them all in the same bag".

Alex Becker

@PeterTamaroff That's a good way to think about it.

Peter Tamaroff

@AlexBecker The good thing is that $f^{-1}$ will still behave like the initial $f^{-1}$, because of how we set things up.

Alex Becker

Indeed.

Peter Tamaroff

So that this new $f$ will be in some sense the most similar continuous function to the original $f$

That's I think, the point he's making with $\mathbb I$ being the weakest topology.

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