Mathematics - 2012-04-10 (page 3 of 5)

Benjamin Lim

12:00

@BrianMScott is what I said not correct?

Brian M. Scott

I don’t see its relevance.

Matt N.

@BrianMScott But I won't do it again.

Benjamin Lim

@BrianMScott to prove continuity it suffices to prove the preimage of every basis element is open

Brian M. Scott

Oy! Read Asaf’s comment on Allen’s latest!

Benjamin Lim

so proving that $\pi$ is an open map is equivalent to proving that it's inverse is a continuous function

Brian M. Scott

12:02

@BenjaminLim Continuity is a given: the product topology is defined to be the coarsest topology that makes the projection maps continuous.

Daniil

Hi

Brian M. Scott

@BenjaminLim It doesn’t have an inverse: it’s not 1-1.

Benjamin Lim

@BrianMScott I am talking about the specific case above

Brian M. Scott

@Daniil Morning!

Benjamin Lim

of the projection map $\pi : X \times y_0 \rightarrow X$

Brian M. Scott

12:03

@BenjaminLim Why bother to prove that its restriction to $X\times \{y_0\}$ is open when you can prove that the map itself is open?

Benjamin Lim

because that's all we need in that case here no?

Because david and I were talking about the tube lemma

just needed to see that $X \times y_0 \cong X$

anyway you're right we should do it in full generality

why build a sidewinder missile when you can build an intercontinental ballistic missile

Brian M. Scott

@BenjaminLim Okay. Yes, that’s true, and the restriction of the projection is a homeomorphism.

Kannappan Sampath

Does page 193-194 happen to show up for someone?

Benjamin Lim

@BrianMScott I concur that we should prove it is an open map in full generality

Brian M. Scott

@KannappanSampath I don’t get 192, but I do get 193-4.

Benjamin Lim

12:10

@BrianMScott Right this is dead easy. Suppose $U$ is open in $X \times Y$

take $x \in \pi[U]$

Then there is a $y\in Y$ such that the pair $(x,y) \in U$

Now by definition of the basis for the product topology, $U$ open means that there is a basis element $B \times W$ such that $(x,y) \in B \times W \subset X \times Y$

Matt N.

Look! Didier was wrong! Hail to the mean king!

Brian M. Scott

@Matt: You’re making me dizzy!

Benjamin Lim

It follows by taking projections that $x \in B \subset \pi[U]$

Matt N.

Although I'm quite sure it won't have any effect on his general arrogance.

Kannappan Sampath

@BrianMScott Oh, I have 192 but not 193-194! But, I have no clue how you'll be able give me those two pages. Do you know of some way? (Ymar did some thing like screenshots but I am not sure it works without doing some trick!)

Matt N.

12:11

@BrianMScott Sorry. Will stop now.

Benjamin Lim

It follows that since $x$ is arbitrary, every point of $\pi[U]$ is an interior point so $U$ is open

@BrianMScott that's correct right?

Brian M. Scott

@MattN Not a problem $-$ just startling.

@BenjaminLim Yes.

Matt N.

@robsquare banzai!

Benjamin Lim

let me get back to tube lemma

Brian M. Scott

@KannappanSampath Let me see whether I can figure out how to get a screen capture.

Kannappan Sampath

12:14

@BrianMScott I'll be grateful if you get that for me. Thank you.

David Wheeler

are we back to the tube lemma yet?

Benjamin Lim

@DavidWheeler yes

since the product topology has a basis

it suffices to consider a covering of $X \times y_0$ by open sets

and by compactness there are a finite number of basis elements $U_1 \times V_1, \ldots U_n \times V_n$ that cover it

The claim now is that $W = U_1 \cap...\cap U_n$

David Wheeler

ok, now just take the finite neighborhoods of $y_0$ in the covering, and intersect them all

Benjamin Lim

this is open

David Wheeler

that is $ V = V_1 \cap \dots \cap V_n$

Benjamin Lim

12:19

crap

I got stuff the other way round

reboot:

David Wheeler

that's a neighborhood of $y_0$, right?

it doesn't matter, Ben, just pick a convention and stick with it

Benjamin Lim

yeah not $U_1 \cap ...\cap U_n$

it should be the $V's$

stick with the V's

David Wheeler

V is a neighborhood of $y_0$ we use to make our "tube"

and notice it is entirely within the open set we have covering $X \times {y_0}$

Benjamin Lim

yes

David Wheeler

and that's essentially the tube lemma part...so now we use that the show the projection is closed

Benjamin Lim

12:23

wait wait

let me digress everything slowly

Brian M. Scott

@KannappanSampath It’ll be a rough and ready job, but I should have it in a few minutes.

David Wheeler

well, you see how we got the tube, right?

Benjamin Lim

slow

Kannappan Sampath

@BrianMScott Sure thank you. Take your time.

Benjamin Lim

wait

@DavidWheeler Let me get this right

David Wheeler

12:25

the tube lemma says: if we have an open set in the product containing a "slice" there is an open subset of it containing a "tube"

Benjamin Lim

because $N$ is an open set containing $X \times y_0$

David Wheeler

right N contains a slice

Benjamin Lim

why can we choose $U_i \times V_i$ that contain $X \times y_0$ and at the same time are completely contained in $N$?

David Wheeler

we haven't used this yet, and it's been kind of a detour, but we'll get there

because Y is compact

Benjamin Lim

you mean $X$

David Wheeler

12:27

Ben,if we are going to use the slice lemma for a slice by $y_0$ we use the compactness of Y

if we are assuming that $X$ is compact, we need to consider ${x_0} \times Y$

like i said before, pick a convention and stick with it

Benjamin Lim

No I think you've got it the other way round

Nimza

Good evening!

Brian M. Scott

@KannappanSampath Look here; there are four files, each half a page. Let me know when you have them, and I’ll delete the folder.

David Wheeler

ok, let's start over...which set do you want, now and forever, to consider compact?

Benjamin Lim

$X$

so we need to consider the slice $X \times y_0$

you see

Nimza

12:30

$\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is called Beta-function $B(x,y)$. Is there some name for function, when we use $x_1,...,x_n$ instead of $x,y$?

Benjamin Lim

we want the fact that $X \times y_0 \cong X$ that is compact

Kannappan Sampath

@BrianMScott Thank you. Downloading...Please wait.

Benjamin Lim

so that $X \times y_0$ is compact too

@DavidWheeler see?

David Wheeler

ok, then we want to create the $U_i \times V_i$ that cover $x_0 \times Y$

Benjamin Lim

for...?

I don't understand the switch from $X \times y_0$ to $x_0 \times Y$

David Wheeler

12:31

now i'm getting confused...ok X is compact.

Benjamin Lim

yes

Kannappan Sampath

@BrianMScott Thank you done now. :)

Benjamin Lim

we want to look at $X \times y_0$

Brian M. Scott

@KannappanSampath They came through okay?

t.b.

> Nowadays, inner-product spaces and Hilbert spaces have their place in undergraduate courses, where the principal difficulty that occurs is teaching the correct pronunciation of Cauchy and the correct spelling of Schwarz. (Garling)

Benjamin Lim

12:32

there is no point looking at $x_0 \times Y$ because how can we deduce this is compact?

David Wheeler

and we have a set N containing $X \times y_0$

Benjamin Lim

yes

David Wheeler

and we found a finite subcover of $X \times y_0$ contained in N

Kannappan Sampath

@BrianMScott Sure they did. They are also useful in that: a proposition for a file. I must thank you for being patient and doing this for mu sake. Thank you so much.

Brian M. Scott

@KannappanSampath No problem: I’m glad to find out how to do it!

Benjamin Lim

12:33

@DavidWheeler LIke I said how can we find a subcover such that each element of the subcover is contained in $N$?

@DavidWheeler Like this:

David Wheeler

N is open, so it is a neighborhood of each of its points

Benjamin Lim

for each $x \in X$ consider the point $(x,y_0)$ in $N$

WhatsInAName

Hello

Benjamin Lim

$N$ open means that there is a basis element say $U \times V$ about it completely contained in $N$

now take the union over all $x \in X$

WhatsInAName

is there an another way to write (a mod (x*y))?

David Wheeler

12:35

so for each point $(x,y_0) \in X \times y_0$ there is a basis element contained in $N$ containing $(x,y_0)$

Brian M. Scott

@WhatsInAName Well, I’d just write $a\bmod xy$.

Benjamin Lim

@DavidWheeler Like this we get a subcover of $(X \times y_0$ such that each element of the subcover is completely contained in $N$

David Wheeler

right, so now we have a (possibly infinite) open cover of $X \times y_0$

Benjamin Lim

done.

Brian M. Scott

@DavidWheeler Gesundheit!

Benjamin Lim

12:36

@DavidWheeler But we know that $X \times y_0 \cong X$ so there is a finite subcover

and by our construction above each element of the subcover is completely contained in $N$

WhatsInAName

I can't write a mod x*y because x*y is too large to fit in a data type

David Wheeler

so we have a finite open subcover $U_1 \times V_1 \cup \dots U_n \times V_n$

Benjamin Lim

yes

David Wheeler

so take V to be the intersection of all the $V_i$

Benjamin Lim

yes

@DavidWheeler sorry for doing a jordan before

David Wheeler

12:38

then $X \times V$ is our tube

Benjamin Lim

yes

David Wheeler

ok, now the fun part

let's pick a closed set in $X \times Y$

Benjamin Lim

don't spoil all the fun just yet!!!!

David Wheeler

i will call this set S, for the sheer joy of it

Brian M. Scott

@WhatsInAName I’m not clear: is this a question about notation, or a question about how to express the number $a\bmod xy$ in terms of $a,x$, and $y$ in a way that doesn’t require using excessively large numbers?

David Wheeler

12:41

since S is closed $(X \times Y)\setminus S$ is open

Benjamin Lim

@DavidWheeler Let me figure this out on my own

WhatsInAName

yes

one that doesn't use large numbers

a way to break up a mod xy

David Wheeler

pick an element y in $Y \setminus \pi(S)$

Brian M. Scott

$a\bmod xy$ can be uniquely reconstructed from $a\bmod x$ and $a\bmod y$, though the details are a little messy.

For that you want the Chinese remainder theorem, which has a decent article in Wikipedia.

Benjamin Lim

@DavidWheeler Let me figure this out on my own please

David Wheeler

12:44

@BenjaminLim that's all the terms i was going to define....knock yourself out

Benjamin Lim

thanks

now I just need to follow my nose

David Wheeler

@BenjaminLim if you get stuck, i'll give you a hint

t.b.

@Nimza I think you're looking for the multinomial Beta function

Brian M. Scott

It’s 0845 here, and I’m starting to fade; I’m off to bed. I’ll see you folks later.

Kannappan Sampath

@BrianMScott See you, Brian. Bye.

WhatsInAName

12:48

This is part of the Chinese Remainder Theorem

that's why this is so annoying

the CRT itself is overflowing

the problem is that my modulus is a huge semiprime

int64_t modPow(int64_t a, int64_t x) {
    int64_t res = 1;
    while(x > 0) {
        if( x % 2 != 0) {
            res = ModM(res * a);
        }
        a = ModM(a*a);  //right here

        x /= 2;
    }
    return res;
}

the problem is that a*a overflows

David Wheeler

@BenjaminLim how are you coming along?

Benjamin Lim

wait

it's getting there

i'll tell you where i'm up to

by the tube lemma there is an open neighbourhood $V$ about $y$ such that $S \subset (X \times V)^{c}$

David Wheeler

i would actually look at the pre-image (under $\pi$) of y instead, and then use the tube lemma

do you see why?

Benjamin Lim

wait

well because then there exists an $x \in X$ such that $\pi^{-1}(y) = (x,y) \notin S$

David Wheeler

13:04

what is $\pi^{-1}(y)$?

hint: it's not a point

isn't it $X \times y$?

Benjamin Lim

yeah

sorry

why did you say preimage under $\pi$ of $y$?

what did that mean anyway?

recall we are projecting onto $X$

@DavidWheeler

David Wheeler

all i meant was $\pi^{-1}(y)$, sorry if that was confusing

Benjamin Lim

how is that even defined?

$\pi: X \times Y \rightarrow X$

David Wheeler

if we are taking X to be our compact set, then the projection that is closed, is the projection onto Y

Benjamin Lim

shit

David Wheeler

13:12

that's why i wanted to be clear on which one we were doing

Benjamin Lim

this means the proof before on the graph of the function being compact is incorrect

David Wheeler

stay calm, don't freak out...just think about the set $X \times y$

Benjamin Lim

@DavidWheeler i'm not freaking out just yet :D

David Wheeler

ben....forget about that for a second

Benjamin Lim

hhahahahahhahahahaha

love you man, no homo :D

ok so now we deal with $X$ compact, $\pi : X \times Y \rightarrow Y$

David Wheeler

13:15

what is $(X \times \{y\}) \cap S$?

Benjamin Lim

not rendering the latex

David Wheeler

shoot..have an error

Benjamin Lim

not right

Kannappan Sampath

Wonderful!!!! I just checked out a combinatorics book. All Hail Erdos.

Benjamin Lim

$(X \times \{y\})$ is not open

David Wheeler

13:16

there....that's what i meant

Benjamin Lim

it's closed

David Wheeler

yes, it's closed, but which closed set is it?

how did we pick y?

Benjamin Lim

$ y\in Y\setminus \pi(S)$

David Wheeler

right

so y isn't an image of ANY element of S

Benjamin Lim

this forces $\pi(X \times \{y\}) =\emptyset$?

David Wheeler

13:20

that is: $X \times \{y\}$ and $S$ have no "Y-coordinates in common"

correct

Benjamin Lim

yes

so $\pi$ of that is the empty set

David Wheeler

therefore, $X \times y$ must lie entirely within the complement of S, right?

do you see that $(X \times \{y}) \cap S = \emptyset$?

Benjamin Lim

$y \in \pi(S^{c})$ that is open

yes

i see the above

@DavidWheeler I am going to have to go soon I am really tired

David Wheeler

now apply the tube lemma to $X \times y$

Benjamin Lim

Ok the tube lemma tells us that if we have an open set $N$ containing $X \times y$

there is a neighbourhood $V$ about $y$ such that $X \times y \subset X \times V \subset N$

David Wheeler

13:27

so there is a neighborhood $V$ of $y$ so that $S^c$ contains a tube $X \times V$

Benjamin Lim

and that $N$ is actually $S^c$

by the result above

David Wheeler

it's the tube we want

because what is the projection of this tube?

Benjamin Lim

it's just $V$

David Wheeler

right, so every point y of V contains a neighborhood of that y contained in $(\pi(S))^c$

Benjamin Lim

$\pi(S)^{c} = \pi(S^{c})$?

David Wheeler

13:31

which means $(\pi(S))^c$ is open

which means $\pi(S)$ is closed...done

Benjamin Lim

@DavidWheeler wait why is $\pi(S)^{c} = \pi(S^{c})$?

David Wheeler

i meant every point y of $Y\setminus\pi(S)$

in general they aren't equal

t.b.

Still proving that projections are closed?

Benjamin Lim

@tb yes

t.b.

oh, boy :)

Benjamin Lim

13:34

@tb doesn't matter, knowledge is all that matters

David Wheeler

the general structure is this: S is closed. take the complement. look at the complement of $\pi(S)$, and pick an arbitrary point in it.

the pre-image of that point is a slice in the complement of S

Benjamin Lim

@DavidWheeler I know that.

David Wheeler

the complement of S is open, so we can apply the tube lemma to get a tube in the complement

Benjamin Lim

The problem now is that we know that $X \times V \subset S^{c}$

so taking the projections of both sides

David Wheeler

project the tube, we get a neighborhood of y.

Benjamin Lim

13:36

we get that $y \in V \subset \pi(S^{c})$

perhaps $\pi(S^{c}) \subset \pi(S)^{c}$

that would do it

David Wheeler

do this for every y in $\pi(S)^c$

we then get a whole bunch of neighborhoods of y, union them all it's open

which means that the SET $\pi(S)^c$ is open

Benjamin Lim

@DavidWheeler You don't get me

David Wheeler

so $\pi(S)$ is closed

Benjamin Lim

all we have shown now is that $\pi(S^{c})$ is open

not $\pi(S)^{c}$

t.b.

one of my first answers here was precisely this lemma...

David Wheeler

13:39

when we project the tube to Y, we get a neighborhood of y

Benjamin Lim

@DavidWheeler you agree with me that $X \times V \subset S^c$?

so that taking the projections on both sides gives $V \subset \pi(S^{c})$?

@tb see the problem I'm talking about?

the problem david is that you only have shown that $\pi(S^{c})$ is open

t.b.

@BenjaminLim No, I haven't followed the discussion.

Benjamin Lim

that is not the same as showing that $\pi(S)^{c}$ is open

David Wheeler

forget about $\pi(S^c)$

Benjamin Lim

ok

t.b.

13:42

But maybe it would help if you recap from the beginning.

Benjamin Lim

Ok

t.b.

(I'm not saying you must, but it might help)

Benjamin Lim

So we want to show that the projection map $\pi : X \times Y \rightarrow Y$ for $X$ compact is a closed map

David Wheeler

you agree that for each y, the neighborhood $V_y$ we get is in $(\pi(S))^c$ right?

Benjamin Lim

@DavidWheeler No.

@tb Let $S \subset X \times Y$ be closed.

t.b.

13:43

ok

Benjamin Lim

Now let us take $y \in Y$ such that $y \in Y \setminus \pi(S)$

t.b.

why is there such a beast?

Benjamin Lim

Then david has argued that $X \times \{y\}$ is completely disjoined from $S$

That means that $X \times \{y\} \subsetneqq S^c$

t.b.

but fine. I'm with you

David Wheeler

i will say it again $y$ is IN $(\pi(S))^c$

Benjamin Lim

13:44

@DavidWheeler Ok

t.b.

Well, if $S$ contained a point $(x,y)$ then $y \in \pi(S)$, right?

Benjamin Lim

@tb So by the tube lemma, we have a neighbourhood $V$ about $y$ such that $X \times V \subset S^c$

t.b.

whatever the tube lemma is, yes.

Benjamin Lim

so therefore from here, taking projection on both sides gives that $V \subset \pi(S^{c})$

@DavidWheeler Do you get what I'm saying?

David Wheeler

yes i do, but i don't think you're seeing what i am saying

Benjamin Lim

13:46

I know that $y \in \pi(S)^{c}$

I just don't get how $V$ is completely contained in $\pi(S)^{c}$ at the end

from the tube lemma

t.b.

why do you apply a lemma? what does the tube lemma say, actually?

Benjamin Lim

@tb Let $X$ be compact, and suppose $y\in Y$

David Wheeler

@tb the tube lemma: let X,Y be topological spaces, with X compact.

Benjamin Lim

If $N$ is an open set containing the slice $X \times y$

then there is a neighbourhood $V$ about $y$ such that $N$ contains the entire set $X \times V$

David Wheeler

basically, if an open set contains a slice, it contains a tube

t.b.

13:49

yes, of course. But then you're done already.

(you find a tube disjoint from $S$, project it down then the image is disjoint from $\pi (S)$, open and contains $y$, hence $y$ is in the interior of the complement. Since $y$ was arbitrary $Y \smallsetminus \pi(S)$ is open.)

Benjamin Lim

Now that is 100% clear.

David Wheeler

right...there's no way anything from the complement of S can project down to the projection of S

Benjamin Lim

no doubts

David Wheeler

which is what i've been saying the set V is outside of $\pi(S)$

Benjamin Lim

@DavidWheeler ok now I get it

David Wheeler

13:53

and all those V's put together union to the complement, since we've covered every point and none of the neighborhoods are outside it

Benjamin Lim

pheww

now the problem about the graph of $f$

Let $B$ be a closed set in $Y$

David Wheeler

ok, now we'll have to change our X's and Y's

Benjamin Lim

Look at $X \times B$ that is closed in the product topology

David Wheeler

but there is an obvious homeomorphism between $X \times Y$ and $Y \times X$

Benjamin Lim

then $(X \times B) \cap f$ is a closed set

$f$ is the graph of the function

@DavidWheeler So you're saying that

David Wheeler

13:56

we just have to remember to project PARALLEL to the compact set

Benjamin Lim

so you mean like if $X$ is compact we project to $Y$

David Wheeler

right

Benjamin Lim

So like now

to use the result above

I would have to project $(X \times B) \cap f$ onto $Y$

what I actually want to do is project on to $X$

god dammit

t.b.

You're not going to use that projection lemma. You're going to use that the graph of $f$ is compact.

Transcript for

$Mathematics$ Mathematics