Mathematics - 2012-07-25 (page 1 of 2)

Suppose $x\in X\cap Y$ such that $h(x)=t\in T$. If $x\in X$ then $x\in f^{-1}(t)$; if $x\in Y$ then $x\in g^{-1}(t)$. It must be the case that $x$ is in at least one of $X$ or $Y$ (or both).

BenjaLim

@anon yes ?

anon

$$\begin{array}{c l} x\in h^{-1}(T) & \iff h(x)\in T \\ & \iff (x\in X\wedge f(x)\in T)\vee(x\in Y\wedge g(x)\in T) \\ & \iff (x\in f^{-1}(T))\vee(x\in g^{-1}(T)) \\ & \iff x\in f^{-1}(T)\cup g^{-1}(T) \end{array}$$

BenjaLim

that's so long

Peter Tamaroff

2:59 AM

@BenjaLim Depends on what you compare it to.

anon

your math?

Peter Tamaroff

@BenjaLim You always remove stuff before I can see it!

BenjaLim

@anon right so it is the case that $h^{-1}(V) = f^{-1}(V)\cup g^{-1}(V)$

and hence is open in $X \cup Y$

anon

woops I guess I meant to use the letter V instead of T

BenjaLim

nvm I got it

Peter Tamaroff

3:07 AM

Can there be sets neither open nor closed?

anon

yes

Peter Tamaroff

Dread. OK.

anon

eg (0,1] in Euclidean topology on R

Peter Tamaroff

Yes. OK.

CLarue

Are there spaces in which all sets are either open or closed?

Peter Tamaroff

3:09 AM

I'm proving $\operatorname{int}(A)$, $\partial A$ and $\operatorname{int}(X\setminus A)$ are mutually disjoint.

anon

obviously. the nontrivial version of the question is "what is the name or characterization of these spaces?"

Peter Tamaroff

@CLarue I think the discrete topology is an example of that.

BenjaLim

@CLarue Put the discrete topology on any set.

anon

wait, do you mean exlcusive or? if so, you're going to have to ignore the empty set and its complement

BenjaLim

@PeterTamaroff the two interiors being disjoint is clear.

Peter Tamaroff

3:11 AM

@BenjaLim I already proved that

I'm trying to prove $\operatorname{int} (A)$ and $\partial A$ are disjoint now.

anon

also, $\partial A= \partial (X\setminus A)$, so you only need to prove one of the other two disjointnesses

Peter Tamaroff

@anon Yes.

Well, I think I got it.

Reductio ad absurdum.

Suppose $x\in \partial A$ and $x\in \operatorname{int}A$

Then $x\in \overline A$ and $x\in \overline{X\setminus A}$

But $x\in \operatorname{int} A$ means $A$ is a nbhd of $x$.

Then $x\in O\subset A$ for some open set $O$.

But since $O\cap (X\setminus A)=\varnothing$

Drop the proof by contradiction.

Just use that $O\cap (X\setminus A)=\varnothing$

Mark Dominus

3:29 AM

It's funny that there is no compact notation à la "$\forall$" that means "for all but finitely many".

Peter Tamaroff

@MarkDominus We can invent it.

$\forall_{\not{\infty}}$

Mark Dominus

Yes, I was thinking of something like that.

anon

maybe $\exists n\in \Bbb N: \exists ! \{x_1,\cdots, x_n\}\subseteq X: \neg P(x_i),i=1,\cdots,n$

Mark Dominus

That is a very generous notion of "compact" that you have there.

BenjaLim

@anon Do you know what is the topology on $S^{\infty}$?

Peter Tamaroff

3:31 AM

@anon Never feed a dog with a tinfoil hat chocolate and put it in a microwave.

Also, never do that.

anon

@BenjaLim is it metrical?

Mark Dominus

And you don't need the ! after the $\exists$ anyway.

anon

@MarkDominus yes you do

BenjaLim

@anon the topology on $S^n$ is coming from the euclidean topology

@anon but S infinity Idk

Peter Tamaroff

@MarkDominus $!$ means "unique" or something of the sort.

@BenjaLim Have you tried thinking about $S^{\infty}$ as a subspace of a Hilbert Space?

BenjaLim

3:32 AM

I don't know what is a hilbert space

Peter Tamaroff

@BenjaLim It is not that hard of a definition:

Mark Dominus

If there is a finite sequence $x_1\ldots x_n$ for which $P(x_i)$ does not hold, then it does not matter if some of the $x_i$ are equal, since $P()$ still fails to hold for only a finite set.

anon

a vector space that is complete wrt an inner product I believe

Mark Dominus

So you don't need the !.

Peter Tamaroff

Let $H$ be the set of all sequences $\{x_n\}$ of real numbers such that $\sum_{n=0}^\infty x_n^2$ converges.

anon

3:33 AM

@MarkDominus if you don't have a $!$, then it doesn't bound the number of counterexamples to $P(\cdot)$...

BenjaLim

@anon inner product gives a norm that gives a metric that is complete, such is a hilbert space?

anon

yes

Peter Tamaroff

Define a metric on $H$ by $d(x,y)=\sqrt{\sum_{n=0}^\infty (x_n-y_n)^2}$

BenjaLim

@PeterTamaroff There is no issue of convergence: All but finitely many terms are zero.

Peter Tamaroff

Then $(H,d)$ is a metric space.

Mark Dominus

3:34 AM

It doesn't bound the counterexamples with or without the !. You need to claim that $\lnot P(x)\to x\in\{x_1\ldots x_n\}$.

Peter Tamaroff

I call it Hilbert space.

@BenjaLim You mean in $S^\infty$?

BenjaLim

yes.

anon

@MarkDominus the $!$ statement would be false if the number of counterexamples is unbounded, whereas (if we add the stipulation that $x_i=x_j\iff i=j$) it holds true if the number of counterexamples is finite. however yours is more compact.

Peter Tamaroff

@BenjaLim I was just defining the Hilbert space-

Mark Dominus

Dilbert space.

Peter Tamaroff

3:37 AM

You can then think about the topological space that arises from $H$

And consider $S^\infty$ as a subspace of both the metric and topological space $H$.

@MarkDominus What?

CLarue

$\{x\in S:\neg P(x)\}\in[S]^{<\omega}$

Peter Tamaroff

I don't like to go around tale taling people, but this user seems to be providing wrong answers everywhere.

@CLarue What is that¿

anon

I noticed that. There's also someone whose name starts with "lab" that's similar but not as egregious.

Dylan Moreland

He also rolled back an edit that made it look better, that's odd.

Mark Dominus

Yeah, his answers are quite awful. Happily, the site takes care of that automatically.

Peter Tamaroff

3:41 AM

It's like his answers are algorithmical or something.

CLarue

@PeterTamaroff the notation on the right stands for the set of all sets in $S$ with cardinality less than the exponent.

Peter Tamaroff

@CLarue And the left?

Dylan Moreland

He seems to be one of those folks that writes answers that are so long and without any sort of organizing comments that I'm too lazy to read any of it.

But setting $y'$ equal to a constant I can catch, I guess.

Peter Tamaroff

PGCIM (Post Graduate Certificate in Industrial Management) from XISS, Ranchi with an aggregate of 70 %.

Bachelors in Electrical Engineering from R.I.T (now N.I.T) Jamshedpur with an aggregate of 68 %.

Class XII from VIDYA BHARATI CHINMAYA VIDYALAYA (CBSE), Jamshedpur with an aggregate of 67 %.

Class X from HILL TOP SCHOOL (ICSE), Jamshedpur with an aggregate of 69 %.

@DylanMoreland =D

CLarue

@PeterTamaroff The set of all elements $x$ of $S$ for which the proposition $P(x)$ fails.

Mark Dominus

3:42 AM

I would like to know who is voting them up. The answer of his that I first noticed as being insane now has +2/0.

Peter Tamaroff

@MarkDominus Link?

What is all that aggregate stuff?

Mark Dominus

This: math.stackexchange.com/a/174196/25554

Peter Tamaroff

He had 71 rep a while ago.

Now he has 53

Now 51

Mark Dominus

"$17=2^3+3^2$
the sum of the exponents of 2 and 3 is (3+2)
i.e. the sum of the exponents is 5"

BenjaLim

@PeterTamaroff His posts are rubbish

Peter Tamaroff

3:45 AM

@BenjaLim I side with you on that.

Mark Dominus

The bottom half of that answer is reasonable, but the top half is rubbish.

Peter Tamaroff

So are you thinking about $S^\infty$?

anon

33 for Generic's answer? wtf!?

Peter Tamaroff

I didn't upvote I swear!

Mark Dominus

I upvoted it because it was simple and to the point, but it was nowhere near 33 when i did that.

Peter Tamaroff

3:47 AM

Everytime I feel like giving up on an exercise I put this full screen.

She's like staring right at your soul.

anon

I don't think I've seen the full, nonshopped version before

Mark Dominus

Have you seen the "after" picture?

Here

Peter Tamaroff

@MarkDominus I hate you.

Mark Dominus

Notice also that she is now wearing a niqab.

ngm.nationalgeographic.com/2002/04/afghan-girl/index-text

Peter Tamaroff

@MarkDominus I'll choose bliss and ignorance.

@BenjaLim

I need some hint in proving $\partial A \cap A^o=\varnothing$

BenjaLim

3:55 AM

yes

Peter Tamaroff

I wrote something up there

Basically

BenjaLim

ys?

Peter Tamaroff

If $x\in A^o$ then $x\in O\subset A$ with $O$ an open set.

Then $O\cap (X\setminus A)=\varnothing$

BenjaLim

well yes that has to be true

@PeterTamaroff $\partial A = \bar{A} \cap \overline{X - A}$ YES?

Peter Tamaroff

I prefer $X\setminus A$

But sure.

anon

3:58 AM

$X\setminus(\overline{X\setminus S})~\cap~\overline{S}\cap \overline{X\setminus S}$. Note that the first intersects $\overline{X\setminus S}$ trivially.

Peter Tamaroff

@anon Should there be an $=$ there?

anon

you can put an = sign on the LHS of everything if you want

BenjaLim

@PeterTamaroff The problem has an easy interpretation

Peter Tamaroff

@anon I don't understand what you wrote there.

What is $S$?

anon

@PeterTamaroff I mean $A$, whatever. It's the intersection of boundary and interior.

Peter Tamaroff

4:00 AM

@BenjaLim No no

I'm with topological spaces now.

BenjaLim

Doesn't matter just replace ball with open set

Peter Tamaroff

@anon Oh, sorry-

BenjaLim

Every open set about $x \in \partial A$ will contain some point in $A$ or the complement yes?

Peter Tamaroff

Ja.

BenjaLim

Now if for all open sets $U$ about $x$, $U \cap (X - A) \neq \emptyset$ we have a contradiction yes?

Peter Tamaroff

4:03 AM

$x\in U$ and $U\cap (X\setminus A)\neq \varnothing$

BenjaLim

@PeterTamaroff But what does it mean for $x \in \overline{X - A}$?

Every open set $U$ about $x$ will intersect $X- A$ yes?

Peter Tamaroff

@BenjaLim Yes.

BenjaLim

In other words there will not exist an open set $U$ about $x$ that is completely contained in $A$ yes?

Peter Tamaroff

Every nbhd of $x$ will intersect it.

BenjaLim

In other words $x$ cannot be in $A^{\circ}$ yes?

Peter Tamaroff

4:04 AM

@BenjaLim Ja.

BenjaLim

@PeterTamaroff Hence $\partial A \cap A^{\circ} = \emptyset$

@PeterTamaroff helped?

Peter Tamaroff

@BenjaLim Da.

BenjaLim

@PeterTamaroff so the problem was easy yes?

@PeterTamaroff Now here's another problem:

anon

says the one with the inverse image question

BenjaLim

We define the closure of $A$ to be the smallest closed set containing $A$

Peter Tamaroff

4:06 AM

@anon HAHAHHHA mean!

@BenjaLim I know what the closure is!

BenjaLim

@PeterTamaroff Prove that $x$ is in the closure iff every $U$ open about $x$ intersects $A$

@anon OK OK.

anon

:)

Antonio Vargas

Ohhhhh man. I'm tired.

Peter Tamaroff

@anon I wonder if you can tell us your name. Or the initial at least.

@AntonioVargas I want a pony.

Antonio Vargas

@PeterTamaroff Tired Santa says "Maybe next year".

BenjaLim

4:07 AM

bye guys

bye!!

anon

nah. I saw DKR recently.

later Benja

Antonio Vargas

@BenjaLim cya

Peter Tamaroff

@BenjaLim Bye Benjamín!

anon

'jamin

Peter Tamaroff

@anon OH MY GOD.

@BenjaLim

You do realize you name contains the word "Jammin" in it?

How awesome is that ?

@AntonioVargas Angry infant kidnaps Rudolph.

Antonio Vargas

4:09 AM

@PeterTamaroff The elves go on a manhunt.

You may think the North Pole is some kind of wonderland, but they slaughter 3 or 4 children a year trying to sneak into the compound.

Peter Tamaroff

@AntonioVargas Stupid elves are stupid. Angry infant lives in Congo. Elves die of malarya and skin cancer.

Remaining group returns to North Pole.

Antonio Vargas

@PeterTamaroff First Blood & Diapers

The elves build a funeral pyre for their fallen comrades, paint all of the toys black.

this is madness.

Peter Tamaroff

@AntonioVargas Elves worship Mick Jagger.

@AntonioVargas No. This. Is.

anon

c-c-c-ombo breaker

Peter Tamaroff

come at me bro

I'm leaving

Byes

Antonio Vargas

4:23 AM

@PeterTamaroff later

Well I guess I'm out too. Cya @anon

robjohn

5:48 AM

@anon: when I read this question, I was thinking of Arturo's method. I guess I need to do more with symmetric polys. I did post a proof of Newton-Girard last year, but that is about it.

anon

I also thought of Arturo's when I read it, but decided to go the more general route

robjohn

@anon I always appreciate different methods. That's why I upvoted both.

user19161

6:19 AM

Hey @rob! How's the trip?

robjohn

@JasperLoy It went very well. We got into town about 6:30, unpacked, and picked up dinner.

user19161

@PeterTamaroff You can always tell yourself that the nice picture is another, imaginary person. So that it is not the original person. So there is bliss and knowledge, not bliss and ignorance. That's why when I put up a pic of myself, I always use the real thing, no photoshopping.

user19161

6:33 AM

@BenjaLim Of course, that definition only makes sense because it refers to the intersection of all closed sets containing A which is again closed.

Jonas Teuwen

7:50 AM

Blub.

Old John

blub?

anon

inblublibly.

Jonas Teuwen

Yep. @OldJohn Just saw your e-mail :-).

Old John

@JonasTeuwen OK

Jonas Teuwen

@OldJohn Yes, same Federer. I will reply soon.

Old John

7:56 AM

@JonasTeuwen OK

user19161

@JonasTeuwen Herbert Federer, my idol?

Jonas Teuwen

Not sure? I mean the writer of phone books.

user19161

So did you sleep well last night @jonas?

David Wheeler

@OldJohn is your gravatar a newton's method fractal?

Jonas Teuwen

@JasperLoy No. Like 4h.

Old John

8:00 AM

@DavidWheeler Yep - the three colours represent which of the 3 roots of $z^3=1$ a given starting value converges to (I think)

user19161

@JonasTeuwen Phone books! Then yes!

Old John

@JasperLoy A bit like Doob in that respect?

Jonas Teuwen

Excellent.

Unwantedly I've got into a discussion about non-commutativity in quantum mechanics with a professor. And he is much better in it... apparently 8-).

user19161

@OldJohn Maybe. I saw Doob's measure theory book once. But Federer is quite unbeatable in conciseness! Well, maybe I can beat him, but we'll have to wait another decade or so for any of my books to be published...

Jonas Teuwen

@JasperLoy You want to write a book nobody reads? 8-).

Old John

8:04 AM

@JasperLoy I was thinking of Doob's book on Potential theory

user19161

@OldJohn Oh I have not seen that.

Old John

@JasperLoy 846 pages ...

user19161

@JonasTeuwen Hmm, they might become the new Bourbaki! Just wait and see...

Jonas Teuwen

@JasperLoy So five books nobody reads?

user19161

@JonasTeuwen Five? I will write three according to my current plans. And Bourbaki has more books than that actually. Actually, it depends on how you count the Bourbaki ones.

Jonas Teuwen

8:06 AM

Yeah, I know...

David Wheeler

@JasperLoy if you write better books than Bourbaki, i'll buy first editions :P

user19161

Speaking of Bourbaki, I am really not sure how many books there are now. There are all kinds of info on the internet. In fact I remember I was quite confused when I checked out Springer website itself.

user19161

And you know what? Amazon seems to have more updated info than Springer sometimes. For example, Lee's Smooth Manifolds second edition was announced for pre-order first on Amazon and then on Springer.

Old John

Speaking of Amazon, I recently saw a maths paperback available for £82 (new) or £91 (used) - seemed a bit daft :)

Jonas Teuwen

8:27 AM

Mm. A good day :-).

Old John

9:12 AM

Good day here too - 2 new books arrived by post :)

Jonas Teuwen

Cool, which ones?

Two more things that look nice on the book shelf?

Old John

Gouvea (p-adic numbers) and Cassels (Geometry of numbers)

@JonasTeuwen Might have to think of getting a bigger shelf :)

Jonas Teuwen

@OldJohn Hmm, I put many of them in my office, I have two big book shelves :-).

Old John

@JonasTeuwen One day, you will need more

Jonas Teuwen

@OldJohn But Erdos had very little books... eh?

Old John

9:18 AM

@JonasTeuwen but a very large brain

Old John

10:20 AM

good grief - while I have been online this morning, there have been 6 questions from the same person ...

Zhen Lin

oh?

robjohn

I don't see notifications that there are new/changed questions on the main site, nor new comments or changes to questions or answers. I've only noticed this since I restarted my computer after our trip. Does anyone else see this?

Old John

I still see notifications the same as always

Zhen Lin

The notifications are unreliable for me.

Jonas Teuwen

@OldJohn Do you know Richard Gill? He was in Cambridge around the time you were there, perhaps a few years earlier.

Old John

10:30 AM

@JonasTeuwen No - I never came acroos him, I don't think

Jonas Teuwen

@OldJohn en.wikipedia.org/wiki/Richard_D._Gill

user19161

@OldJohn And also silk underwear.

Old John

@JonasTeuwen So - he was there at the same time I was. But due to the college system there, I met very few others apart from the 10 students at my college

@JasperLoy ??

Jonas Teuwen

He is pretty cool and vocal.

user19161

@OldJohn Well, he had a skin disease so he has to wear silk underwear and not other kinds.

Old John

10:33 AM

I was a pretty hopeless student in my undergraduate days :(

@JasperLoy Ah OK!

Jonas Teuwen

@OldJohn But you were at Cambridge, so hopeless?

Old John

@JonasTeuwen I might have had some small ability, but I was distracted by many other things going on in my life at the time, and came out with a poor degree - hence the interest in doing a PhD later in life

Jonas Teuwen

@OldJohn But it ended up quite okay ;-).

Old John

@JonasTeuwen Yes - I am happy with the eventual outcome :)

Jonas Teuwen

In an interview Gill stated he didn't really wanted to do a PhD but that it ended up that way.

user19161

10:37 AM

@JonasTeuwen Indeed. Life takes mysterious turns.

Old John

@JonasTeuwen curious

Jonas Teuwen

Quite a peculiar guy.

Zhen Lin

I've heard claims that, in today's job climate, people take Ph.D.s to avoid having to fight for jobs...

Jonas Teuwen

Hmm... so they have to fight harder 4 years later? 8-). Postpone fighting! Get Permanent head Damage!

Old John

@JonasTeuwen Many mathematicians are peculiar (but pure ones more so than statisticians, in my experience)

user19161

10:39 AM

@OldJohn And many of them are nutcases.

Jonas Teuwen

@OldJohn He is quite... mixed in pure/applied. Very pure statistics, so to speak.

Old John

@JasperLoy Yes - but most of them don't get into the real world much :)

user19161

@JonasTeuwen Statistics can be very theoretical, not that I know any.

user19161

@OldJohn Because mathematics is more real than $R^3$.

Old John

@JasperLoy Maybe. Speaking of getting out more, I must get outdoors sometime today

user19161

10:42 AM

@OldJohn Yes, walking is the best exercise for the body, according to an ad I saw ten years ago with the model Linda Evangelista.

Old John

@JasperLoy Agreed - I went hill walking yesterday and really enjoyed the exercise

Jonas Teuwen

It is very warm outside.

user19161

@JonasTeuwen It's worse here bro!

Jonas Teuwen

Possible. It is very humid here.

user19161

Hello @matt!

Jonas Teuwen

10:44 AM

About 30 deg C. That isn't so bad but the bloody humidity...

Matt N.

I don't want to take any oral exams :,(

user19161

@JonasTeuwen Yes, it's the humidity you feel. After all, body temperature is 37.

user19161

@MattN. Well, try to get it over and done with.

Old John

Right - time to return to reality for some exercise - back later, folks

Jonas Teuwen

@MattN. You will do fine. The worst thing is your anxiety, not the exam.

@OldJohn Good luck!

user19161

10:47 AM

Oh, he's gone.

user19161

@JonasTeuwen You don't need luck for exercise!

Jonas Teuwen

Then you are not making it hard enough.

The cliché: no pain, no gain certainly applies. Where you can see "pain" in quite some broad context.

user19161

I used to go gym once a week. I used to have a Taylor Lautner physique, but not now.

Jonas Teuwen

Say in handling anxiety for example. Your mind is the boss.

user19161

@JonasTeuwen Though in this case, the pain and the gain are not so well-defined.

Jonas Teuwen

10:49 AM

Yes, Sir. That doesn't matter, the idea is clear.

user19161

@JonasTeuwen That's why I need help from that Buddha guy.

Jonas Teuwen

You don't need his help. Support is what you need!

user19161

@JonasTeuwen Thanks Madam.

Jonas Teuwen

That Buddha guy surely wants you to do it yourself and would support that if you want to, no?

user19161

Meow. Now I am also a cat lady.

Jonas Teuwen

10:51 AM

A crazy cat lady? Have to go to a MSc thesis defense to ask annoying questions. Be back later.

Ilya

12:29 PM

@Jonas: do you actually know, how can I try to talk to did?

anon

unfortunately, the belief one can do everything at the last moment doesn't apply to dentist trips. panics

Matt N.

@JonasTeuwen Thanks Jonas.

@Ilya Have you tried changing your name and avatar? Or if that didn't work, creating a new account?

@anon ...?

Ilya

@Matt: you think, otherwise it wouldn't work?

Matt N.

@Ilya I don't know but I thought he ignored you. So this seems like the only way.

But he doesn't strike me as someone who wants to talk to people.

bbl

Ilya

@anon for you

robjohn

12:48 PM

@Ilya Go here and invite him to chat?

Ilya

@robjohn a good idea - let me try

t.b.

spam flags/votes for deletion please. Sorry gotta run again.

ami

@jon

@JonasTeuwen there?

@JonasTeuwen sent you the email.

Jonas Teuwen

1:06 PM

@Ilya Who?

user19161

@Ilya Do you actually know if Matt is ignoring me? It seems to me so, but I wonder why...

user19161

Anyway, never mind. I don't think I did anything bad to him ever. At least not that I know of.

user19161

1:32 PM

@mattN. I am not sure where I have upset you in the past. I am sorry if I did, but if you don't respond to this, I will take it that you are ignoring me and I won't talk to you from now.