Let $f$ be a continuous and proper. We need to prove that $f$ is closed!

Let $V$ be a closed set in $X$.

We need to prove $f(V)$ is closed.

To prove this: Consider a convergent sequence in $y_n$ in $f(V)$

$X$, $Y$ are metric spaces? (I assume)

I have "fixed" my problem!

Yes, they are. Sorry should have said that!

I wrote my own numerical integrator for mathematica 8-).

The set $\{y_n\} \cup \{y \mid y_n \to y\}$ is compact in $Y$.

Am I right?

I think you want to say: "Let $y_n$ be a sequence in $f(V)$. We want to find a convergent subsequence."

@JonasTeuwen Here is my congratulations although I don't how hard this is to accomplish

Well, not so hard actually.

mathematica.stackexchange.com/questions/2356/… check out my edit here.

@JonasTeuwen Gauss quadrature? : )

No.

@MattN We want to prove $f$ is a closed map. Right?

So, closed sets are taken to closed maps by $f$ is what we claim.

@KannappanSampath Yes. And for this you want to show that $f(V)$ is closed.

And I think your plan is to show that it's sequentially closed.

So, if we have a convergent sequence in $f(V)$ then, I would like to show that the limit is in $f(V)$. Is that fine?

Yes.

Hello bear : )

Hi @tb. How has your weekend been?

@KannappanSampath (Just so you know: I don't see the proof yet)

Hi the two of you.

@KannappanSampath nice, how was yours?

@tb Pretty much fine except for those physics troubles. But fine yeah!

@KannappanSampath physics troubles? Did CERN send wormholes again?

@tb No, my physics teacher sent in butterflies "into" my stomach

Sweet : )

Oh why did I claim that

I don't know.

@Kannappan: My answer here is related to a previous question of yours: a metrizable space is compact if and only if all its compatible metrics are bounded.

@KannappanSampath So what's your definition of a proper map?

I think I wanted to get a compact set that contains $f(V)$ or something!

One whose inverse sends compact sets to compact sets: chat.stackexchange.com/transcript/message/3584463#3584463

Beat me to it!

And you want to prove that its image is closed, right?

Yes.

(Continuous proper maps are closed)

So take $y_n = f(x_n) \in f(X)$ and suppose that $y_n \to y \in Y$. You want to show that $y \in f(X)$. The set $K = \{y_n\} \cup \{y\}$ is compact because it has only one limit point. Now $f^{-1}K$ is compact and $x_n \in f^{-1}K$. Can you take it from here?

@tb: greetings :-) just got back from walking Lilly

Hi, robjohn! :)

Hi @tb!

@robjohn I assume that Smoky has taken over your job of babysitting Lilly so you're free to chat?

Hi, Jonas!

@tb Lilly is resting after the walk. Smoky is in her crate for the time being.

Ah, but the result is the same, you're here :)

@tb Yes. I will get some breakfast in a bit, but that won't be too much of an interruption.

@KannappanSampath so, got the argument?

Not yet. Seemingly, the contrapositive is also hard to argue! : (

You agree that $K$ is compact, right?

@tb why is $f^{-1}(K)$ compact?

Because the map is proper.

@tb You wrote something down? I was thinking it on my own!

@robjohn by assumption $f$ is proper in the sense that it is continuous and pre-images of compact sets are compact.

Oh, by assumption then

@KannappanSampath yes, just before robjohn entered the scene :)

@robjohn exactly.

So, happy I claimed the right thing. I did not name it $K$ though

Well I named it $K$...

@KannappanSampath aren't all closed sets called $K$? :-)

@robjohn They are; but I am trying to say how quickly I lose arguing that way because I don't see a way out so quickly, in a rhetoric way. :-)

No, sometimes $U, V, W$, I would name a compact set $K$ 8-).

Or $F$ for a closed set and $K$ for a compact set.

@JonasTeuwen $U,V,W$ is very bad notation for closed sets...

Oh, why? 8-).

@tb I keep them for open sets.

Oh, $F, G$, sorry.

Better :)

So, we have $x_n$ come from a compact set. Their image is compact, right?

@KannappanSampath well, $x_n$ is a sequence in a compact set, so....

It has a convergent subsequence?

exactly. Now take such a subsequence and...

Say $x_{n_k} \to x$

okay

then $f(x_{n_k}) \to f(x) \in f(V)$

and? :)

Now finish up!

Our aim was to show $y \in f(V)$.

and what can you say about $f(x_{n_k})$?

Well This $y=f(x)$

why exactly?

@tb It is a subsequence of a convergent sequence and hence the limit of the sequence equals the limit of this convergent subsequence..

I wanted to hear $y_{n_k}= f(x_{n_k})$ and thus $y_{n_k} \to y = f(x)$, hence $y \in f(X)$.

Alright. Thank you for guiding me through!

with ample prompts

too many hints?

No, that's fine. I felt like a child treading on a soft cotton :-)

Never bleeding when those thorns pricked me!

: D Teddy bear is made of soft cotton, that's probably why.

ah, teddy bear :)

hi all

@Srivatsan Hello there!

Hi @Srivatsan How have you been?

Hey Srivatsan! how are you doing?

i am doing fine.

How have you guys been doing?

@Srivatsan Are you back?

@Srivatsan!!!!1111ONE

@KannappanSampath I haven't seen the solution to the other exercise, so here's what I would do: take a sequence $x_n$ that has no convergent subsequence. wlog we may assume that the $x_n$ are pairwise distinct. Now each $x_n$ has a nbhd $B_{r_n}(x_n)$ containing no other point from the sequence. On $B_{r_n}(x_n)$ define $f_n (x) = \frac{n}{r_n} \max{\{r_{n} - d(x,x_n),0\}}$. Then $f = \max{\{f_n\}}$ is a continuous unbounded function on $X$ (since $f(x_n) = n$.)

@Srivatsan Because of too much time loss, I assume.

@Srivatsan You won't miss much, there aren't too many interesting questions.

@tb Reading it through. :-)

Maybe I should start asking interesting questions!

But I think that there is hardly anybody with expertise in the esoteric things which I do 8-).

@JonasTeuwen well, how do you think I feel?

Heh.

@MattN That plus something else: I was wondering if I was becoming too dependent on mse... :-)

@Srivatsan Yes, I know what you mean. : S

@tb Good to hear. =)

Did B. come back btw?

Yes.

Ah, that is nice.

@Srivatsan Yes, he didn't take the opportunity to learn how not to mistreat MathJaX :)

@tb Why should there be a neighbourhood that contains no other points of the sequence?

there could still be finitely many terms right?

Yes, but they are all at a positive distance, so make $r_n$ smaller than the least distance to one of those finitely many points.

@tb Ah, I see. ;)

@tb Oh, fine! Sweet : )

@Srivatsan I was not able to respond on the other day as I had to go AFK. I am sorry!

No, don't worry about it.

@tb I don't see it though!

And, I am working this summer with Prof D S Nagaraj of the IMSc @Srivatsan

@KannappanSampath what do you mean? My argument or the MathJaX abuse by BD?

@KannappanSampath I was thinking about you today. I have a question in basic group theory, and I am not able to get a complete solution. Do you want to try your hand at it?

MathJaX thingy @tb

@Srivatsan Yes. Sure!

Have you ever looked at the source of his posts?

@tb No, let me do it right away

@Srivatsan Are you still home in India?

@tb Actually, I have run across a few of his posts that are in NormalTeX (that is, with normal italics fonts). I was tempted to change it to \rm.

Yes, I am in India.

@Srivatsan :)

Of course, I wouldn't dare touch it.

@Srivatsan :-)

What's that group theory problem @Srivatsan?

@Kannappan Question: Show that every group of order 35 contains an element of order 7. Catch: You may not use Sylow's theorem, conjugation, or the class equation. Essentially, you can use only the more basic notions: subgroups, cyclic subgroups, order, cosets, Lagrange's theorem, homomorphisms.

What is that?

Cauchy's theorem should be good enough :)

Oh, I didn't know it by that name. I am pretty certain you don't need it. This is an exercise from Artin. And he introduces Sylow's theorems (and this as a corollary) only muuuuch later.

Well, proof 1 looks pretty simple and specializing it should give a pretty straightforward argument...

Well $G/Z(G)$ is never non-trivially cyclic.

Using this we can rule out that $Z(G)$ is either $5$ or $7$, right?

I think so, but maybe that's already too hard a thing to use :)

@tb Proof is pretty simple no?

@tb Hm, perhaps. Of course, there's a fine line between what is allowed and what's forbidden.

:)

Yes, but it seems like Kannappan's argument is even simpler.

@KannappanSampath I don't get it. What if Z is trivial?

And what if the group is abelian? (I do agree that Z is either trivial or the whole group.)

Well, forget it, it seems to be getting complicated.

Well using Lagrange's theorem, we can prove that two distinct groups of prime order intersect trivially.

Well, I guess if we know that the group is abelian, one can "easily" show that it is C_5 x C_7. :)

It is clear that $G$ has an element of non-trivial order.

tb: In the mean time, let me look at proof 1.

If every non-trivial element has order $5$, then, ...

$4n+1=35$ for some $n \in \mathbb N$ which is a contradiction @Srivatsan

Sorry, I didn't follow the last line, Kannappan.

If there is no element of order $7$, then the group of order $35$ is union of cyclic subgroups of order $5$.

They either intersect trivially or they are the same subgroups.

@Srivatsan Is that fine?

Ah, nice.

This will probably generalize, no?

Hey, Srivatsan, I should be going pretty soon. I hope to see you very soon!

@KannappanSampath nice :)

@KannappanSampath Only now did I see why you wrote 35 = 4n+1 :)

Not sure if it will generalise. :)

I guess the same argument works for groups of order $pq$ with $p,q$ odd primes, no?

@tb Take care. See you around.

@tb No. It works when $p-1$ does not divide $pq-1$. I think, not sure.

well, I leave it to you guys to figure it out. I never liked numbers, and the feeling's mutual :)

pq -1 = q -1 modulo (p-1), so that is equivalent to "p-1 does not divide q-1".

I think that works when $pq \not\equiv 1$ mod $q-1$

Bye, teddy bear.

Bye Matt!

Bye to the rest of you.

Now I need a bear...err beer.

@mk The same as I had previously written.

Oh snap! @Srivatsan!

@KannappanSampath: Ah, I didn't notice

@AsafKaragila Hey Asaf.

@Srivatsan Yeah, this is the finer version, methinks.

@JM I got Generalist. Which wager did I win/lose? :-)

How have you been, Hindu?

@KannappanSampath Will you be doing any more topology? I'm asking because I think I'll have a beer but after that I won't be of much use anymore.

BTW, the same argument works to give an element of order $5$ in this group.

@KannappanSampath I can get that in another way as well.

The lecture went fine, in case you guys wonder.

@MattN I will be. But, let me not dissturb you for some time.

@KannappanSampath How much time? : ) (You don't disturb me)

I didn't manage to prove the Stein theorem, but I did overview whatever was possible within the confines of 45 minutes.

@AsafKaragila How did your presentation go?

@MattN for 2 or 3 hours, I'll work with things and I'll get back with some doubts. Is that fine?

@Asaf: My favorites page in mse contains less than 10 items now... :)

Wow!!!

@KannappanSampath Yes, let's try that.

@AsafKaragila I have been doing ok. How are you doing?

@AsafKaragila Sorry, I missed that.

@Srivatsan What's your favourite page on mse?

@AsafKaragila Did you have time to talk about the actual problem contrary to your wondering if you could?

Nothing!

@Srivatsan Mostly tired. Lots of work, I have to finish my coursework and time is running out.

@Srivatsan How else then in a elementary fashion?

@KannappanSampath I ended up presenting the problem, with basic definitions and some basic claims. I proved one theorem and just gave an overview on two more. I couldn't really go into details about how the set theoretical independence comes into play when you want to consider the $\aleph_1$-case.

@Srivatsan Is that why I got some votes for old questions several times?

Oh, thanks for thinking that I'll understand!

@AsafKaragila Perhaps. :)

Because a few of those times that would happen after I capped out! :-P

@Srivatsan Are you around?

@AsafKaragila That is unlikely because the time I would vote would be a few hours after the start of the MSE day. =)

@KannappanSampath Right now? [Yes, I am...]

Err, you mentioned of another way to get an element of order $5$ right?

@Srivatsan Right. I knew that! :-)

Oh -- favourite*s* page! Not your favourite page. It's not my day today. At all. I need a shovel and a coffin.

@MattN Ah, don't get frantic!

I am so tired.

@KannappanSampath Ah, that one. Suppose every element has order 7. Pick an arbitrary x, not the identity. That generates a subgroup of order 7. Pick a second item "y" not in the subgroup. Then the set ${x^i y^j}$ is a set of at least 49 distinct elements in the group. (This needs proof.) Contradiction.

@Srivatsan: hey! how are you, man

Hey, Ilya. How are you doing?

@Srivatsan You need to show they are distinct, which is easy and follows from the fact they are of order $7$ and this fails for an element of order $7$, right!

@Srivatsan when will you be back?

Permanently? Not for a month or two.

</3

=)

@Srivatsan I am fine, what about you? Are you in India (sorry if I've misremembered)

That's bad AND TOO LONG!

^I can relate to that feeling.

@Ilya You didn't; I am in India...

Are you in Chennai now? or elsewhere?

Hello, guys.

@Dan: hi

@Daniil Hi. How are you?

@Srivatsan: hope you're fine (you didn't answer though)

@MattN :) Reassuring to hear that actually.

@Matt: I didn't say hi to you. My apologizes - but I won't :)

@Ilya Now you have, sort of.

@Srivatsan : )

@MattN that's why I won't

@Ilya Oh sorry. Hard to keep up with the simultaneous chats. :) I am doing fine.

@Srivatsan a small piece of this information was lacking for my great plan of the world conquest

@Ilya Too bad I quit duelling... : )

@MattN yeah, fortunately you've quit duelling not in the way Galois did it

@Ilya At the moment I'm not so sure how fortunate that is.

@KannappanSampath Ah, I missed this. I am in chennai.

@Srivatsan I realise you're way too busy :-)

@Srivatsan why don't you guys meet? India is not too big as I remember (comparing to Russia :-p)

Right, India is not as big. :)

@MattN I wonder about your quiff now

I am going to sleep for a while. See you later. I hope to see you soon for tea, Sriva.

Ah. Chennai-Bangalore is only about 5 hours by car.

....And Bangalore is also a place I like for some strange reason. I want to visit it every time I am in India.

That is interesting if I'll get to meet you : )

Can somebody please check my automata theory/language theory proof: mathbin.net/89742

@Sriva: I don't remember, have you ever lived in India for some time (like a couple of years)?

@Dan: pump it~

@Ilya Yes, he did atleast during 4 years of his undergrad at Indian Institute of Technology -Madras(Now Chennai)

@Ilya :D

Hey guys!

@Ilya I am from India; I lived there for over twenty years =)

@Ilya Hm. This is kind of rude. I'm sure that was not your intention : D

@Srivatsan I didn't remember that - now I do, obviously

@MattN :D:D:D

@Ilya Or was this what you wanted to say?

I have to count the integer solutions of $x^2+y^2 \le 25$, I am getting $81$, but woframalpha says it's $79$.

@KannappanSampath Sure, I was going to suggest that myself (once I know when I will come there).

@MattN Quiff apparently is a hair style, no?

I don't understand why is it not taking care of the case $(\pm 5, 0)$

@Srivatsan : D Anxious to meet you but will wait patiently!

@MattN Kannappan is right

@KannappanSampath That as well apparently but urban dictionary tells me that it means female genital.

@MattN LOL, there's always a right perspective. :-)

@MattN well, you choose what to answer - I meant a hairstyle

I see.

@Ilya :-) This is $\ldots$!!

@Ilya I don't have a quiff hairstyle.

@MattN but you are not sure if you are happy that you are still alive, right?

@Foool It has counted them but seems to still say less!

Ok, I have to go offline soon (my brother is back from work).

@Srivatsan: see you later, good luck

@Foool One good way to count is to let $x$ take on values from $-5$ to $+5$ and count possible $y$

@Srivatsan See You Later. Take Care :)

I am getting 81 FWIW

@Srivatsan ????

I am getting 81 as the answer to Foool's question.

@Srivatsan: which answer?

And, Wolfram says that too.

@Ilya to Foool's question.

Wolfram wants to make money these days :/

Ok, I got to go. I will see you guys soon. Bye!

@Ilya Yes. That's right.

@Srivatsan Byee! See you soon!

@Matt: but why? I was about to think you're emo

@Ilya : )

@Matt: I've lust one stud and I was looking for it, so I haven't seen your answer, sorry

I'm a bit sad - it was a present

@Ilya : ( If it just dropped on the floor now it can't be too far, right?

@MattN I've already repeated the way to the busstop, nothing

there is no sense to look it further - I was to many places today

@Ilya You need to rent a metal detector.

This site is the best place ever.

@MattN joke?

@Ilya Well if you rent one it would be easier to find the stud, right? : )

@ymar It is : ) (starred)

@MattN I don't think so

Today's visit was too short. </3

@MattN Thanks, my first star.

@ymar You're welcome : )

Important exam tomorrow. If I don't pass I'll have to decide whether I want to stay one more year or drop out.

@ymar You want to stay one more year.

@Ilya Is it not a metal stud?

@MattN it is- but I've been to different cities today, and took about 6 trains/trams

@MattN I don't know. I'd like to move to another country and work. But I don't have to think about it now.

@Ilya : (

@Ilya And what's up apart from losing your stud?

@MattN what do you mean?

@Ilya Well are there any more interesting events in your life? : )

^This is not sarcastic, in case you are wondering!

I didn't

sorry, I was away

and will be now, sorry again

Np.

@Matt: see you and good luck with recovering. Watch some good movie, take a break - try to enjoy the life, it's not that bad some part of the time (there is no sarcasm as well)

@Ilya Thanks : ) See you!

What do you guys think about this book: amazon.com/Mathematical-Logic-Undergraduate-Texts-Mathematics/… ?

I have not read it, sorry.

Hm, what is the difference between a formula and a term?

A formula consists of terms.

o, true

thanks

A term is either a variable or a constant or if t_1, ..., t_n are terms and f is a function symbol then f(t_1, ..., t_n) is again a term.

@Daniil Np : )

@Matt: are you here?

@Ilya Yes. What's up?

Ok.

nice, thanks :)

Never mind : )

no-no. I was kidding

Are you planning to join Matt's universe? : )

hm, maybe - depends on when I will start learning set theory

@Ilya Done.

@Ilya How about today? : ) The first few chapters are very basic, you won't want to stab yourself in the guts.

@MattN maybe I'll read it in a train tomorrow, now I will cook a dinner

Sounds good : ) (both)

@Matt: ah, no I won't, I've forgotten the charger in the office

It won't run away : ) (the book)