As, $y_n \to y$ , we must have that, $f(x_{n_k}) \to y$

Yes.

But, as $x_{n_k}$ converges to $x'$, we know that, $f(x_{n_k}) \to f(x')$ as $f$ is continuous.

@KannappanSampath Heya

@N3buchadnezzar Hey!

Perhaps a wierd question

@KannappanSampath Yes.

And, this proves that $f(x')=y$

Yes.

How do I show that $\sqrt[a-1]{a}$ and $\sqrt[a-1]{a^a}$ always is irrational when $a$ is a positive integer greater than one ?

So, we are through! @Matt. OK?

@N3buchadnezzar Perhaps, you'll know one follows from the other!

@KannappanSampath The problem was to solve $x^y = y^x$ when $x$ and $y$ are positve integers and $x \neq y$

@KannappanSampath Ok, good. Where in the proof did we use the assumption that $G$ is closed?

@MattN Seemingly nowhere!!!!

@KannappanSampath Indeed. So it seems that there must be something wrong. ; )

I used logarithms and obtained the solutions $x = \sqrt[a-1]{a}$ and $y = \sqrt[a-1]{a^a}$ where $a$ is a positive number. Now my problem is showing that the only integer solutions occurs when $a=2$ ^^

@N3buchadnezzar I am preparing for an exam. I am sorry. Not now.

@MattN To my knowledge, our argument is pretty airtight!

Okai =)

@KannappanSampath Let me think about it for a minute.

As of now I can't fix your proof but let me post mine:

(Then you can tell me where we went wrong in the proof we constructed just now.)

Let $y = f(x)$ be a limit point of $f(G)$. We want to show $y \in f(G)$.

Let $y_n = f(x_n)$ be a sequence converging to $y$.

Ok.

We have shown that $K := f^{-1}(\{y_n\} \cup \{y\}) = \{x_n\}\cup\{x\}$ is compact.

Yes.

@MattN Are you sure, we constructed a wrong proof? : (

@KannappanSampath No. Let's see.

@MattN I think I figured out, where we went wrong. The linked message seemingly is the step where we went wrong.

Hence since $X$ is a metric space we know that it is sequentially compact hence we can find a convergent subsequence $x_{n_k} \to x^\prime$. Then $x_{n_k} \in G$. $G$ is closed and $x^\prime$ is a limit point of $G$ hence $x^\prime \in G$.

See, let $y$ be a limit point of $f(G)$

Why do we think, $y=f(x)$ for some $x$ at all?

Hence $f(x^\prime) \in f(G)$ and $f(x_{n_k})$ is a subsequence of $f(x_n)$...

@KannappanSampath Good question.

@KannappanSampath That's wrong but given that I never use it in the proof it makes no difference : )

I think even I made that mistake!

And, made use of it in the proof too, methinks!

My computer seems to be struggling with the rendering. I'm going to reboot. BRB

Sure!

Just a quick question, is the problem y^x = x^y asked on the site? I tried searching, for it it. But I do not know exactly what to search after.

@mk Your group theory question was pretty nice. I didn't go through the details of the answer there but seems like it achieves a general construction!

@N3buchadnezzar looks like you'd be the first one to ask if you decide to : D

I will ask then, I always feel so silly for asking questions that has been asked before

@N3buchadnezzar Good SE'r

Back I am.

@SidharthIyer I think your dropping into the room was not captured by my eyes. Hello!

@MattN Me too.

The way to fix my proof will be to do:

Consider the inverse image of the sequence with its limit. It is compact, so, it has a convergent subsequence converging in $K$.

Now,

Now apply $f$ to the convergent subsequence, it will converge in $f(G)$.

Even now, I don't seem to need the closedness of $G$. : (

@KannappanSampath Yes.

I think TeX now helps. I should TeX my proof and send it to you. I think we don't need closedness!

I think I'll ask teddy bear about this the next time he shows up.

@KannappanSampath Btw, I really liked what you wrote about soft cotton and thorns yesterday. He makes me feel exactly the same way.

@MattN He's a lovely person, waiting for the right moment to do the right thing. I admire his online persona. : D

I might be his fan in real life too. : D

math.stackexchange.com/questions/114453/solve-xy-y2 =)

@KannappanSampath Same here. : )

:-)

: )

@N3buchadnezzar LaTeX your title please. And, is it $y^x$ or $y^2$?

:-)

@KannappanSampath Fikex

@N3buchadnezzar :-)

@KannappanSampath Gl on your exam

@N3buchadnezzar Thank you. Two more to go (counting this exam, and I'll be back to writing posts for SE and my blog)

I have 5 this year :/

I have 2 more and a BSc's thesis : (

@KannappanSampath Topology and what else? And when is the last?

People, two more exams for the mid semester to get over. My feeling is all of you are writing about the TOTAL exams to be held this year!

Speaking of which... : )

Hello teddy bear!

Hi guys!

@tb Ahoy!

Long time no see... : )

@MattN Friday is the last! The other one is Probability theory!

@tb Hi @tb

@MattN yes, almost 24 hours :)

ahoy cap'n

Not topology, it is Metric spaces and only that @MattN

@tb Maybe in teddy bear time. But in hoomins (and Matt's) time this was almost a week.

@KannappanSampath My exams are on 22 june 30 june, 2 july, 4 july, 6 july, that is going to be fun =)

Do I move that quickly? I don't think so...

: ) No. But when you're away time moves much slower and when you're here it moves faster.

:-) Truly, slowly!

So, what did you want to ask me?

For one: we have a potentially broken proof. The one with $f$ continuous and proper implies $f$ closed.

We don't seem to be using that the set is closed anywhere. At least in one version we don't.

@KannappanSampath Fancy elaborating?

That is, we don't seem to need to start with a closed set to show its image is closed!

@MattN I was not doing that, but typing in like an ant moving over the keys! slowly

Well, yesterday I only argued that the whole image is closed.

But since the restriction of a proper map to a closed set is again proper, this is enough.

Good morning/afternoon/evening all :-)

But you can do it directly:

@tb But that's not what the original question is asking so I redid the whole thing this morning...

(@robjohn Hello there : ))

So, given $C$ take $x_n \in C$ such that $f(x_n) \to y$. We want to show that $y = f(x)$ for some $x \in C$. Given that $K = \{f(x_n)\} \cup \{y\}$ is compact, we know that the pre-image $f^{-1}K$ is compact. Hence so is $f^{-1}K \cap C$ (because $C$ is closed!). Now observe that $x_n \in f^{-1}K \cap C$, so we can pass to a convergent subsequence and its limit $x$ will map to $y$.

@robjohn Good Morning :-)

@MattN but it follows, as I said: if $f$ is proper then $f|_{C}$ is proper for every closed subset $C$ of $X$.

(the argument is exactly what I used in the third sentence of my previous post.)

@robjohn Hello there!

@tb nice, standard convergence argument.

@tb how are things today?

@robjohn Sunny this morning, but I did a whole lot of house cleaning, looking forward to enjoying the nice weather when I'm done. Now it's all cloudy again, but at least I have a clean apartment :)

@tb Isn't it true that $x_n$ are in $f^{-1}K$ and since this is compact, there is a convergent subsequence $x_{n_k}$.

@KannappanSampath Yes, but you want to show that $y \in f(C)$, so you need to intersect $f^{-1}K$ with $C$, which is still compact, so you can guarantee that the limit is in $C$.

Is it that the point of intersecting it with $C$ is to ensure limit of the convergent subsequence is in $C$?

But aren't the $y_n$ in $K \subset f(C)$ and so $x_n \in f^{-1}K \subset C$?

@KannappanSampath Exactly. Then $x \in C$ and $f(x) = \lim f(x_n) = \lim y_n = y$

@MattN there might be preimages of $y_n$ outside $C$ and you want to show that $y \in C$ and $y \in K$!

Oh right. Doh. Thanks : )

@tb Today it is supposed to be cold (for here, not breaking 60°F is cold). Saturday it is supposed to be in the mid 70s (F).

shooting sparrows with a cannon...

@robjohn sheesh, I wish it were that warm around here :)

Don't inverse images preserve containments?

But you don't know that $K \subset f(C)$ since you don't know that $y \in f(C)$ (that's what you want to show!)

@tb Heh, I had to look up what a complemented subspace was.

@tb Right, I suck!

@robjohn well, the theorem by Lindenstrauss Tzafriri is very nice but it is very hard.

@KannappanSampath No you don't : (

@tb I figured that was the sparrows/cannon thing :-)

@tb Norbert doesn't necessarily do things the easiest way. You seem to have a simpler suggestion :-)

@robjohn But I found it tremendously surprising. A Banach space is isomorphic to a Hilbert space if and only if every closed subspace is complemented.

@robjohn Norbert only commented. The answer is by Philip Brooker (his Gravatar is anice picture of Stefan Banach by the way!)

Is that a pen or a cigar? I'd assume a pen.

A cigar, of course.

It looks like a pen.

It looks like a cigarette in a holder if you look at a blow up

And here's a picture that show that he smoked cigars like this: kielich.amu.edu.pl/Stefan_Banach/jpg/best/15x.jpg

@tb That still looks like a cigarette in a holder to me

@robjohn I'd go with this!

@robjohn okay, some tobacco he smoked and not a pen :)

(but you're probably right).

I never know what to do when a question I've just answered has just been closed because it is a duplicate. In the past, I've copied my answer and noted each on the other.

@tb Yes, definitely not a pen. Hey, guys, take a picture of me chewing on a pen

@robjohn To me that's equally ridiculous as hey guys, look I smoke

@MattN back then, that was not so. Thankfully, today, it is ridiculous to most people.

I'm always surprised when I come across another manifestation of irrationality and/or non-logic in professional (successful) mathematicians.

@robjohn see the discussion between Mariano and Pierre-Yves here

@MattN thank god, they're human!

However, the holder could have a filter in it.

@robjohn What use could that be of?

@tb Well but it's not all black and white. The funny thing about the examples I'm talking about is that they are extreme: very successful mathematicians do very illogical things. Take for example Gödel starving to death: one tad of logic applied to his own situation and he'd still be alive (ok, maybe)

@MattN genius and madness...

@MattN well, paranoia is a psychosis, how can you expect any sort of rationality there?

@KannappanSampath The filters usually filtered out some of the tars and nicotine.

@tb What puzzles me is how someone can be logical most of the time and then completely illogical some of the time.

It did nothing for the second-hand smoke effects.

I am sad to tell you guys that smoking is not banned in our campus! : (

Can we change subject, please?

Sure.

Sure. But, sorry if that was because of me!

@tb Have you solved the carrots puzzle? : )

@MattN I know the odds and I still buy (a small number of) Lotto tickets.

Anyway, I probably mooted a point that was potentially going to lead to discussion, but, only later did I notice that it was viral!

@MattN No, I haven't even understood that it was supposed to be a puzzle... But now that you mention it, I suspect that it is an allusion to the fact that bears like carrots

@tb Wrong : )

I haven't paid enough attention, I just saw that you mentioned carrots twice.

@tb Three times, one way back and immediately deleted. : )

But don't think about it too much, the answer to the puzzle is not worth it.

I didn't intend to (think too hard). In fact, when a deleted comment is involved, I have no chance, because there are too many of them...

@tb Why don't you pick one? It's hard to tell whether the smoking or the irrationality discussion upset or bored you or both.

Sorry, I was distracted. I wasn't bored and I wasn't upset, it's just that mental illness and smoking aren't among my favorite topics to discuss.

I added them on the list, right below duels and your mother jokes.

Seriously. : ) (It's just a mental list though)

Well, I'm hitting a dead end with the carrots thing, I guess. I give up

@tb The deleted carrots message was right after this and said something like "I prefer carrots." : )

Anyway: don't think about : )

@KannappanSampath Are you still there?

@MattN Okay, I won't... Hey, I'm going to be off for a few minutes, but I should be back in ten to fifteen minutes.

@MattN Yes. I am.

@tb I'll be here.

I think I don't have a proof of complete intersection property for compact collections. So, I am writing one on my own, now.

Sounds good!

@Matt: what's up?

Shall I tell you once I get it?

Sure!

@Ilya Nothing much. Been doing topology with Kannappan and been enjoying myself. Now reading about complemented subspaces. What about you?

@MattN math is not going well with the current state of my mind. Meanwhile we discussed candidates from students for the iGEM project

@tb When you come back you could tell me what your favourite film(s) is/are. I only seem to know which ones you don't like and I can't think of good films to watch so I'm keen to broaden my horizon : )

@Ilya What is your state of mind?

@Ilya Did you get any interesting applications?

@Ilya How come a Math guy is involved in selecting candidates for Biology stuff?

@MattN confused, hard to focus

@MattN yeah, there are some. one guy from India, another from China :)

@Ilya Did anything happen? Are you upset about losing that stud?

@MattN Oh, that's a hard question. I guess if I have to name just one then it'll be My life without me.... But I also like lighter stuff...

@KannappanSampath I support them on the modelling side, as an advisor

@tb Yes, looks good but too heavy to digest for me. (You were quick : ))

it's really hard for me to watch movies where any fatal disease is involved

Ah, Sarah Polley. Of course...

@tb That discussion is helpful and that seems like the best thing to do.

My computer shut down and I am in the process of sending a report to Apple.

@robjohn You report every crash?

@robjohn good luck then

@KannappanSampath No, and I don't get many. My laptop is sometimes on for months at a time. However, this one was completely strange. A grey veil came down over my display and a message came up saying that I needed to restart my computer.

@robjohn I see. Truly strange.

Well, I was watching a lot of Almodóvar movies around that time, and Isabel Coixet is a protégé of his, so it was natural to follow up on it, and I was blown away.

@tb what about "The skin I live in"? I've just seen the beginning of it in the plain when was flying the last time, but then didn't manage to watch it till the end (we landed namely)

@tb I see. I'm not that keen on Almodóvar (ok, la mala educación was not too bad). Do you have any non-sad films that you like? Otherwise I'll end up watching this and be depressed for a day or two afterwards.

@Ilya I haven't seen that one, yet. I haven't been to the movies in a very long time. (I think last time was when I saw the abysmally bad Slumdog Millionaire)

Still glad I watched that at home. That was boring.

@tb I agree to that comment about SDM!

I just wonder what that director has with people diving into feces...

I was surprised that the director of Pan's Labyrinth made Hellboy and that kind of stuff, though I haven't watched the latter

(if one is looking for any connection between that post and previous ones: there is no. I told, that I'm a bit unfocused)

@MattN Well, it definitely isn't a feel-good movie. But I'm not so much into those. The only one that comes to my mind right now is Bridget Jones, but I'm sure you've seen that..

@tb Yes have : )

I conclude that your life is too happy : ) Otherwise you wouldn't be watching all the feel-bad films. (<- what an outrageous thing to say.)

Well, how about some of the later stuff by Clint Eastwood, sort of everything after (and including) The Bridges of Madison County with a few exceptions is pretty brilliant.

@Matt: you're not the only person with such point of view I met ever

@tb Ok. Thanks : )

Well, here is my proof. (Not so brilliant though!)

I want to prove by contradiction.

That's not to say that some of the earlier work like Unforgiven wasn't brilliant either, but that's also not too light stuff.

@MattN I think that this is a non sequitur :) You don't have to be happy to like sad movies...

@KannappanSampath what do you want to prove?

I am writing out the proof of complete intersection property for compact collections.

complete intersection property = finite intersection property?

Yes.

Okay, shoot!

Let $\{K_\alpha\}_{\alpha \in \Lambda}$ be a collection of compact sets in the metric space $X$. Fix $K_{\alpha_0}$. Our claim about complete intersection is equivalent to proving that $\exists k \in K_{\alpha_0}$ such that $k \in K_\alpha$ for every $\alpha \in \Lambda$

@tb (Possibly not. But for me the reason to watch a film or to read a book is to escape all this. So if it comes haunting me in the book or the film then this is missing the point for me.)

Suppose not.

Then, for each $k \in K_{\alpha_o}$, there is an $\alpha \in \Lambda$ such that $k \in K_\alpha^c$.

So, $\{K_\alpha^c \mid \alpha \in \Lambda\}$ is a open covering for $K_{\alpha_0}$

Since, the set $K_{\alpha_0}$ is compact, there is a finite subcover.

Sorry to interrupt, but what exactly do you want to prove? Could you please state it?

Suppose $\{K_\alpha\}_{\alpha \in \Lambda}$ is a collection of compact sets in a metric space $X$ with the property that any finite collection intersect non-trivially, then, then, their complete intersection is non-trivial.

Okay, thanks!

Go ahead, then!

(i.e.) $$\bigcup_{\alpha \in \Lambda} {K_\alpha} \neq\varnothing$$

This means, $\exists \alpha_1,\alpha_2, \cdots, \alpha_n$ such that $\displaystyle \bigcup_{i=1}^n K_{\alpha_i}^c \supseteq K_{\alpha_0}$

So, you're done :)

(well, spell out the concluding line, perhaps)

I wrote down a wrong thing in display style there but none pointed out.

: (

Oh, sorry I thought you meant $\supseteq K_{\alpha_0}$ instead of $\neq \varnothing$.

(bad at multitasking...)

(I have to pop out for groceries. Teddy, will you still be here in about half an hour or does this mean goodbye for today?)

This means, $\bigcap_{i=0}^n K_{\alpha_i} =\varnothing$

Hi @tb

Hi, Asaf!

What's up?

@MattN I'll probably be around then.

(most likely)

Yay! Nice : )

@AsafKaragila Kannappan explains the finite intersection property to me.

@tb I was almost sure that you have seen it before, seeing how you have some knowledge about metric spaces. :-)

@KannappanSampath exactly. That's the only argument I know.

Yes. My teacher gave a direct proof but I seem to noted it wrongly, that I see noway of fixing it!

Well, if you only have a sequence of sets, you can do it with a diagonal argument. But I don't see how to do it with sequences directly for an arbitrary family.

Mmmm... diagonal arguments....

@AsafKaragila Hey, can you enlighten me about diagonal intersections of subsets of a cardinal? I'm stumped by an equality that's always claimed...

(I feel a bit silly for asking this, but I must be missing something).

Oh lookie lookie, another update of Texmaker! 3.3.1

@tb Yeah, it's a really magical argument (because it depends on choice of order of the sets)

Thanks for the annnouncement. I use TeXmaker.

So, Jech defines the diagonal intersection of $\langle X_\alpha, \alpha \lt \xi\rangle$ by $$\mathop{\Delta}\limits_{\alpha \lt \kappa}X_{\alpha} = \{\xi \lt \kappa\,:\,\xi \in \bigcap_{\alpha \lt \xi} X_{\alpha}\}$$

I know you did, @Kannappan. This is why I made that announcement.

@AsafKaragila I suspect it's more basic than that.

@tb No. It's black magic I tell ya!

@tb So he does. What should I clarify?

and then he claims that $\Delta X_\alpha= \bigcap_{\alpha} (X_\alpha \cup \{\xi\,:\,\xi \leq \alpha\})$.

Why is that?

@AsafKaragila Thank you. I am happy you remember every speck of details from the chat room them. I should appreciate your Macalauian memory.

Hey guys. I'm a 4/5 year undergraduate student, and I'm starting to think about where I want to go for grad school...So! Straightforward question: Would a PhD with research in Mathematical Logic in the United States be tantamount to career suicide?

:-)

I couldn't really think of a way to phrase that question more nicely. Lol

I am laughing out aloud!

@AsafKaragila According to me, the right hand side always includes $0$.

And this obviously can't be.

But what if $0\notin X_\alpha$ for all $\alpha$?

@KannappanSampath Was the laughter at my question?

@AsafKaragila Well, note the union and the second set always contains $0$.

In principal, you go along the diagonal and ask who belongs to the intersection "so far"?

@DanMKatz About your career suicide!

Oh oh, I missed the second one.

@AsafKaragila that's how I understand the first definition, and that's perfectly fine with me, but I've seen this claimed at various points in the literature, and I just don't see it.

Oh wait. Actually $0$ is in the first definition as well. Vacuously.

@KannappanSampath Haha can you be more specific? I meant for the question to be somewhat humorous, but there is a serious side to it. I find myself deeply interested in mathematical logic, but I'm a bit concerned as to whether I would find myself unemployable if I decide not to go into academia after getting my degree. Any insight?