Oh.

@MattN Oh, sorry didn't see you type in something.

This is an application of the supremum property, rite?

@JonasTeuwen Read as not going to $0$.

Weird proof.

Let me think.

So, we can have a finite subcover in terms of intervals of $K_1$, and of $K_2$ and so on, then there should be sets that intersect, no? So, we can find a decreasing sequences of intervals, then the intersection is non-empty.

I got the $\LaTeX$ thingy for that: $\nrightarrow$

@JonasTeuwen Yes, we have that sets will intersect non-trivially and hence there's atleast one element is clear.

Good.

So now you want two points?

Yes.

Or at least.

Atleast two points.

@Matt Did you see my "fix" for your proof?

Right, that it is non-empty is obvious from the finite intersection property.

@JonasTeuwen Matt does not like the word obvious but yes we can infer that from the property you mention.

@KannappanSampath No but now I have. But that's not the function I want to swap the limit with in the proof. : ) The function in the proof is not constant.

@KannappanSampath : )

Okay, let's see, $K_n = [-10^{-n}, 10^{-n}]$ for $n \geqslant 1$.

What is $\bigcap K_n$?

$\{0\}$?

Right, which are our two points then?

I seem to misunderstand the question.

Shit! I was doing union or something strange!

@JonasTeuwen No, that's exactly what we want.

Does the diam not go to $0$.

Oh..., you want to show that there is only one such point!

The claim we want to prove is that if the intersection only contains one element then the diameter of it is $0$.

yes!

(In the limit)

I proved it here and the only bit that's missing is a justification why I can swap the limit with the diameter function.

The function is on $\Bbb N \to \Bbb R$, $n \mapsto diam (\cap_{i=1}^n K_i)$

Attempt: Let $a_n = \min K_n$ and $b_n = \max K_n$. Let $A$ be the set of all $a_n$, this set is non-empty and bounded above, say by $b_1$, hence it has a supremum, say $x$. If $m, n$ are positive integers, then we must have that $a_n \leqslant a_{m + n} \leqslant b_{m + n} \leqslant b_m$. We can thus conclude that $x \leqslant b_m$ for each integer $m$. Clearly $a_m \leqslant x$.

And now limits are unique...

Okay, maybe there is some gap 8-).

$\max$ and $\min$ of $K_n$?

Yes, our set is compact right?

yes.

So, how to make sense of these operators on a set in a abstract metric space?

Hah.

@JonasTeuwen ?

Okay, it is an abstract metric space. Let me see.

Is it evil of me to hold my answer to this question until 4 PM PST?

Since I've capped today? Perhaps I should just add it now.

@robjohn Your votes will be ethereal. And the OP is asking a soft question he can wait to know the answer for. but still I leave it to oyu to decide!

Each $K_n$ is compact therefore so is its intersection. Pick a $k_n$ from each $K_n$. Then the sequence $k_n$ is in $K_1$, right? Now, $K_1$ is compact, so we have a convergent subsequence again denoted by $k_n$ with limit in $K_1$. But this subsequence, except maybe the first term is also in $K_2$, since $K_2$ is compact this limit is also in $K_2$ and so on, so the limit is also in the intersection. Now we should argue why it is unique.

Ah, what are we arguing by the way for?

That the intersection of a decreasing sequence of compacts sets whose diameter goes to $0$ consists is a singleton.

Now, what does this $\textrm{diam}$ mean for generic metric spaces?

Is it something like $\sup_{x, y \in M} d(x, y)$?

Yes!

(Too quick)

I guess I could enter and delete it and wait until 0h UTC to undelete it :-)

@robjohn Sure!

Can 10K+ people vote on deleted answers?

I'd think not!

No they can't. I just tried.

Okay, say we have two limits $x, y$, we want to compute $d(x, y)$. We know that $\textrm{diam} K_1 \geqslant \textrm{diam} K_2$ and so on. We also have $\textrm{diam} K_1 \geqslant \textrm{diam} K_2 \cap K_1 \geqslant \textrm{diam} K_3 \cap K_2 \cap K_1$ and so on. So now pick in those sets two elements, $x_1$ and $y_1$ from $K_1$, then $d(x_1, y_1) \leqslant \textrm{diam} K_1$. Okay, I'm messing this up, I need a piece of paper 8-).

And a drink.

@Matt @Jonas I tend to believe : The exercise should have been more like if $diam \to 0$, then a single point.

Yes, you basically say that the limit is unique.

So even Hausdorffness of your space is sufficient.

@JonasTeuwen I don't get what you mean by saying limit is unique. Usually in any limit, this has always been the case. :-(

You make a limit $a_n = d(x_n, y_n)$ where $x_n \to x$ and $y_n \to y$ and $a_n \to 0$ means that diameter thingie.

Where $a = d(x, y)$ (the limit of $a_n$).

@KannappanSampath Is it not from a homework sheet?

@MattN No, it was asked in one of the previous exams.

You take two elements $x, y$ in your intersection and prove that $d(x, y) = 0$, you do this by using that diameter condition.

@robjohn Or am I talking bullshit? 8-).

$$\Huge{\mathbf{I\ HATE\ DIAGRAM\ CHASING!!!!}}$$

I'm not sure, long day.

@AsafKaragila Wow, you need a drink.

@JonasTeuwen where is this?

@AsafKaragila ouch! Not so loud, I'm napping ;-)

@robjohn Here.

@KannappanSampath It's bed time for me. Will you and Jonas & co be alright with this?

@JonasTeuwen where? Ah, up the page :-)

@robjohn This chatroom. The question is: We have a decreasing set of compact sets in a generic metric space whose diameter go to zero. The intersection consists of one element.

@Matt So soon. Sure. I'll catch up with you once my exams are over. Thanks for all the help, man. :-)

Have a good night's sleep!

@JonasTeuwen That is each of the decreasing compact sets contain that one element, but not only that one point?

@robjohn No, the intersection of those sets.

@KannappanSampath I thought the exam was on Thursday so I assumed you'd be studying again tomorrow : )

@KannappanSampath Anyway: Thanks. I'll be here again tomorrow.

@MattN No, its on wednesday. But never mind.

@JonasTeuwen Um, if the intersection of the decreasing compact sets contain that one element, then each set would contain it as well.

@robjohn Yes, the question is not if that element exists, but that there is only one.

But oh well, nevermind 8-).

@JonasTeuwen Sure: the diameters go to zero.

@KannappanSampath Sorry : / But I think you'll do well. You're well prepared : )

@JonasTeuwen Oh, you don't know it is the only point that is in all of them?

And look who's here : ) Brian can definitely help you better than I can : )

Well, that is the question, and I was saying that this basically is because limits are unique... and I was asking if that was bullshit!

Good night folks! : ) What a nice day today. : )

@JonasTeuwen Just to be safe, since I only just got up: what exactly is the question.

@MattN Thank you. I am just hoping not to get tensed, and I have $2$ choices. Thanks for your wishes! Good night once again!

G’night, Matt.

The diameters are monotone, but do you know they go to $0$?

@robjohn Yes.

Oh, then yes to what Brian said.

@BrianMScott Oeh, no, it would a waste of precious brain power if another person thinks about it 8-).

@KannappanSampath I don't know what tensed means, I'll ask you tomorrow. But make sure to ask Brian with whatever you need to know : )

(He's as "soft cotton" as teddy . ))

Now, I seem to have a spun an argument out of thin air!

Say, we have two points in the intersection and $diam \to 0$.

Let those two points be $x,y$.

Then, $d(x,y)>0$

@MattN I agree here! :-)

So, there is some $N$ such that $diam K_n < d(x,y)$ for all $n>N$.

Can this be used to get a contradiction @Jonas

@KannappanSampath True, but overkill: you just need one $K_n$ whose diameter is less than $d(x,y)$.

Where is the argument?

@BrianMScott Since $diameter$ is the supremum of those quantities, it cannot be strictly less than one of those quantities over which supremum is taken, right?

You can’t have $x,y\in\bigcap_k K_k$ if $\operatorname{diam}K_n<d(x,y)$.

@KannappanSampath Exactly.

Ah! Finally I did it with all your help!

Who invented diagram chasing?

I need his name, the date in which the method was invented and a time machine.

@AsafKaragila I don't know, but if he's American I would sue that bastard.

@AsafKaragila You should read more science fiction: then you’d know that that never works!

@AsafKaragila But certainly not favoured by people like Serre. They seem to hate it. They claim that is not the right way to write Mathematics!

@BrianMScott I am willing to destroy the universe. I don't care.

@Jonas How do you think are my teacher's paper? Are they referred often or mark making paper or some such?

I'm not sure, you told me not to click the links.

In my institute, people claim he is a good researcher and a horribly demanding teacher/guide!

@KannappanSampath Weird stuff.

@JonasTeuwen ??? Too short to make out!

@KannappanSampath I don't know that stuff, is it even harmonic analysis?

@JonasTeuwen Did you see his research interests?

@Brian Can you tell me some interesting exercises in Dense sets?

I solved that $X\setminus\{x\}$ is dense in $X$ if $x$ is a limit point!

@AsafKaragila That doesn't all fit in the sidebar on my computer.

@KannappanSampath Now I did.

There he claims so....

@KannappanSampath Can you find a minimal dense set?

@robjohn That's what she said.

As in, every other dense set has this set as a proper subset.

In a metric space, do you want me to construct a minimal dense set?

@JonasTeuwen As in every set is dense in itself :-)

@KannappanSampath Find a minimal dense set of $\mathbf R$.

@KannappanSampath Ooh, this should be fun to watch :-)

@KannappanSampath Do you know the definition of totally bounded metric space?

@robjohn I understand from here, there is no general construction.

@BrianMScott Yes. I do. Those sets that can be covered by finitely many $\epsilon-$balls.

@KannappanSampath Worse yet: there generally isn’t a minimal dense set.

@KannappanSampath For every $\epsilon>0$.

@KannappanSampath the question is, is there a minimal dense subset of $\mathbb{R}$?

@BrianMScott Hey, don't give the answer 8-).

Okay: can you prove that a totally bounded metric space is separable? That’s a nice exercise in dense sets.

First let me answer Jonas as I seem to be closer from the hints given

@BrianMScott and hard for dense students :-)

@BrianMScott With or without AC?

@JonasTeuwen You need only countable AC.

@robjohn True, true.

A minimal dense set is also dense in $\mathbb R$.

Yes.

@KannappanSampath That is why it is called a minimal dense set.

Can it be of any use if I track the usage of AC in my theorems?

@JonasTeuwen Doubtful, I’d say.

Okay, but maybe someday we reject AC! 8-).

And then my proofs are rejection-proof...

Note that $\mathbb R \setminus \{x\}$ is dense in $\mathbb R$ for every $x \in \mathbb R$. @Jonas

@KannappanSampath Yes.

So?

Who said what about AC?

Clearly $ \bigcap \mathbb R \setminus \{x\}$ is empty.

There is no minimal dense set in $\mathbb R$.

@KannappanSampath Yes, so?

@AsafKaragila But, a minimal dense was supposed to be in every dense set, but the intersection is empty thus forcing us to conclude that minimal dense det is empty! @Jonas Sorry I pinged the wrong person.

Which is not dense, therefore there is no such set.

Then you should still argue why this is equivalent with being minimal.

@KannappanSampath Oh, I was thinking that A minimal dense set is one in which no proper subset was dense.

Right, you have the idea correct.

Of course there is none no matter which definition you take for minimal.

@Kannappan: Your argument slightly resembles to this "Every number in $(0,1)$ is positive, but you can get arbitrarily close to $0$. Therefore $0$ is the minimal element of $(0,1)$!"

Claim: Minimal dense set by definition of Rob is the intersection of all dense sets.

But a minimal element need not always exist!

The empty set equals no set?

Furthermore, minimal is not minimum.

The intersection of all dense set would have to give us the minimum.

So, I am all wrong then?

The set containing the intersection has cardinality 1, where as the set that contains my minimal set has cardinality 0

@KannappanSampath Since the intersection of all dense subsets of $\mathbb{R}$ is empty, and $\varnothing$ is not dense in $\mathbb{R}$, your argument shows that $\mathbb{R}$ has no minimum dense set.

@robjohn I don't understand this!

@KannappanSampath There is a set that is the intersection of all dense sets. There is no dense set so that no proper subset is dense.

To show that $\mathbb{R}$ has no minimal dense subset, you must show that if $D\subseteq\mathbb{R}$ is dense, then some proper subset of $D$ is also dense.

@BrianMScott That's what I'm on about! :-)

@Kannappan: The difference between two integers is an integer; there is no least integer. Your argument says "Since we can subtract more and more, $-\infty$ has to be an integer!".

Strange! I am a duffer! I don't understand the difference b/n minimal and minimum Thanks @Brian for pointing that out!

Actually no. This is not what it shows.

It shows that you don't understand that a subset of $P(\mathbb R)$ need not be a complete lattice.

@AsafKaragila Your analogies have not made my understanding clearer though. I am sorry to tell you this!

Just as well. I need to get back to chasing after the Zig-zag lemma.

@AsafKaragila Be careful: it’s easy to zig when you should have zagged.

$$\Huge\stackrel{\bullet\ \odot}{\Huge\frown}$$

Hey @Asaf. Is your diagram running?

No? Well then you better go chase it!

@anon Bite my shiny metal paper clip.

Wait, that joke didn't work. Oh well. I was trying too hard.

The MS Word paper clip?

I don't have MS Word. Or Windows for that matter.

Alright. Let $D$ be a dense set in $\mathbb R$. We need to produce a subset of $D$ that is dense, right @Brian

I have your mother, though. She says hello.

Can you guide me through? @Brian

@AsafKaragila teddy does not like your mother jokes FWIW

@KannappanSampath Yes. Do what you did for $\mathbb{R}$: what happens if you remove one point from $D$?

@KannappanSampath I do like Your Mother (jokes) FYI.

@AsafKaragila Hey, Dedekind called. He is going to cut you up real good. He was very rational about it (yes, it is yours).

Yes, I remember these things because I admire you.

Like nothing else.

@AsafKaragila Ooh, it’s a hung-over frog!

@Brian Is that also dense?

@AsafKaragila shouldn't that be $\text{Your Mother}^®$?

@robjohn No. It shouldn't be that.

@robjohn By the way, can't your bookmarklet just load an old version of MathJax where it did work?

@JonasTeuwen It's not MathJax it's the chat.

Oh, right.

WireShark®?

@KannappanSampath What do you think?

I think so.

@JonasTeuwen It is not the MathJax that is the problem, it is the Ajax supporting the chat that has changed that is preventing the bookmarklets from working.

@KannappanSampath And you’re right. Now, can you say why?

Besides, I think that the old MathJax is no longer working.

Okay, I'll shut up 8-).

@Dylan: Who invented diagram chasing?

@JonasTeuwen Please don't. Some comment might prove useful :-)

@Brian Let $D$ be a dense subset of $R$. So, $cl(D)=\Bbb R$. So, We claim $D\setminus \{x\}$ for $x \in D$ is also dense in $\Bbb R$. We need to prove $cl(D\setminus \{x\})=\Bbb R$.

I have two page headers that I use locally, one for the old MathJax and one for the 2.0 MathJax. The header for the old MathJax no longer renders properly.

@KannappanSampath Okay.

Well one inclusion that $D\setminus \{x\} \subset \Bbb R$ is obvious.

@KannappanSampath as long as $x$ is not isolated.

@robjohn of what set do you mean?

Okay. Screw the Zig-zag lemma. I'll get someone to help me chase that diagram tomorrow. I'll go write the proof for the Hurewicz homomorphism.

@KannappanSampath Think on it. It is a hint.

@Brian To prove the other inclusion, we'd like to prove that, $r \in cl(D \setminus \{x\})$

For every $r\in\mathbb{R}$.

@AsafKaragila No idea. I never really thought of that as something that was "invented" in the sense that schemes were invented.

Whoever started homological algebra is a good candidate. Leray?

@DylanMoreland I should go back in time and just execute all those homological assholes before they get the chance to make my life so damn miserable.

To prove that, we need to prove there is a sequence in $D \setminus\{x\}$ that converges to $r$.

Well, you don’t need to, but that’s one way to do it.

Well, since $D$ was dense in $\Bbb R$, we have a sequence in $D$ that converges to $r$.

@BrianMScott : ( Yes!

@KannappanSampath or that for any $\epsilon>0$ there is a $d\in D$ so that $\mathrm{d}(d,r)<\epsilon$

@KannappanSampath Correct.

If finitely many of those terms were $r$, we could as well drop it!

Holy cow, is this still my question? 8-).

From my viewpoint, it's just some tool. You use it a few times and it becomes second nature. I don't think I've ever checked that, say, derived functors have to do what they do. So my only interactions with them are positive.

Then maybe the proof that I have in my head is wrong...?

If infinitely many of those terms were $r$, then the sequence itself would have converged to $r$.

@KannappanSampath and make sure that the proof includes $x$ in $cl(D\setminus \{x\})$

@JonasTeuwen what is your question?

Of the minimal dense set.

Ah, no I don't have a proof. Screw this question for now. Later, I'll think!

The proof I am thinking of requires a slight variant for $r=x$ than for $r\not=x$

@JonasTeuwen The only time you have a minimal dense set in a $T_1$-space is when you have a dense set of isolated points.

Yes, the minimal dense subset of $\mathbf R$.

That was my question, right?

Or is that resolved...?

@JonasTeuwen yes, Kannappan is trying to show that there is no minimal dense subset of $\mathbb{R}$.

but it appears that we have taken a break today.

Already gave up. :/

Okay, what is all this fuss with sequences and closures?

Oh, you gave up?

: (

Can't we just say that if $A$ is dense, then so is $A \setminus \{x\}$?

@JonasTeuwen twice already :-)

@JonasTeuwen that is what we were trying to elicit.

I'm confused now. Maybe I don't understand it myself then 8-).

It does require proof.

Oh, that requires proof.

@JonasTeuwen not a proof from you

Yes, I know, I gave the exercise.

@Jonas: I just wanted to preclude a spoiler.

@AsafKaragila Maybe you should switch to harmonic or functional analysis, no homological algebra for you then!

Ah.

Because Kannappan is not getting off that easy :-)

Yes :-).

@JonasTeuwen Indeed, I have never chased a diagram for Harmonic Analysis

@JonasTeuwen I have never chased a diagram in set theory either.

I asked my supervisor if I should learn homological algebra, he was frowning and said: You will never use that.

The point is that because I have to take four courses in set theory (being a set theory undergrad) and my university is a lot more algebraically inclined, I ended up having to take three algebra courses instead of just one.

Can't you take analysis courses then?

Well, I could have taken functional analysis. Good thing I didn't. They all failed in the exam.

Two of which are very sharp students. Much sharper than me.

Cool! Must have been a great course. Who was the professor?

Ah!

I once took a functional analysis class. It was horrible.

Why the "Ah!"?

Almost all of Conway's book. Homework: The last two exercises of each section.

Because I remember you talking about him.

Or mentioning his name.

Possible.

He taught me measure theory, so I might have talked about that before.

How was the flunking ratio there?

Oh, is it your assignment to chase diagrams? That's bad.

youtube.com/watch?v=-ZYa8p-bRyw This is one cool Greek.

@DylanMoreland I have a handful of questions to solve. One of which is to prove the 5-lemma and the zig-zag lemma. Both require extensive diagram chasing.

Well, there's an easy way to make someone hate the subject, I guess.

I also had to chase a few diagrams when I prepared the Whitehead Problem lecture... it's a horrible week for me.

I pity the fool.

@JasperLoy :-p

@JasperLoy Ack! I leave the chat for a minute and all of these quotes are gone!

@JasperLoy instant rimshot

@JasperLoy at least we know they were removed by the person making them. If a mod removes them, we don't see a trace.

@Jasper: wow, you just bumped yourself from the chat :-)

@JasperLoy Perhaps they can do either, because a few weeks ago there was a whole thread that was all removed and no trace was left.

@JasperLoy Your gravatar was going out and coming in at the same time.

to and from the same spot on the gravatar bar

Spooky.

@JonasTeuwen very

@JasperLoy Ah, that might have been it. I thought I looked back in the log, but it might have been the transcript.

Is it some kind of midlife crisis?

@JonasTeuwen he is 5 and quite a prodigy ;-)

Any age can be midlife!

Can 100 be midlife?

@JonasTeuwen he got a spiffed out trike to try to resolve his midlife crisis :-D

@JasperLoy I am sorry. I hope my comments are not out of line.

:D.

@JonasTeuwen Okay, okay; I admit that I was thinking more about the other end of the scale.

I think I won't do well, because I could not think about a simple dense sets question!

I'm looking for some properties of directed halin graph, Is there any good reference? like wiki article about undirected case:en.wikipedia.org/wiki/Halin_graph

@JasperLoy well perhaps there might be some understanding of my humor as long as it is not a sense of great humor.

@SaeedAmiri I can see the natural direction for the tree edges, but for the edges connecting the leaves, how is the direction defined?

@JasperLoy The Bridge?

@JasperLoy Ah, I assume yoda hangs out there.

@robjohn In fact my halin directed version doesn't have any special feature, but we can restrict it to directed cycle (for outer loop)

@SaeedAmiri not that I know anything beyond what I just read in the Wiki article you linked to.

@JasperLoy There must not be many here, since I don't think I've received one here.

Oh, no, I take that back... The blue circles...

I have seen a few here.

blue circles?

@robjohn I saw my first one earlier today, in fact.

I know that the @-pings are different colors in different chats, are the flag-pings always blue?

@JasperLoy They appear on the left instead of the @-pings which appear on the right.

one minute left...

@JasperLoy and then it is tomorrow :-)