Why am I the only vote to close here? What is everyone doing on a Friday evening?

@DylanMoreland I'm watching Regular Show!

Guys, how to post a link in the chat? Same way as in the questions?

is made to solve the equation*

@GustavoBandeira for instance [google](www.google.com)

got it?

@leo Oh, its the same way as in the forum. Thanks.

@GustavoBandeira You are welcome! :-)

@BillDubuque Congrats on the diamond! Who the other new mod?

@PeterTamaroff Eric Naslund...

hey! does anyone have a quick example of a pair of adjoint functors that does not determine a unique adjunction?

Wikipedia states "However, the adjoint functors F and G alone are in general not sufficient to determine the adjunction." with no examples :P

doesn't this contradict proposition 19.9 of the joy of cats? "adjoint situations associated with a given adjoint functor are essentially unique"?

@BrunoStonek, consider the identity functor $I:Vect\to Vect$ from the category of real vector spaces to itself, which is its own adjoint. Then the map $\hom(IV,W)\to\hom(V,IW)$ which is given by multiplication by $\lambda\in\mathbb R\setminus0$ gives an adjuntion.

Since there are many real numbers, there are many adjunctions

Votes for close please.

And am I the only one who thinks this is uncanny?

@robjohn are you there, mean square?

@robjohn Did RI really make the same mistake as me here? Or am I missing something?

@MattN That's a very cool theorem! (in the comment)

@MattN Now I am here!

'evening everyone

Hi

@MattN nope. Robert never divides by $0$ except inside $\min(1, u/(2 |g_2|))$, where it doesn't matter.

@robjohn: Hi, does the last part of my answer make sense to you?

@Gigili I have commented on your answer.

@MarianoSuárezAlvarez: thanks! I guess I didn't think of going through the most trivial examples of adjunctions (identities) and tilting the isomorphisms a little bit...

This question should be closed.

@robjohn: could you be so kind as to take a look at my answer to the circle inversion question with compass only? I always have the impression that I'm not making myself perfectly clear when talking about geometry in English.

@tb Okay, you win :-)

@tb It might be nice to mention that by similar triangles, we have $\dfrac{\overline{OX}/2}{\overline{OP}}=\dfrac{\overline{OP'}/2}{\overline{OX}}$

late for the park. bbl

@tb Hi :-)).

@robjohn Thanks, that's a good suggestion. I added it to the answer. See you around!

@JonasTeuwen Hi, Jonas

@JonasTeuwen Did you see my question? It was partly inspired by my writing the answer to your Alaoglu vs. AC question. So you're not the only one who's nuts in this respect :)

Okay, I should be doing some stuff. Take care!

@tb Kickass!

@tb I was thinking about the same question some time ago!

@tb I saw this proof in Conway which does not use Baire... but maybe it is "under the surface"?

@JonasTeuwen Hm... What proof in Conway? He proves the open mapping theorem from Baire. Do you mean the proof of uniform boundedness? This proof doesn't use Baire, but it involves a plain application of dependent choice, as does the much nicer approach by Alan Sokal.

@tb Oops, I mean PUB yes.

Look at how much nicer you can put it if your name is Sokal :) As I argue in my question PUB is equivalent to DC, so this doesn't help much.

Sorry I meant the Baire category theorem instead of "PUB", of course.

@tb I had to show that equation to make sure the point was the inverse, so I thought it would be nice to have it there.

@tb Yea, so it is the same below the surface...

Just undressed.

Looks like I'm late and missed my favourite teddy bear.

@JonasTeuwen Heh. And what are you going to do now? : )

Seems to cold to run around naked outside.

@robjohn Shouldn't it be divided by the total amount?

@Gigili The total amount is $1.00$

@MattN 8). Yes.

@robjohn The total amount of that probability, not the whole.

@Gigili and on that problem, why wouldn't the total amount be $.47$ instead of $.34$?

@Gigili of what probability? they are asking what the probability of $R$ and $Y$ is, not the probability of $R$ given that $Y$ is true.

@robjohn I know, why not the total probability of $r$ and $Y$ divided by the total probability of $R$.

I recall we always did it this way.

@Gigili That is the way you compute the conditional probability. $P(A|B)=P(A\&B)/P(B)$

The amounts in the table are not probability, are they?

I'm so confused.

The amounts in the table look like relative amounts, and since they sum to 1, they are indeed probabilities.

Okay. Seems you're right.

What about the last part, without the denominator?

Okay, thank you @robjohn.

They would be complements if they don't intersect and their union is the whole. Their union is the whole, but $(R\text{ or }Y)\,\cap(R'\text{ or }Y')=(R\cap Y')\text{ or }(R'\cap Y)$

Duplicate alert.

@JonasTeuwen We call that Banach-Steinhaus : )

@robjohn Poor me. :,( I want to get better at this.

@MattN It takes practice and perspective. You'll get there.

@MattN Yes, but there is this corollary which I would call PUB 8-).

Nothing : )

@MattN ?

How is it this doesn't get closed as "too localized"...

Call for votes.

@ZhenLin If I understood it, maybe I could comment :-)

@MattN I understand enough to realize it is close :-)

@robjohn I didn't realise it had 4 votes now. I was wondering what all these people were up to on a Saturday night!

This simplicity guy annoys me greatly.

@MattN people might have lives :-)

@MattN Relax. Have a beer.

@robjohn Are you infering I don't have a life? 8-).

@MattN what about him annoys you?

@robjohn Outside maths and math exchange? No. : )

@JonasTeuwen whistling to the middle distance

@robjohn You would be quite right. 8-).

@MattN It seems unlikely, doesn't it?

@robjohn He's an idiot. Exhibit A Exhibit B

Exhibit C (mainly because I doubt he could do the question, even knowing what $C(X)$ was and also $C(X)$ is completely standard for continuous functions $X \to R$ or $X \to C$)

@MattN I see :-)

@robjohn </anger vent>

@JonasTeuwen All relaxed now, even without beer : )

@MattN Surprising.

@JonasTeuwen This isn't real life so it can't annoy me properly : )

@MattN Excellent!

@robjohn It does! : )

It needs one more.

@robjohn Simplicity is outrageous so I needed to relieve myself. Though pots shouldn't call kettles black.

Makes me wonder if he also has zero branes or just a weak background : )

I should go to bed. Good night!

zero branes :)

Is it possible for an inverse system of distinct groups to have an inverse limit isomorphic to the direct product of said groups? I would guess "no" but I don't have an idea how to prove this..

The obvious embedding of the limit into the product won't be an isomorphism, that much is clear.

isomorphic to the product under what isomorphism?

@robjohn Do you think it is wise to allow one user to call another an "idiot" here? Broken windows galore.

isomorphic under any isomorphism

then surely one can construct examples

I wish I knew who the One is, so I could ask them this question.

:)

For example, let $G$ be a fixed group

let $K_n=\underbrace{G\times\cdots\times G}_{\text{$n$ factors}}$

oh that is totally cheating. clever bastard.

and let, for each $n\geq1$, $f_n:K_{n+1}\to K_n$ be the projection on the first $n$ coordinates

waitaminute...

yeah that works.

The limit of the system $$\cdots\to K_{n+1}\xrightarrow{f_n} K_n\to\cdots\to K_2\to K_1$$ is the isomorphic to the countable product $G^{\mathbb N}$

which also happens to be isomorphic to the product of all the $K_n$

this shows that such a phenomenon is pretty uninteresting

Now to segue into the original question I was pondering. First, a sanity check: any subset of a partially ordered set inherits the partial order and becomes a poset in its own right, correct?

prove it

@BillDubuque calling what someone does idiotic vs calling the person an idiot are different things, but I would try to curb the name-calling, but sticks and stones...

@BillDubuque the context is important, I guess.

@robjohn The person wrote "He's an idiot". Are civility standards lower here than the main site?

@MarianoSuárezAlvarez It seems so obvious that the axioms would carryover I don't even know how to put it into words. Bleh.

prove it, then :)

check that the restricted relation is a partial order

Are civility standards lower here than the main site? If this is an empirical question rather than rhetorical, my answer is: yup.

«I don't even know how to put it into words.» is almost indistinguishable from «I have not tried»

:)

Indeed, it explicitly says I can't figure how to try! But lemme see. Say $J\subseteq I$. If $a\le a$ in $I$ and $a\in J$ then $a\le a$ in $J$ checks out. Same for $a\le b,b\le c\implies a\le c$ (as long as all of $a,b,c\in J$). I guess that's what I'm supposed to say..

Don't write «if $a\leq a$ in $I$ and $a\in J$, then $a\leq a$ in $J$»

it is pretty messy

@robjohn Could you give me the link to the CHatJax script? the one I drag into favourites?

@BillDubuque Ah, that was borderline, but my spidey-senses were not tingling at the time.

Please =)

you want to check reflexivity, that is, that «for all $a\in J$ we have $a\leq_J a$».

It's in his profile Peter!

Reflexive spaces?

@anon Ok!

So: let $a\in J$. Since $J\subseteq I$, then $a\in I$ and since the original relation is reflexive, $a\leq a$. Now $a\in J$, so $a\leq_J a$.

@PeterTamaroff The Chat Rules has this link.

And same for the other axioms, I see.

Starting with «if $a\leq a$ in $I$...» is not great because that is always true

Anyone up for a (probably easy) question?=D

I'll ask Mar my question first :)

(Before I look at yours)

@BillDubuque use your judgement... it's what you're paid for :-)

Oh go ahead, I didn't men to interrupt you

@BillDubuque and yes, I might have flagged it on the main site, so perhaps the "verbal" vs "written" standards are a bit different. I don't know.

@PeterTamaroff So you have only countably many times $\frac1n$.

For some $n$.

@robjohn Being the room owner, I thought I'd check with you first. I think it is best to discourage these kind of remarks. That helps ensure that sparks don't turn into flames.

@PeterTamaroff Yes, so you know...

@BillDubuque I agree, and I try to keep my "verbal" comments as calm as my "written" ones, but I know that editing is harder when you're speaking.

Suppose $I$ is a poset and $J\subseteq I$ inherits the partial order. Suppose $(G_i)_{i\in I}$ is a family of groups indexed by $I$ and has morphisms so it comprises an inverse system. Call the inverse limit $A$. We can then delete many morphisms to make $(G_j)_{j\in J}$ and inverse system; call the limit of this $B$. When can we say $A$ and $B$ are isomorphic? First off, the obvious map $(x_i)_{i\in I}\mapsto(x_j)_{j\in J}$ (simply delete those $x_k$ with $k\in I\backslash J$) (I think) is an

@JonasTeuwen Right. But I can't close it. I can't prove it!

isomorphism iff: for all $i\in I$, there exists $\ell\ge i$ in $I$ such that the map $$G_\ell\to \prod_{j\le\ell,j\in J}G_j$$ is injective. However, even if this "obvious map" is not an isomorphism, is it still possible for the two limits to be isomorphic?

Oh

Activity in my mathchat?

there is a natural map from one of the limits to the other

that map is an isomorphism if $J$ is cofinal in $I$

@PeterTamaroff So basically you have for each $x$ just either $0$ or a permutated sequence $(\frac1n)$.

(there may be other maps which are isomorphisms, just like there was the fluke above with the product...)

For example, suppose $I=\mathbb N$. Then if $J$ is the subset of even numbers, you will have an isomorphism

if $J$ is finite, on the other hand, with the same $I$m then you will generally not have an iso

naturals with the usual order (rather than, say, the divisibility order)

I get the (in)formal observation. I want to find the delta that will work, to prove that

yes

Well, I spent way too long on this today.

@PeterTamaroff That is an easy formal observation. Just wrote down the assumption.

$$|f(x)|< 1/n$$ whenever $$|x-a|<\delta$$.

I was discussing systems with Zhen and we figured out that cofinal is sufficient but not necessary for the "obvious map" to be an isomorphism.

@PeterTamaroff Does it depend on $a$?

@JonasTeuwen It seems not. Why do you ask? THe statement is $\forall a$, so it is independent of the $a$ chosen.

@PeterTamaroff So why do you care about $a$?

@anon, I doubt you can get a necessary condition

@JonasTeuwen Because I want to prove the limit for $x \to a$ is $0$.

(it is also a rather useless thing :) one never needs that!)

That doubt also crept over me, which is why I'm now interested in what a counterexample to the usual-state-of-affairs might look like.

@PeterTamaroff Yes, but the question is basically what I wrote, right?

I'm going to ask on the mainsite for an explicit example!

@JonasTeuwen I presume. It kills me that I can see the function in my head, understand why it is $0$, but I can't put it into a formal symbolic/literal proof.

for an example where you the canonical map is an iso but $J$ is not cofinal?

@Peter: I think the limit only exists for those $a\in[0,1]$ that are not an accumulation point of the union of the $A_n$...

well, wait, no

nevermind that.

Can't you just say: Which terms $a_n$ are such that $a_n \geq \epsilon$ for say $\epsilon = 0.5$? So, which $1/n$ can be larger than $0.5$? That is $1/1$... Finitely many.

@anon ^_^ OK!

So that sequence converges like five monkeys.

pick any directed $I$, any $J\subseteq I$ which is nonempty, directed and is not cofinal, and pick any direct system indexed by $I$ such that all maps are isomorphisms. The canonical map will be an iso.

@JonasTeuwen I'm learning, dude, I'm learning.

@PeterTamaroff I know, that is why I don't give you a full argument bro :-).

I don't understand anon's remark.

@MarianoSuárezAlvarez Nah, I already found such an example. Take the profinite completion of the integers, indexed by naturals ordered by divisibility. We can let $J$ be the set of (a) powers of 2, (b) powers of 3, and (c) all numbers coprime with 6. This is not cofinal (for there is no $j\in J$ for which $6\le j$) yet the limit is the same. I seek an example where the canonical map is not an iso, but the two limits are iso nonetheless.

that's an awfully complicated example :D

I had the profinite integers and the chinese remainder theorem in mind when I was talking with Zhen.

take $I$ to be countable and discrete (so that no distinct elements are comparable)

then the limit is just the product

suppose all the groups are isomorphic, and take $J$ to be any proper infinite subset of $I$

the canonical map is not an iso

but the two limits are isomorphic

@MarianoSuárezAlvarez Very nice!

@robjohn I'm not quite sure what that means: did you have to make that computation to convince yourself that my argument was correct? In that case, what's lacking in my description?

@tb Good to see you around.

@tb Good to see you around.

If you want a directed example, construct a «self similar» I which contains a non-cofinal copy of itself. This can be done with a little ingenuity.

@Mariano: Indeed, this is anticlimactic. Now what if I add the condition that in $I$, every two elements has at least one lower bound? (Ie, for all $a,b\in I$, there exists a $c\in I$ such that $c\le a$ and $c\le b$)

that is what I meant by «a directed example»

@DylanMoreland Hi, Dylan, how's life? Apart from the fact of having spent too much time on finite vs. quasi-finite morphisms?

(I was typing before your comment came.)

:)

Hmm, self-similar with a noncofinal copy. I suppose an infinite binary tree, where $J$ is one branch down (and therefore disjoint from its mirror image) would work.

if you pick a directed system with a minimal element (so: if the tree has a root) then the limit is just the group attached to it

if you are OK with such an example, then your binary tree works

Yes, I see.

Alright, now what if we further stipulate that the groups be distinct :D

put a group G at the root and on its left child

and let all the other groups be zero

@tb what was the criterion you were using to show that $P'$ was the $c$-inverse of $P$?

you have fun with that, @Jonas :)

@anon Wrong window...

I can only imagine what the other window contains.

@robjohn I was just saying that the red dashed line x' was mapped to the red circle x (that's obvious: they share two points on C and O is mapped to infinity)

Some guy asked me if the food he made was too spicy...

and that the line OP intersects the circle where it should.

@MarianoSuárezAlvarez Is this in response to my "groups be distinct comment" ?

yes

in any case, you see that pretty much anything can happen...

I don't understand how, if a whole bunch of groups are zero (meaning the trivial group?) they can be distinct.

I guess I should say "nonisomorphic"

the system I suggested had two distinct groups

zero and the other one

you want them all to be distinct?

Yes!

@tb Ah! When I saw reflection, I did not understand that it meant inversion. That is where my confusion lay. With that understanding, then your answer is quite fine.

well, honestly, the problem interests me very little :)

I am sure you can produce an example, though

fine with me. :) this line of questioning got more complicated than I hoped for.

it is not so much the complication as the pointlessness :D

Really, what one should do is to take the original diagram, put a suitable Grothendieck topology on it, and study the corresponding sheaf topos. :p

On the contrary, if we go into topological groups we very well do have points.

(ba dum tss)

@robjohn Oh, I see. I thought that circle inversion was synonymous with reflection in a circle. Is that wrong?

@tb So using similar triangles, we don't need to use inversive-geometry at all, it seems :-)

@tb It's alright! I managed to mangle my hands while out running with my dog and can't write, but typing is okay and I'm learning interesting things about modular forms.

@tb I had never used 'reflection' that way, which doesn't mean it is wrong. I usually think of reflection as being across a line, but I can accept the usage with a circle. I've never had a course in inversive-geometry, so the usage might be good.

Woops, sorry for the double message up there. I went back to see my family in Ann Arbor and my dad's spare laptop is one of those old metal MacBooks that is designed to keep wireless signals out. It's really wonky.

@robjohn apart from the very definition :)

@robjohn Geogebra uses it, and I saw it on cut-the-knot, too. Not the two most trustworthy sources, but I might prefer the term because in German it's called Kreisspiegelung.

@DylanMoreland Does the double message persist if you refresh?

@anon: up here

oh

@tb okay, I will try to remember to think inversion when I see reflection. I didn't think that you were using inversion at all when I read your answer at first :-)

@tb I just looked at your diagram and used regular geometry with no inversion other than $\overline{OP}\;\overline{OP'}=\overline{OX}^2$, the definition, as you say :-)

This Makoto norm thread is degenerating.

Oho, so it's a seminorm?

@robjohn I changed my answer from reflection to inversion. Does that read better now?

@DylanMoreland Kato reminds me of Perturbation Theory...

@tb I now peg that as using inversion, so yes, it reads better to me. It may be better for those who don't know about inversion, since they will know that something different than equal-distance reflection is going on.

@tb However, I can't upvote it again :-)

@robjohn Thanks a lot for your input. I never really read a geometry book in English, so my vocabulary is a bit shaky... :) That's why I asked earlier today.

@tb and I completely missed the point. I am glad that you pressed me so that I realized what you were actually saying. I like that there are two ways to look at your diagram to get two different proofs!

morning

It's almost 1am here, but sure, yes, morning.

@ZhenLin It's almost 5 PM here, so no where near morning :-)

@BenjaminLim I'll say g'day!

@robjohn Yes, I liked your observation, because it didn't occur to me. Otherwise I'd probably have drawn the segment $XP'$ :)

You've gotten similar-triangles in my inversion! You've gotten inversion on my similar-triangles!

hello

@robjohn not a lot of people say "g'day" nowadays....

@tb Do they have Reese's Peanut Butter Cups there?

@BenjaminLim That's too bad... I think of it as terribly Australian.

Or "Strine", as it is known locally. :p

@robjohn At least in Germany they do. See here

I never saw them here, but that doesn't mean much. After all, the sweet genes were transmitted to my brother, not to me :)

@robjohn Yeah that's what they show you TV most of the time perhaps....

@tb Their commercials is where the phrase "You've gotten peanut butter on my chocolate! You've gotten chocolate in my peanut butter!" came from.