Ah, there we go... I can upvote again!

Anybody here? I need somebody to jog my memory...

@J.M. Yes, I'm (still) around. What's up?

The coefficients of the product of two generating functions is a convolution, yes?

I remember that there's a formula for the coefficients of a product of n functions as a single sum, but I've forgotten the exact formula...

Yes, it's called that way, sometimes. \sum a_n x^n * \sum b_n x^n = \sum c_n x^n, where c_n = \sum_k a_k b_{n-k}

Anyone here?

Yep. I'm trying to remember the form when you're multiplying, say, three functions.

@yayu: yes?

something like d_n = \sum_{i+j+k = n} a_i b_j c_k

i got downvoted 7 times in the past three days

its not a lot and it isnt a big deal, i know

but i was wondering

@yayu: I've seen your meta post, yes.

what exactly is my problem

@t.b. I think that's it. Thanks. :)

was wondering if you senior guys had any opinions

Have you flagged for a mod to look into it?

yeah, i did

as I said on that meta thread I can't see anything immediately wrong with your posts.

Also, since you're still concerned about it, I presume the script for detecting anomalies didn't see it as anomalous?

it didnt

Just to cover everything: did you also e-mail [email protected]?

that's weird, you have a couple of different user names?

I did not mail them, just the flag

no i have just one account

@Jack: apparently the chat account doesn't sync with the main account for a rename... ;)

@yayu: you might want to e-mail just in case...

@J. M. see the rep page the meta thread was opened on Oct 10.

ahh i see. yeah i changed my username..

neither is my real name though.. my current username means "nil" in my native language

which is a close approximation to my progress in mathematics

@yayu: I gathered as much; Aryabhata used something similar in his profile page... ;)

@yaya: the vector laplacian one looked fine. the linear algebra one "omitted a lot of context" that made answering it hard. Including more context and a link to the original would fix any problems I would have had with it had I read it before now.

@yayu: even

the combinatorial one strikes me as neither good nor bad. kind of long. uses the word "combinatorial" but has a subtraction sign, which i've been told is a no-no.

After understanding the proof for schur's lemma, I realize that my question was out-of context... so that probably deserves a downvote

but easily fixed

@Jack: BTW... thanks for these!

oh no problem. irritatingly i have to use them myself now.

@Jack I agree. Though it seems someone has taken it upon himself (or herself) to punish my bygone sins

@yayu: oh weird, there were some earlier downvotes too

at any rate of the first three, had I been in a voting mood, would have been +1 for vector laplacian (because it invited a very good answer and had a very clear description of your misunderstanding) +0 for the combinatorics one, and -1 for the linear algebra one. I went ahead and gave the +1 on the vector laplacian one which gives more rep than the two undeserved downvotes :P

direct sums of matrices was on my reading list, never got to read it til now

+1 with SN's addition of context

I counted all my downvotes before this incident, and it totals to five. Statistically, seven upvotes in 3 days should stand out.

Well, the script, being dumb like most scripts, only accounts for a day...

Thus, you really should get a human to look into it; that is, send the e-mail.

@Jack thanks. What would you have done on the remaining: this and this

@yayu: for your direct sum question, you can get the same conclusion with the Kronecker product instead the direct sum, but Jyrki's explanation seems best

@J.M yes I will, thanks for suggesting

@jack can you see some kind of a generalization... when the eigenvalues of $D_1\oplus D_2$ would be the same as $\pi \circ (D_1\oplus D_2)$ I tried but couldnt conclude anything

@yayu: I ironed out a few kinks in your partials question and upvoted.

@yayu: first one is fishy to me. I mean, it is pretty common in pde to use the pde in question to simplify expressions, but you didn't include that. I would like your question more if it said "In a PDE text, they claimed this. It doesn't seem true in general, only for the DAlembert solution. Are they using that special form or is it true more generally?" Your current question just gave a brief sort of dull answer.

@J.M thanks ...

@J.M btw talking about scripts.. I tried implementing one of your animations in sage (trochoid one).. right now I've gotten to a cycloid on a plane and on a curve.. would post it when Im done

@yayu: for direct sum: I think Jyrki's method conjugates by pi, unlike Arturo's multiplication by. Conjugation will preserve eigenvalues and conjugate eigenvectors, so is very good. Multiplication can do fairly arbitrarily bad things to eigenvalues.

@yayu: Sounds quite nice. :) As I did say in the thread about the parabola, you can generalize to any curve, as long as you have a convenient arclength expression for the base curve...

(but the answer given to the PDE question is exactly correct and answers the question; hence the problem with the dullness is the question. Dull = +0 for me, I don't see why someone would downvote it.)

@Jack the question was from a physics based text. I think I should have included that.

There you go... :D

i admit, i am at fault regarding this. I realized it and added the postscript and accepted the answer posted (as it was right, ofcourse, while my question omitted crucial context)

Can I acquire the assistance of someone who would like to work on an abstract algebra problem?

@Lily: What's up?

@Lily: still on the Cayley tables?

hey mix.

@Lily: I guess you mean this question?

hiya J.M.

I'm actually finished with that question and onto another one. :)

I believe this problem uses the cayley table. Shall I just type up the problem?

@Lily: if you're finished, don't forget to accept the most helpful answer... :)

If the problem's quite long, the main site might be a better place than chat...

Okay, I'll type the problem and see if it's too long for here.

@mix: I have a good feeling you'll be part of the 10k club in at most two days...

If H and K are subgroups of a group G, then HK denotes the set of all elements of G that can be written in the form hk, with h exists in H and k exists in K. Find subgroups H and K of S3 such that HK is not a subgroup of S3. What condition can you impose on the group G to guarantee that HK is always a subgroup?

speaking of abstract algebra: are there any typical/easy examples of groups that are not converse-to-lagrange-theorem (so they have order a*b but no subgroup of order a), other than the first one, Alt(4)?

So far, I have determined that the group G must be abelian. HK = KH.

I

@Lily That one can be answered by just fiddling around with elements. Take H and K to be cyclic subgroups, and multiply them. (Your abelian condition seems like the most reasonable answer to the second part)

I'm hoping so, J.M. -

@mixedmath: I'm quite sure you'll manage it today :)

@Lily Lagrange can help a little too

What groups did you guys learn about in abstract algebra? cyclic, dihedral, Q8, A4, S4

@Jack: Looking at the Cayley table for S3, I can just multiply the elements?

Or rather see which ones prevent it from being a subgroup?

*sigh* I wish I could be more helpful... my only exposure to group theory is crystallographic point groups... so not that abstract.

@Lily Yup. If H = { (), (1,2) } then HK is going to be all k and k*(1,2) for k in K

Ah. Which elements of S3 would be in H and which in K? Or would they both have the same elements?

My first algebra class used lots and lots of Dihedral groups and symmetric groups (good to be really familiar with)

@Lily: try around!

Fortunately, there are only 4 subgroups of S4 (that aren't trivial)

so there aren't many options - that's why they can just say to do it

By do it, I mean find such an H and K without much intuition, perhaps

The trivial subgroups are the identity and the whole group itself, right?

i.e. brute force is very useful here... :D

@Lily: yes.

so mix means there are four others you need to look for... or should know already.

Well, I'm dealing with S3 which includes {e, a, a^2, b, ab, (a^2)b}

Do you know the permutation way of thinking of S3?

I've done permutations in stats, but not in abstract, yet.

Well, that or consider the symmetries of an equilateral triangle...

@Lily when I said "(1,2)" I meant "b" and when I said "()" I meant "e"

Oh, well then it might seem strange for us to immediately know how many subgroups there are - but you'll get to that later

oh, S3 is dihedral - yeah, cool

@JM I've been trying to learn more about those. Been reading a very good book. Title something like "Symmetry". It was hard going because it only handled 3D, and 3D is a little complicated. I'm actually trying to understand "Frieze groups", a nearly 1D type (with a little 2D allowed).

(and the long delay in reply is because I've apparently lost the book, and cant remember the author)

@Jack: Funny, I still haven't gotten the hand of permutations myself. :D I'm a chemist, so I'm more used to viewing groups as seeing what "moves" leave a shape invariant...

Ah okay. So when asked to find the subgroups H and K such that HK is not a subgroup of S3, I'm still a bit confused as to what exactly I'm looking for...what makes it non-abelian?

You should look at the product of ab and ba

that is, a*b and b*a

@Lily: if you take a look at it as a triangle, you'll note that reflecting and rotating don't commute...

@mixedmath what is a in cyclic notation?

(123)

I interpret a as a rotation, b as the reflection

a = (123) and b = (23) in my little world

Ah, I read a book called "symmetry," but it had very shallow touching of math.

i have a question since we're on this topic...

@JM I've been trying to introduce people to the tetrahedral group (rather than A4) lately. it helps to have several tetrahedra lying around, though a combinatorialist did show me how to make a tetrahedron from two pencils (and no paper)

@JackS: I have been using your JS bookmarks. They are wonderful.

Break them into pieces?

Oh okay. :) A_4, eh? I'm used to T_d myself...

The shock for most people is showing that tetrahedra can be embedded into cubes, which is apparently easier for people to grasp...

my friend knows a physics professor who's a particle physicist and i get an appointment. I told him that i was studying abstract algebra from artin. he said that if I wanted to learn the physics then I should get to representation theory and lie groups. he said his students usually started there, even non-math ones, as physicists are primarily concerned with representation theory, and the only finite groups in particle physics was z2 (parity,charge conjugation etc)

so i short circuited to a group theory in physics book after reading the chapter on groups from artin

does anyone know if physics programs do this regulalry?

i like the SU(2) books on the hydrogen atom

@jack any names?

that you might recall?

karen somebody i think for one of them, 2 of them were from the library. one was an ebook

Is someone willing to start another chat room with me?

@mix: Okay, I was pessimistic back there; you will get to 10k today.

@Lily: You'd have to explain what for.

Can I get 10K, too?

@anon It's for the problem that I posted above.

@yayu: Stephanie Frank Singer

@rob: assuming my extrapolation's valid, you need at least a month and a half...

Ah, get to 10K.

Or I could just continue asking questions here.

@Lily: that is perfectly acceptable.

coincidentally, i stumbled on the same book in the library.. though in the flap she said that she assumed the reader was familiar with math at the level of artin and physics at the level of feynman

Okay! So I have the group S3. and I have the cyclic subgroups of all the elements within S3.

What do I do next?

i have the physics covered to some extent.. but artin is a monstrosity to someone who has been taught "engineering math" most of his life

@J.M.: my rep gain is bumpy.

Hence the hedging word "assuming"... ;)

look at me, I'm dry on most days and killing it on others...

That has been my experience of late.

There were two other good ones. They both focused on SU(2) and SU(3) and used hydrogen a lot

@Jack: gassy?

explosively so

:-)

the library i have access to has this

Heh. Well, helium's a bit more of a headache to look at... so everybody's happy with hydrogen.

and i see she has written a couple

Anyone have an idea?

well, time for big bang theory and bed... thanks @john and @j.m for the comments on my problem.

Do both H and K have all five elements of S3?

Is anyone here?

@Lily S3 has six elements. I think you want H and K small. Try cyclic subgroups. Cyclic subgroup of a is { a, aa, e } (just keep multiplying by a until you get to e). Cyclic subgroup of b is { b, e }.

Oh, right. I have the cyclic subgroups.

What do I do with them?

The symmetry book was by Roy McWeeny. Not much luck finding the SU stuff.

@Lily: see if they work as the subgroups

@JackSchmidt I'm not exactly sure how to do that. I know ab^-1 must be in the group...

If we let H={a,aa,e} and K={b,e} then HK={a*b, a*e, aa*b,aa*e, e*b, e*e}. You look those up on the Cayley table and see if it could be a subgroup. If it is not, yay, you are done, as an example answers this problem. If it is a subgroup, then oops, try different cyclic subgroups.

@Lily consider the following elements...

(1)(23) and (13)(2)

construct subgroups

H={(),(1)(23)}

K={(),(13)(2)}

ok

Now find out HK

HK is the whole group.

no

I've never worked with numbers like that before

but in Jack's example, HK is the whole group.

HK={(),(1)(23),(13)(2),(123)}

ok

@Lily: yup, that is correct. HK has 6 different elements in it. H has 3, and K had 2. HK will have a divisor of |H| * |K| as its order

this is not a subgroup of S3

as it has four elements

ah

@Lily: So 3 and 2 was not a good choice, but yayu has found a good choice. His H and K each 2 elements. Then HK has 4 elements. But a group of order 6 does not have a subgroup of order 4 by Lagrang

i had just started watching big bang theory.. and your question came back.

oh so we're trying to get a number that does not divide 6?

@lily: i think that is a good strategy, yeah

ok so we could do <ab> and <b>

@lily: yup, exactly

So because 4 does not divide 6, HK is not a subgroup?>

By Lagrange, yes.

Also { ab, e } {b,e} = {ab*b, ab*e, e*b, e*e } = { a, ab, b, e } is missing a*a, even though it has a in it

(but lagrange is easier :-)

Oh, I did the reverse. {b,e} {ab, e} = {e, b, ab, a^2} ...but it's missing a, so that's why it is wrong?

@lily: yup. a^2 * a^2 is supposed to be in the "subgroup", but a^2 * a^2 = a is missing

Chemistry crystal question: in commons.wikimedia.org/wiki/File:Benzene-xtal-3D-vdW.png are the bulbous black things in the center 6 carbon atoms in an overlapping arrangement?

ahhh

or ab*b = a too, i guess

@Jack: Not really; it's just a model. :)

Why does it matter what's missing? (I know that prevents it from being a subgroup.)

@Jack: the truth is slightly more complicated, so for "space-filling" models, we pretend that the atoms are squished together to represent the bonding.

@lily: it doesn't matter much, i just like to be specific. once it has x and y but x*y, then it is definitely not a subgroup. lagrange is fancy stuff, multiplying ab and b is basic stuff. if you ever doubt what you are doing, try basic stuff.

@JM: i was looking at some "orbital" pictures too, but i couldn't figure them out either. do the electron move around between benzene molecules when it is in a crystal?

(if you don't doubt, then use lagrange, it is easy and clear)

Did you mean not x*y

Oh, for benzene, you have what is called "delocalization". The six carbons in benzene have a "swirling mass" (in quotes since it isn't rigorous) of electrons shared among them.

@lily: yes, sorry. been huffing benzene (pictures)

hah, okay, thanks

You have helped immensely. Thanks so much!

@JM: but the swirling mass in the crystal isn't suddenly 6 million carbons all sharing electrons, they mostly stay inside their own private 6-carbon swirls?

@lily no prob

Oh, no. Only metals do the "we share evrything!!1!" bit.

"Is benzene like copper?" might be my question

cool

silicon shares it a little, right? or is it only when it is doped?

With benzene, each electron has three electrons to contribute to the ring (the fourth one's stuck with hydrogen).

@JackSchmidt Silicon is way more complicated. :D

and doping is a further complication.

but i should imagine carbon is more or less nothing like copper

Yes, neither diamond nor graphite, nor buckyballs act anything like copper.

cool

Graphite conducts electricity like copper, but with a different mechanism.

In any event: the fact that there is no "we share electrons with everybody" in a benzene crystal is why electricity won't pass through it; solid benzene is an insulator.

Maybe you'll jiggle the electrons within the rings, but that's it.

@JM: thanks. Is there anything besides electrical charge that makes sure benzene lines up into its crystal shape? No magnetism or "spin" or anything?

(ice is formed by electrical charge, right? hydrogen of one H20 attracted to the oxygen of another, right?)

(trying to leverage my 8th grade chemistry to understand crystallography, thanks for any help)

"lines up" - that depends on how you solidified it. If you solidify benzene too quickly, you end up with something like glass (amorphous).

i.e. you didn't give them enough time to line up properly.

Okay, I'm off. Thanks again. I'll probably be back soon. :)

@JackSchmidt Well, the oxygen shares its electrons with the two hydrogens

for ice, you have water arranging itself so that the oxygen of one molecule is not too near that of another one.

which nets you the hexagons you see in snowflakes...

oh, more of a + hates + thing, rather than + likes - thing

More or less. :) Oxygen's electron-rich, so it stands to reason why the oxygen end of a water molecule will try to keep away from the oxygen of another one.

Also, since the oxygen tends to draw electrons away from the hydrogens it is attached to, the hydrogens are slightly positive, which makes them more attractive to the oxygens of neighboring molecules.

haha, ok, apparently I learned it all wrong. Oxygen has a negative charge in H20?

big atoms steal electrons from small atoms?

It isn't about "big" or "small", there are elements that like giving, and there are elements that like taking.

oxygen is one of those takers.

Are takers on the right hand side of the periodic table, and givers on the left? (and inert on the far right?)

Yes. Your chemistry isn't that bad if you remember that. :D

(If we have to be technical, we say the ones on the right are more electronegative.)

and fluorine is even more of a bigger "taker" than oxygen.

ok, this is good so far. back to reading crystallography. Thanks JM!

no worries. :)

Anyway, I must leave for work. See you!

@Jack: One last thing before I really leave... if you're serious about pursuing the chemistry-group theory connection, you'll need to look at this book. The notation might not be what you're accustomed to, but I think translation won't be too hard.

...and now I really have to go. :)

what's up?

depends on your coordinate system

True

yayu came into the room, so I was greeting

Sure. 20^(1/2+1/4+1/8+...)=20^1=20.

or prod 20^(2^-i) if you want

it equals the first

@tb: are you here?

phew, that was interesting a bit

@Gortaur Now I'm here. What was interesting?

@tb interesting was this question: it seems easier from the first glance. I hope, I gave a correct proof now - but I'm not satisfied with its complexity, especially in the notation: math.stackexchange.com/questions/72247/…

oh, you've disappeared

I still have a question to you )

yes?

so have you done your PhD in ETH Zurich?

Yes.

Willie is in Switzerland right now.

Are you guys meeting for drinks?

I know. No I'm in Zurich and he is in Lausanne (about 3 hours by train). No, we haven't met IRL

Oh, shame.

I have an opportunity to go for 2-3 months to the other university during my PhD

Okay? And you're free to choose where to go?

my supervisor wishes to send me to Berkeley while I myself was considering ETH to be honest

Whom would you want to talk to in Zurich? Why does your supervisor think that Berkeley would be better?

I mean, he was studying in Berkeley and Stanford so he is amazed by these universities. On the other hand he is more focused on hybrid systems and formal verification though with a lot of stochastics

when I've started working with him I just switched to the general case which also catches hybrid models - so he is happy about it

in Zurich there is a group of J. Lygeros within the same project as us, so that was my first idea

I see. Let me put it this way. If you have contact to some people, ETH may be very profitable. I don't know people from electrical engineering so I don't know how the culture is there.

on the other hand, in my research I almost finished with what I wanted from the discrete time. Now I have 3 years which I would like to devote to the continuous time stochastic processes - that's why I think some months within strong stochastic school will be an advantage

that's my question. I am not a engineering person

since you're experienced in stochastic stuff I thought that maybe you can tell me if you know any people at ETH of that kind

Oh, I have no idea about stochastics, to be honest. Most of the people working on stochastics at the math department at ETH are working in Finance, as far as I know.

that's very good in fact, I've done MSc in Financial Mathematics and still have some interest in it

my problem is that at my department I'people are doing systems and control stuff

my supervisor is cool in the sense he gives me a lot of freedom and happy about my results - but direction of my research now goes quite outside of his area

Well, here's the list of people working in probability and stochastics: math.ethz.ch/assistant_groups/gr3

I'll be back in a few minutes.

@tb thank you

@Gortaur No problem. I don't know whom to contact for an exchange. But consider the following: your supervisor probably has some contacts in Berkeley, so you'll immediately be integrated in some group and knowing some people from the beginning may make your stay much more profitable and enjoyable. Just a thought.

@tb: sorry, he just called me for a meeting (

@Gortaur no problem. we can chat later. see you

@tb you see, I have the same thoughts and also my supervisor told me that going there it wouldn't be a problem because he has contacts. Unfortunately, my research interests are very different with their since they're mostly from Computer Science. Moreover, I have now about 1-1.5 years to arrange this trip so I think it is reasonable to try to find the school where I maybe will fit better

thank you very much for pointing to the right group in ETH, I think I shouldn't bother you more

@Gortaur That's really no problem... I don't really know whom to contact at ETH for exchanges. I always had the feeling that as soon as you know somebody things become far easier than taking the official administrative road. Maybe you met somebody at a conference whom you could contact, maybe somebody in Lygeros's group is interested in your work or projects or knows your supervisor and could help you along. That's about all the general advice I can give.

@tb I see your point, thanks for the advice.

I also mentioned that I happen to reply to you just after your gravatar disappears from the chat list )

yeah, you've got good timing, man :)

btw. where do these minimalistic emoticons come from? never seen them except from you

@tb Perhaps he has no eyes.

Also, nicely done on bumping that thread in meta.

1. heh :) 2. I hope the comments by ShrevatsaR make my point...

@tb sometimes we use it at least in runet. as for me, ) refers to the moderate smile )) to the ordinary one.

)))) and ))))<put some more brackets here>)) refer to the funny joke

I gathered as much, but thanks for the explanation :)

'I gathered as much': sorry, didn't understand

btw, as for me I don't see any reason to give eyes to emoticons. emotions are blind )

Sorry: except for the runet part, that's what I figured out myself

got it. but how do they blink? ;)

blindly? )

for the blinking emoticons I use ;) as well or sometimes $) when do a misprint

shift+4 corresponds to ; in russian keybord

btw, am I the only person here who see crossed 69 in $ ?

never thought about dollars this way... Maybe a crossed cancer ♋ ?

man, forgive me for that, but I see 69 in the cancer sign as well )

so $ mean 'no to cancer'?

or 'no to 69'

no smoking, maybe ? odd, afaict the tobacco industry is all about $$$

maybe. anytime you see this sign it implicitly tells you not to smole

I start to editing 50% of my posts. do I repeat you?

seems so. I'm always a little fast when it comes to clicking submit or send

with papers as well? )

maybe. but usually I let them lie around long enough that I'm quite happy with them when I submit. Then the reviewing process takes so long that I'm disgusted and would prefer to rewrite the paper from scratch...

I will submit my first journal one in a couple of days, so we'll see how it goes

good luck!

thank you!

I also rarely use '!' since it seems to me that I shout that way in chat

ah I remember a deadline you mentioned a few days ago. That should be around now, shouldn't it?

If I'd shout IT WOULD LOOK LIKE THIS but less civilized words, probably

Shouting is fun.

@tb yes, but it was delayed by one more week, so with regards to the conference paper, I have some time to read it and fix some typos. Journal one I will also read on this weekend for the last time. I almost finished the second journal paper on financial applications - need to look for the journal to send it

figuring out where to submit is surely the hardest part of the submission process...

I wonder when I will finish this stuff and finally can focus on research and studies. I promised to myself to recall continuous time stochastic processes, analysis on manifolds. Learn modern algebra a bit and read all posts by @Asaf about AC )

Good luck with that :D

you won't finish. he writes faster than I can read.

Haha, I write fast and if there's really one thing I know enough to write good answers about it is axiom of choice.

well, sometimes I'm getting a bit tired of this whole choice debating on this site. it's fine with me if the question asks about it, but oftentimes it looks like beating a dead horse to me (no offense intended)

I can imagine ) but yeah, I mostly would like to put effort on algebra - in my university I was told only about matrices and linear stuff (and I thought, such a bore). As I can see from MSE, algebra is quite different, they lied to me

I'm pretty much with Theo J-F here: meta.mathoverflow.net/discussion/381/…

Theo, I completely agree really. I do think it is important that it is being used.

In my freshman year the professor who usually heads the introductory course in set theory took a sabbatical. So my teacher (who's now my advisor) changed it a bit and added that a countable union of countable sets is countable.

Recently I had a conversation with a friend from my class, and he was surprised to see that the axiom of choice was involved.

yeah, that Fefferman-Levy model where R is a countable union of countable sets is a complete shocker

I completely agree that the axiom of choice is now taken for granted almost as much as the power set and union axioms. It means that it's fine to use it blindly, but it is also important to make note where you do use it (in general, at least) since people unexperienced with these things usually think things are constructive when they are not.

You think that's shocking? the Gitik model in which all uncountable aleph numbers are countable unions of smaller sets... now that is shocking.

well, but look at when the things were done. Fefferman-Levy was back in the early sixties!

I know.

Gitik model was from 83 or so.

I agree that Gitik is shocking as well, but given how forcing exploded in the twenty previous years, I still think the surprise at FL was bigger...

The problem I have with this choice thing is that there are canonical places where people insist. Quite often you see them insisting on bases, maximal ideals, Hahn-Banach, etc. But if you look a bit closer, they have no problem taking it for granted that the reals are separable.

or that various definitions of compactness coincide for metric spaces

I'm sorry for the dummy question - but if any fact from the theory of limits of sequences in R does not holds without AC?

like, any bounded monotonic sequence has a limit

oh, I forgot to address it to you, @Asaf

sorry I should have said subsets of the reals are separable

I find this example quite telling: if you know that x is such for every \epsilon there is a \in A such that |a - x| < \epsilon. Now how do you conclude that there is a sequence from A converging to x?

hm, for each n choose a(n) such that |a-x|<1/n

exactly. You know that A_n = {a \in A : |a - x| < 1/n} is non-empty. To say that there is a sequence (a_n) with a_n \in A_n and thus converging to x is equivalent to asserting that the countable product of the A_n's is non-empty.

but sorry, I should go now. I'll be back in a few hours.

good luck

There is a model in which a subset of the reals is inseparable.

In this model continuity by functions is not equivalent to continuity by sequences.

@tb It's shocking, yes, but remember that the technique they developed was based on permutation models which was well known by then. I do agree that it was completely new. It's fun though. I hope to someday develop such method, it is gonna be fun on the bun!