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12:02 AM
@Mike, Prof. @Ted \o
 
Hello
 
Good night.
 
bon nuite
 
@DanielFischer thanks for your help but this argumentas are new for me, and I have a bad time. I'm better ask the question the truth no longer want to bother you with my stupid questions
 
12:33 AM
Wait @MikeMiller that's not all we need for continuity, is it?
 
for continuity of what
 
We have xy, and x+y, but then how do we construct xy^2?
polynomials
 
let's say $f(x,y) = xy, g(x,y)=x+y$
those are continuous
your latter function is $f(f(x,y),y)$
 
Sure.
Wowe.
 
So (inductively) all monomials are continuous
And then (inductively again) all sums of those (e.g., polynomials) are continuous
 
12:48 AM
<3
 
12:58 AM
@MikeMiller Miiiiiiiike. I have another queeeestion.
 
Are you familiar with Kuratowski's axioms for a closure to specify a topology?
 
sure
 
I'm trying to show that those in fact specify a topology.
And I'm not sure how to go about showing that the arbitrary intersection of closed sets is closed.
 
let $C_i$ be closed; then we want to show $cl(\cap C_i) = \cap cl(C_i)$, yes?
Actually, show me where you're stuck
 
1:06 AM
I just wasn't sure how to go about it.
I mean, I said what I had.
So.
 
Well, you have one inclusion immediately, right?
$\bigcap C_i \subset Cl(\bigcap C_i)$.
 
Oof give me one moment while I think. Sets are blowing my mind.
 
Tell me your axioms.
 
The set and the empty set must be open sets
Wait
wrong ones
cl(empty)=empty
$cl(a) \cup cl(b) = cl(a\cup b)$
c$l(cl(A))=cl(A)$
$A\subset cl(A)$
 
Look at that last one again, and look at the thing I just said
Hurry, naptime is approaching!
 
1:14 AM
Oh sorry, you can go. I'm trying to wrap my head around things.
Those do look very similar though.
 
$A$ is just... a set. I picked a set, and invoked the axiom.
 
Beautiful.
So then the reverse inclusion.
 
OK, so you want $Cl(\bigcap C_i) \subset \bigcap C_i$.
In particular, you want $Cl(\bigcap C_i) \subset C_i$ for all $i$.
Any ideas?
 
Wait why the second line?
 
Definition of intersection.
 
1:17 AM
Oh.
 
Does "Oh" mean "Oh, I see it."?
 
No, it means I get what you meant there, sorry.
 
Okie doke.
 
I'm thinking.
I mean the only useful thing we seem to have is the union.
Which we could flip with demorgan.
 
Don't even think about the axioms for a second. Look at that second line. Say: "If we had another axiom, then we'd have this..."
Then prove the axiom you want to use.
Remember: $C_i$ is closed!
 
1:19 AM
Well I mean we want to show $cl(\cap cl(A_i)) \subset cl(A_i)$
Yeah?
 
Sure.
What can you say about the stuff inside those?
 
Must be the same
 
(What's almost the most trivial thing you can say?)
 
Hurrr
 
1:20 AM
Oh sorry.
Must be contained.
 
The first is in the other.
 
egm;pwerg
 
what is the name for the irreducible polynomial an extension field is created with
 
There's the 'lemma' you want: if $A \subset B$, then $cl(A) \subset cl(B)$.
 
like, $E=F[i]/(p)$, what do we call $p$?
@MikeMiller these are finite fields
 
1:22 AM
Which I'm sure I can prove.
Thanks @MikeMiller.
 
i've never heard terminology for $p$.
 
so in this context that's fine
 
@Anthony Aye
@AlexanderGruber i should have assumed. i forgot you don't believe in infinite sets.
 
@MikeMiller i sure don't
 
@AlexanderGruber do you believe in finite sets of unbounded size?
 
1:24 AM
Wait @MikeMiller You said show that $\cap C_i \subset cl(\cap C_i)$
 
That was the first thing we did, @Anthony
 
@MikeMiller Yeah - so, I believe in countable sets, anyway
 
That's not the same as showing that the intersection of closed sets is closed, is it?
 
I just don't dig on $\mathbb{R}$
 
Oh.
 
1:25 AM
$\mathbb{Z}$ is ok
 
Yes it is.
It's just more specific.
 
@Anthony We want to show $\bigcap C_i = Cl(\bigcap C_i)$, aka, that $\bigcap C_i$ is closed.
 
But it could be the closure of anything.
 
@AlexanderGruber why should you need to believe in infinite sets? it seems perfectly reasonable to me to not be willing to take infinite unions.
@Anthony "A is closed if it's the closure of something" is equivalent to "A is closed if $cl(A) = A$"
 
@MikeMiller I just feel that most of the things I consider reasonable to think about can be approximated to an arbitrary order of magnitude using finite or countable sets
 
1:26 AM
if the former is true the latter is true by the idempotence axiom; if the latter is true, it's the closure of itself, so the former is true
 
googles idempotence
 
@Anthony $cl(A) = cl(cl(A))$
 
I see.
 
(Idempotent means $x^2 = x$)
 
Hmm.
 
1:28 AM
once we start believing in infinite sets we start getting into stuff where I read theorems but i don't care whether they're right or wrong, because the exceptions are goofy. the whole AC thing with well-ordering is a good example.
 
goofiness is cool
which is why i think finitists are cool
 
I need zorn's lemma to feel safe at night.
 
@anon With finitism, you too can feel secure in your effortless, absolute understanding of Zorn's Lemma.
 
Hi, everyone, a challenge to your IQ on a website: Which one of the numbers does not belong in the following series?
2 - 3 - 6 - 7 - 8 - 14 - 15 - 30
 
six - unlike the rest, it's not prime
 
1:33 AM
8 is the only noncubefree one
calls it a day
 
@Victor 2 is the only one that's equal to 2
 
IQ tests are perfect for measuring one's ability to take IQ tests
 
@AlexanderGruber why should I be able to well-order a finite test?
 
because then how will students know which problem to do first?
 
i mean, i'm not asking mathematically, this is a moral question
boo
 
1:37 AM
@MikeMiller well i mean
 
the definition of "finite" is "in bijection with $\{1, \dots, n\}$", which is cheating
 
pick any element and let that be the least one
 
it should be "not in bijection with aproper subset
@AlexanderGruber how do you pick an element?
 
then take it out
@MikeMiller your choice!
 
that's silly
 
1:37 AM
If you have a set you can't take elements of you're probably doing something goofy
like working in a hamel basis
 
but why can't i do goofy things with finite sets?
 
ewww
 
@MikeMiller it's more about the ability to avoid accounting for goofyness. :p
 
I get the feeling nobody but me is enjoying the pedantry, so I'll stop
 
@MikeMiller Sorry to bother you again, how do I know $ cl(A) \subset cl(B)$ implies $A \subset B$?
 
1:40 AM
@Anthony Because within the next 5 minutes you're going to prove it
Wait, no, what?
You're trying to prove $A \subset B \implies cl(A) \subset cl(B)$
 
Oh woops.
I thought we assumed that.
:D
 
what did you want the first thing for?
 
@MikeMiller i have, however, been attempting to branch out
by fixing analysis
 
oh dear
 
I didn't.
 
1:42 AM
ok
 
my secret goal has been to develop a framework in which one can analysis at only reasonable things. An analysis theory where you can't have ridiculous things like cantor sets or Weierstrass functions
 
i'm fairly certain those are called smooth functions
 
@MikeMiller i'm not sure. i can't decide whether I like the step function or not.
 
sigh
 
I guess you can tanh it into something reasonable
it's in the formulation stages
 
1:46 AM
Thanks Mike.
 
someday i will write an amazingly belligerent monograph
 
i hope you get a permanent job before you publish belligerent monographs, because I imagine those can only bode ill on one's career
 
@MikeMiller the joys of tenure.
 
2:34 AM
Don't be too belligerent, @Alex. Remember that you like some of us continuous people!
 
2:47 AM
Species are supposed to categorify combinatorial proofs, but they only do so for "natural" proofs. For instance, the species Lin (linear orders on a set) and Perm (permutations of a set) are not naturally isomorphic, but obviously |Lin X|=|Perm X| for all X via an "unnatural" proof: pick a "basepoint" in Lin X, etc. Notably, Lin x Lin and Perm x Lin are naturally isomorphic. One recalls the notion of an affine space as an originless vector space.
Wonder if the (categorical or model) nature of unnatural combinatorial proofs and "arbitrary choices" has been studied in more detail.
 
 
2 hours later…
4:34 AM
 
Does $1 + 2 + 2^2 + \cdots + 2^{ak-1}$ have some sort of obvious factorization I'm missing if a is an integer and k is in naturals?
 
like the geo sum formula?
it's $2^{ak}-1$, which is divisible by $2^a-1$ and $2^k-1$ for instance
and $2^{{\rm lcm}(a,k)}-1$ if you want the most bang for your buck
 
my whole point was to prove $2^{ak}-1$ had two factors given what I know about $a$ and $k$, and I was told I could do that by expanding the series
and I didn't see how writing it out helped me
nor do I recognize what you invoked to realize $2^{ak}-1$ was divisible by $2^a-1$
For if I did then I would divide it by that and have my solution along with the result of the division
although admittedly this is an algebra question
 
4:49 AM
@anon Can you help me with a hw question?
It involves a fair amount of algebra, and I'm rusty.
 
@anon Do I need to invoke the division algorithm or something?
Am I missing something incredibly obvious?
 
@Anthony okay
@GBeau do you recognize that $x^m-1$ is divisible by $x-1$?
 
@anon no
 
$(x^m-1)=(x-1)(x^{m-1}+x^{m-2}+\cdots+x+1)$, it's the geometric sum formula yo
 
after I show this division, my goal is to then also show that the result would be an integer as well
oh, yes
 
4:54 AM
invoke $(x-1)\mid(x^m-1)$ with $(m,x)=(a,2^k)$ and $(k,2^a)$
 
I just figured it would become useless since $(2-1)=1$
 
@anon it's the 3rd problem here
 
@GBeau I never said $x=2$ :-)
 
that definition of JK isn't correct
 
4:56 AM
Really?
 
JK is generated by the set {ab} of products, but is generally strictly bigger than
 
I see.
Well still, I think we can go on with the question.
 
sure
haha, you have to turn that in tomorrow?
 
Yeah, although part a is optional.
I spent most of today doing the first two questions... I'm not very good at this stuff. :/
 
@anon Oh, I see the way now...(the thing I'm proving is if $2^n-1$ is prime, then $n$ is prime)
the geometric series is easily shown inductively
and I believe resolves the proof
 
5:02 AM
@Anthony part (b) doesn't require me to wake up any sleeping parts of my brain, so I'll help you with that
 
:D
I'll stew over a later.
So the intersection of the ideals is just going to be a product of primes, no?
 
pretty much
 
All the primes in the subset A?
 
mmhmm
 
So why is there no metric for that?
:/
 
5:06 AM
you have to describe the topology first
hmm, I think it is missing the condition that A must be finite
 
Well the topology is just all ideals generated by primes... right?
 
huh?
 
Er.
Oh that's the closure.
 
let A be a set of prime ideals of Z. what is the intersection of the prime ideals of A?
 
So an ideal that's just $ap$, where p is prime, is closed right?
 
5:10 AM
what's $ap$ mean?
 
Let me just try to answer your question.
The intersection of the prime ideals is the ideal with elements that are $p_1p_2....p_i*z\in \mathbb{z}$ right?
 
$\bigcap (p_i)=(\prod p_i)$
 
@anon Since $a\in\mathbb{Z}$, my proof is not resolved since $2^k$ may not represent an integer and my factorization may not represent an integer factorization?
(consider $a<0$...)
 
Yeah.
So then it's complement is open?
 
@GBeau if $a$ is negative then your sum is not an integer. what are you trying to do in that case?
@Anthony what's complement?
so for instance $(p)\cap(q)=(pq)$ and $(p)\cap(q)\cap(r)=(pqr)$, if $p,q,r$ are distinct prime numbers
 
5:14 AM
That is a closure of the set right?
 
@Anthony what set?
 
A
Oh all prime ideals that contain the intersection.
 
@anon Here's my proof so far so you may see what I have done:
 
@Anthony the intersection of A={(p),(q),(r),...} is (pqr...). What prime ideals contain this ideal?
 
My idea was to plug in $ak$ for $n$ and use that to show the theorem
 
5:15 AM
@GBeau proof of what? answer my question: if your sum is not an integer, what are you trying to prove about it?
also, I don't see why you're letting a be negative, that has no bearing on the original problem
 
@anon the elements of A, and all combinations of ideals in A?
So like p, q, r, pq, pr, qp...?
 
@GBeau the contrapositive is that if n=ab with a,b>1 then 2^(ab)-1 is not prime, which is easy: it's divisible by both 2^b-1 and 2^a-1, both >1
@Anthony What prime ideals contain (pqr..)? :-)
read carefully!
 
@anon I wasn't sure it was possible to show the full contrapositive by only considering a,b>1
 
@anon I still really don't have a good intuition for prime ideals... I've never learned about them.
 
@GBeau "only" considering that? either a natural number is prime, or it's a product of two things >1
@Anthony the intuition you want here is one for ideal containment
$(\ell)$ contains $(pqr\cdots)$ iff $(\ell)$ is one of $(p)$, $(q)$, $(r)$, $\cdots$
 
5:20 AM
oh I see...I'm silly...
 
Well that answers the question you asked then...
 
@Anthony so assuming A doesn't contain (0) and is finite, the closure of A is just A. what happens if A contains (0) or has an infinite number of prime ideals?
 
If it has 0 isn't it still just 0?
Oh wait.
My head.
 
"still" ????
 
It's everything.
Right?
 
5:23 AM
yes
 
And if it's everything, it's nothing?
 
what's the intersection of an infinite number of prime ideals of Z?
(more pointedly: what's the only number that has an infinite number of prime divisors?)
 
wait.
 
there is one such number
 
Just 0.
Eh?
 
5:24 AM
mmhmm
so the intersection is (0), and every prime ideal contains (0)
hence the closure of A will be everything again
so the closed sets are Spec(Z) and finite collections of nonzero prime ideals
(I like to think of this as "cofinite mod (0)," because if you forget about (0) this is the cofinite topology)
 
I don't know what that means, I've basically only learned about groups as far as algebra goes.
And I'm just starting topology!
 
finally, we want to show it's not a metric space
in metric spaces, any two points have disjoint open nbhds. is that true for Spec(Z) with the zariski topology?
(why or why not)
 
What is the zariski topology?
Oh what we defined.
 
the topology you just defined
or the author
or really, some guy in history
 
:P
So points in this case are... Rings?
 
5:29 AM
points in this space are prime ideals
which correspond to certain rings (localizations of Z at prime ideals)
this is the scheme POV
 
But so you asked if any two points have disjoint open nbhds.
 
mmhmm
 
Don't I need to know what open sets look like then?
 
yes
if proper closed sets are finite, then nonempty open sets must be cofinite. can two cofinite subsets be disjoint?
 
Proper closed sets are finite?
Just because it's some prime ideals?
finitely many?
What does cofinite mean?
 
5:33 AM
if A is finite and doesn't contain 0, then cl(A)=A is finite. if A is not finite, then cl(A)=Spec(Z) is not proper.
a subset is cofinite if it is "all but finitely" many elements of a set
IOW the complement of a finite subset
 
I see.
And I suppose two cofinite sets cannot be disjoint, else they wouldn't be cofinite.
 
two cofinite subsets of an infinite set cannot be disjoint, and Spec(Z) is infinite
 
I see.
Thank you @anon.
 
wow, apparently there is a classification of multiply transitive groups. I wonder why that doesn't get more attention.
 
1
Q: What is a good word that means "group of groups?"

IsaacBolingerIf you have a bunch of groups of different things, and you group the groups into a bigger group, what do you call the bigger group?

 
5:46 AM
all of those suggestions are stilted and would sound weird in practice
 
class please divide yourselves into groups of five
 
Huy
6:21 AM
Good morning everyone.
 
hi pal
 
 
1 hour later…
7:26 AM
Mystical morning everyone
 
Greetings
 
7:48 AM
I was thinking these days that if I managed to publish a book, then I need to do it in such a way it shouldn't be taken as a contest between book. Ovidiu already has a very nice book, and it will remain like that no matter how many books will be published out there.
$X$ book is better than $Y$ book or $Y$ book is better than $X$ book is already a dangerous area, one might remain isolated, alone. I prefer to cooperate with people not to be in a competition with them.
 
good idea
 
@IceBoy Behind each book there is a crazy effort, and like the clothes, $X$ book might fit better for some and less for others especially if we consider the difficulty level of the problems in the book, the knowledge required.
 
I agree.
 
I like the very crazy problems ... and my tendecy is to add harder stuff in my book.
 
8:06 AM
well then, you are writing for readers like you
 
For instance, I'm thinking now at Au-Yeung series, it's not an easy series, it's actually a difficult one, but the good news is that some research might reveal ways incredibly easy. Then, even if you have harder stuff in your book, adding there some revolutionary solutions is just a very welcome thing in my opinion. You know, the reader might say at first sight "This is a crazy hard question" and then, when he sees the solution "Wow, that's such an easy way!".
 
@Chris'ssis: very nice
 
@robjohn Yeap :D
 
@IceBoy an organization?
 
@robjohn that works
 
8:15 AM
I think each question & answer must be as a good lesson to learn, and this is the hard part of the book. Repetitive ways of solving problems (in your book) is not that helpful for a reader that wants to learn a lot of techniques.
Each question should have something special from which you can learn something, that "wow" element one shouldn't think of at first sight.
 
would you say there is any difference between a technique and a method, in terms of what is learned?
 
"In general method is the general or specific way in which an activity is conducted, while techniques are the various methods and processes developed through knowledge, skill, and experience."
I'm focused on techniques that I got through experience. Moreover, the problems I wanna add I doubt they can be approached by some well-known methods. If we talked about a textbook, well, then we could talk more about methods.
 
So, the primary objective of your books is...?
 
@IceBoy Bringing in front of the reader new problems, very nice ones, that are solved by new techniques. Of course, I'll also have some known problems that are solved by new techniques.
 
Have you written down a list of your primary, secondary, etc. objectives?
 
8:25 AM
This is also my dream when I buy a book, to see something I've never ever seen before, a book with amazing problems, amazing solutions, full of that "wow" element.
@IceBoy Not yet, but I have these things very clear in mind.
 
@Chris'ssis Along with a title for each lesson you want to present.
 
@IceBoy At the moment I only work on some problems I wanna add to my book, so I didn't establish yet other details of the book. This will happen later on.
 
@Chris'ssis How do you plan to present the answers in your book?
 
@rehband As Ovidiu did, in a very concise, clear way.
 
@Chris'ssis Very good. Will every problem come with an answer?
 
8:32 AM
@rehband I think I'll also have some open problems. On the other hand, where is possible, I can refer to results alread obtained from other problems, like some generalizations.
(as Ovidiu did)
 
@Chris'ssis Gotcha, sounds good!
 
@rehband :D
 
 
2 hours later…
10:28 AM
hehe, I'm working on some marvellous things here ... (thinking to create something new though)
 
10:55 AM
@Chris'ssis Hi!
 
@Anastasiya-Romanova Hi
 
Are you busy?
 
@Anastasiya-Romanova Working on my book. I didn't have time to attend that question of yours if it's about that.
 
@Chris'ssis I'm just asking. I won't bother you about that
 
@Anastasiya-Romanova You can ask anything, and when I have time I try to respond.
I'm a bit involved in some weird formulae right now.
@Anastasiya-Romanova like this one math.stackexchange.com/questions/925092/…
 
11:00 AM
@Chris'ssis I hate double sums
 
@Anastasiya-Romanova Oh ... :-(
 
@Chris'ssis Sorry, I can't help. 2-3 years again maybe
Okay, since you're busy. I won't bother you again.
 
@r9m. I hate fillers :(. The last Naruto filler was not worth my time :(.
@r9m. However, it seems that they were trying to replicate the androids of Dragonball Z :D. That was kind of nice. Though, boring none the less.
 

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