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12:18 AM
hi @Ali
 
hi @ZachHauk how are you?
 
tired and stressed
you?
 
not doing too bad
happy with maths atm
 
:P
 
why are you stressed?
 
12:20 AM
i have a lot of things to do
 
school work?
 
yes
life sucks
 
not it doesn't
it just requires you to convince yourself to be happy
 
@AliCaglayan Just studying commutative algebra
 
@Adeek nice, i started hartshorne this week
 
12:28 AM
@AliCaglayan and quadratic forms I would like this semester to build very good understanding in commutative algebra because I want during the summer to study Category theory through Allufi chapter 0 and also algebraic geoemtry.
oh cool how is it btw @AliCaglayan ?
 
I am loving it
 
I am gonna start working through it in the summer.
 
the algebra is slightly intense but there is a good appendix
 
once this semester ends
 
ya'll are too smart for me
i'm still studying galois stuff :(
 
12:29 AM
I finally get why its called algebraic geometry now
@ZachHauk galois stuff is cool
 
you're studying AG?
that's my topic of interest
 
@AliCaglayan Yeah I want to have perfect understanding for commutative algebra. Also, I am slowly moving in allufi chapter 0.
I am in page 131 now.
But I can't do too much because I have to study for my subjects this semester + TA work
 
@Ali are youundergrad?
 
@ZachHauk yes
 
but my goal is to finish allufi chapter 0 during the summer. I am very excited to go to spectral sequence stuff and derived categories.
 
12:30 AM
waht yaer?
 
sorry
i can't write
 
I want to maybe work in mirror symmetry + quadratic forms @AliCaglayan
 
@Adeek I met someone working in mirror symmetry in the past
same day I met Reid
 
@AliCaglayan I know one prof working in mirror symmetry and my other prof working in quadratic forms.
 
12:31 AM
a lot went on that day and I remember so little
 
I think I will ask the other guy to also supervise me as well. This mirror symmetry stuff seems very interesting.
 
So I am looking at Nullstellensatz today
The interplay between geometry and algebra is amazing
 
very cool
 
affine algebraic sets V <--> coordinate rings k[V]
 
cool
 
12:32 AM
points of V <--> maximal ideals of k[V]
affine algebraic subsets in V <--> radical ideals of k[V]
subvarieties in V <--> prime ideals in k[V]
 
oh
very cool.
@AliCaglayan you might be interested in my blog btw
 
and morphisms f : V -> W correspond to k-algebra homomorphisms k[W] -> k[V]
@Adeek throw us a link
 
it been 2 month since I have updated it, but I will update it this week.
Maybe thursday or friday.
 
@Adeek I'll give it a read this week
I have come up with some geometric images for thinking about the correspondances
 
cool.
 
12:35 AM
you can think of affine algebraic sets as some union of circles
think of a venn diagram of 3 circles
 
have you heard of this derived categories @AliCaglayan ?
 
this corresponds to some coordinate ring
 
It seems very interesting
 
@Adeek I have heard of them however I know very little about them
affine algebraic subsets you can think of as 2 of the 3 circles
 
oh I see.
 
12:36 AM
subvarieties you can think of as 1 of the circles
 
oh
btw I was reviewing some stuff in commutative algebra. Maybe in your knowledge of geometry so far you can answer it to me.
 
go on
 
So we used zorn lemma to prove $Rad(R) = \cap P$ where P is taken to be prime ideal.
Is there a geometric way to see this ?
I mean inclusion $Rad(R) \subset \cap P$ is obvious.
but is there geometric way to see the inclusion $\cap P \subset Rad(R)$ ?
here Rad is the radical of R
that is the ideal whose elements are nilpotent elements.
 
hmm I will think about it
 
could you guys maybe talk about something i understand
/s
This looks like commutative algebra, with Radicals and Nilradicals and prime ideals and what not
 
12:44 AM
well we are trying to draw a picture about it
 
How about we talk about something everybody understands, like why can you never divide by 0 in the set of real numbers :P
 
0 doesnt have an inverse
ya goober
that applies to all fields <3
 
Why doesn't it have an inverse?
 
because it can't multiply with anything to give 1
 
@skullpetrol cuz it's a field...
 
12:50 AM
Exactly. @AliCaglayan
 
who was that "exactly" to?
:[
 
yeah @AliCaglayan
yeah @ZachHauk this is commutative algebra with Radical being the set whose elements are nilpotent elements
 
someone talk to me about maths I've already studied
 
I just did :P
 
@Adeek I am not really sure how to think of the nilradical
Geometrically that is
@Adeek actually the nilradical is going to be $\operatorname{rad} I/I$
that is the nilradical of k[x]/I
So for some algebraic set V
the ideal of V is I(V)
the nilpotent radical is rad I(V) / I(V)
 
1:00 AM
very cool @AliCaglayan
 
@Adeek try asking on the main I don't think I can come up with a better picture
 
okay @AliCaglayan
0
Q: Nilradical geometrical way of thinking about it

AdeekI am following Michael Atiyah commutative algebra. In one of the proposition he proved $nil(R) = \cap P$ where P is prime ideal. One can easily see the inclusion $nil(R) \subset P$. However, for the other inclusion we used Zorn lemma. I understand the proof, but I don't really get it conceptually...

@AliCaglayan
 
now we wait for a wizard
 
haha
@AliCaglayan do you have fb btw ?
facebook *
 
1:11 AM
Can I add you ?
 
sure
my name should still be the same hopefully
 
Is it same photo ?
 
I am the one with the dog
I hardly use facebook however
only to occasionally check whose who
 
oh
 
do you have my email?
 
1:13 AM
yeah
let me know once you get it
 
thats probably the better way of communicating
 
okay cool
got it ?
 
@Adeek I have your email
I thought you were talking about fb
I kept refreshing lol
just to make sure I sent you one
 
oh okay we can just discuss stuff through email then :D
 
im back!
not that anyone cares
 
1:21 AM
hi @ZachHauk
 
art is soooo boring
 
I am gonna go back to studying brb I will wait for an answer on mse :D
cya l8er guys @AliCaglayan @ZachHauk
 
alright bye
 
1:44 AM
@skullpetrol it's just me and you
 
whaddya want to talk about
 
I gotta run sorry
 
:(
i thought we were having a moment
 
:(
next time pal
 
1:46 AM
lol
 
2:17 AM
The One (Polynomial) Ring, found it on my hard drive:
10/10 very good quality I know
 
2:46 AM
niemann ist hier lol
 
@DHMO Oh no, the incomprehensible German text is back runs and hides
 
@LegionMammal978 lmao
 
3:15 AM
Challenge/exercise: find a real function $f$ such that $\lim\limits_{|x|\to\infty} f(x) = 0$ and $\limsup\limits_{|x|\to\infty} f'(x) = \infty$
 
3:34 AM
2
Q: Do all Steiner Triple Systems of order greater than 7 satisfy the Parallel Line Postulate?

SAWbladeA Steiner Triple System is a set $\mathcal{S}$ of $v \geq 3$ elements together with a set $\mathcal{B}$ of $3$-subsets (triples) of $\mathcal{S}$ such that every $2$-subset of $\mathcal{S}$ occurs in exactly one triple of $\mathcal{B}$. As an example, the following forms a Steiner Triple System o...

 
3:57 AM
@DHMO $\sin(x^3)/x$
 
@AkivaWeinberger exactly the one I had in mind. Wonerful.
 
 
1 hour later…
5:00 AM
Can you give me an example where Borel Sigma algebra is properly contained in sigma algebra ? i know that Borel sigma algebra is the intersection of all sigma algebras of subsets of R that contains the open sets
any1 !!!
 
@BAYMAX Wouldn't powerset of $\mathbb R$ be a simple counterexample. I mean $P(\mathbb R)$ is a $\sigma$-algebra, isn't it?
 
you mean $B \subset M$ is not possible?
@MartinSleziak
 
I do not know what is $B$ and what is $M$.
Is $M$ a σ-algebra?
 
B is Borel sigma algebra and M is sigma algebra
 
You asked for an example of a sigma algebra properly containing all Borel sets. I suggested $M=\mathcal P(\mathbb R)$ as an example.
 
5:08 AM
I think its possible as Borel Sigma algebra is the intersection of sigma algebra
example where Borel Sigma algebra is properly contained in sigma algebra
 
Once again, $M=\mathcal P(\mathbb R)$ is an example of such sigma-algebra.
 
so what is B here?
 
Or you can take simply any non-borel set and take $M$ as the smallest sigma-algebra containing all Borel sets and this set.
@BAYMAX $B$=set of all Borel sets. At least that is what you said.
Unless I misunderstood your question.
 
yes B = set of all Borel sets
 
I thought that by Borel sigma algebra you meant the sigma-algebra consisting of all Borel sets.
 
5:12 AM
yes
 
And $M$=set of all Lebesgue measurable sets should work too. Also it is probably more difficult to show. Main site: Lebesgue measurable but not Borel measurable
 
As wiki says -- The Borel algebra on X is the smallest σ-algebra containing all open sets (or, equivalently, all closed sets).
 
Still, I think $M=\mathcal P(\mathbb R)$ is probably the easiest example.
 
if $M = P(R)$ i cant figure what B is ?
is it $B =P(R)$ ?
 
No. I meant $B$= Borel subsets of real line (with the usual topology).
I.e., $X=\mathbb R$ in the definition you copied from wiki.
Re: I can't figure what B is. $B$ is defined in the way you described.
As the sigma-algebra generated by open sets.
In this case, open subsets of $\mathbb R$.
 
5:16 AM
ohh nice !!! so is there an element in $M$ which is not in $B$ ?
here $M =P(R)$
@MartinSleziak
?
 
Single to n sets?
single to n sets ?
singleton sets
sorry
 
This can be seen from cardinality argument. Since $|B|=\mathbb c$. Cardinality of Borel sigma algebra
@BAYMAX Singletons are clearly Borel subsets of $\mathbb R$. You can easily get a point as a countable intersection of interval.
 
@MartinSleziak how to show that the Cantor set has Lebesgue measure 0?
 
Hello, can anyone help me evaluate this proposition?
 
5:19 AM
@DHMO By showing that complement has measure one...?
 
@MartinSleziak how?
 
Measure of the complement can be calculated as a sum of series.
 
I've come down to: ((false)-->(false))-->true = True
 
@WillNjundong $((p \land \neg q) \implies (q \land r)) \implies (s \lor \neg q)$
@MartinSleziak oh, thanks
 
$\frac13+\frac2{3^2}+\frac{2^2}{3^3}+\dots=\frac{1/3}{1-2/3}$.
 
5:21 AM
@DHMO sorry that message is riddled with ajax syntax or whatever its called. cant read it. Am I the only one?
 
@MartinSleziak you meant $\frac{1/3}{1-2/3}$
 
Ok thank you @MartinSleziak
 
@DHMO Thanks! Corrected.
 
@WillNjundong ((p∧¬q)⟹(q∧r))⟹(s∨¬q)
≡ (¬(p∧¬q)∨(q∧r))⟹(s∨¬q)
≡ ¬(¬(p∧¬q)∨(q∧r))∨(s∨¬q)
≡ ((p∧¬q)∧(¬q∨¬r))∨(s∨¬q)
≡ ((p∧¬q)∨(p∧¬q∧¬r))∨(s∨¬q)
≡ (p∧¬q∨¬r)∨(s∨¬q)
≡ (p∧¬q)∨(¬q)∨(¬r)∨(s)
≡ (p)∨(¬q)∨(¬r)∨(s)
@WillNjundong ok?
 
@DHMO YES thank you so much
 
 
1 hour later…
6:53 AM
@MikeMiller How did you immediately see that the one-point compactifications of (open) moebius strip x $\Bbb R$ have different $\pi_1$? If I am not wrong nbhd of the point at infinity looks like a cone over the Klein bottle (whereas for solid torus it's just cover over the torus), right?
 
@BalarkaSen @BAYMAX hi
 
I was dealing with a problem as follows, "Let f be a function of a real variable such that it satisfies f(r+s)=f(r)+f(s) for all r,s . Let m and n be integers, then f(m/n) is ?" The answer is (m/n)f(1) and what I assume the solution is that using f(x+y)-f(x)=f(y) we get that the derivative is constant and the function is of the form f(x)=kx for some k. But doing so would require the function to be differentiable wouldn’t it? How do I go ahead with proving that?
 
Hi @DHMO
 
@CompulsiveMathurbator we don't need the function to be differentiable
hint: find f(0) first
 
6:55 AM
@DHMO I think you can find a very nice explanation that Cantor set has measure 0 ,uncountable ,closed in Royden Real Analysis , it has nice lucid explanations there
 
@DHMO I'm sorry I don't see why not.
 
@BAYMAX @BalarkaSen my face reveal will take place two months from now lol
@CompulsiveMathurbator after finding f(0), prove that f(n) = nf(1) where n is an integer
@CompulsiveMathurbator the function doesn't have to be f(x)=kx
 
@DHMO you are a chemical bond now !! he he
 
@BAYMAX is it a book?
 
Yes @DHMO its a book , Royden Real Analysis
Hey @CompulsiveMathurbator are u interested in Chaos theory ??
 
6:57 AM
@BAYMAX then why?
 
we don't have a face anymore. it's all blood and bones under the skin
 
@BAYMAX Interested, not well versed.
 
Ohh sorry , you dont read from books right !!
 
@BalarkaSen indeed
 
ok its nice @CompulsiveMathurbator
you heard of Lyapunov exponents ?@CompulsiveMathurbator
 
7:00 AM
@BAYMAX heard of the stability, not the exponent. Time to google
 
Ok..explore !!! @CompulsiveMathurbator
bye guys!!!
 
@BAYMAX bye
 
Hi @KanwaljitSingh please guide me
A pipe can empty a tank in 40 Minutes , A second Pipe with diameter twice as much as that of the first pipe is also attached to the tank to empty it .The two pipes can together empty the tank is
for this question
 
7:09 AM
@Learninguser it is assume that the rate of draining is directly proportional with the cross-sectional area...
 
@DHMO I have tried

**for first pipe - 1/40**

**second pipe - 1/20**

so

**i have consider as tank volume as 1**

**1-(1/40 + 1/20)t =0**

If i find t,i will get answer, is this approach is correct
 
@Learninguser the area is proportional to the square of the diameter...
 
what i am doing mistake
please provide your answer
 
so the second pipe is 1/10...
 
can you explain How you got 1/10
show me the steps
 
7:19 AM
1/40 x 2^2 ...
 
ok reamining steps are correct
 
I'm trying induction on the rationals. My argument is, given a statement 'P' true on an interval '[a,b]' and two rationals 'x' and 'y' in 'I' . Would showing that 'x' and 'y' satisfying the statement implies that 'x+y' satisfies the statement show the statement to be valid on [a,2b] ? Or am I missing a few values in [b,2b]?
 
@CompulsiveMathurbator wouldn't 'x+y' give you [2a,2b]?
 
@DHMO Sorry [2a,2b], but would it work even if [a,b] was infinitesimal, say [0,2^-n] as n approaches infinity?
 
@CompulsiveMathurbator why not?
 
7:26 AM
Thank you@DHMO
I have done
 
you are welcome
@MikeMiller greetings
 
@BalarkaSen one is RP^2 the other is S^2/S^0
h
 
someone mind reading question 8?
 
one-point compactification of the moebius strip is RP^2, not one point compactification of moebius strip x R? Or am I misunderstanding
 
how can we say that triangle AEB is a right triangle?
because they are using the pu]ythagoreas theorem
 
7:32 AM
@MartianCactus AB is a diameter of the bigger circle
 
so..?
 
angle in a semicircle...
diameter corresponds to a right angle...
 
oh yeah!!
 
@BalarkaSen Sorry, was he saying the mobius strip x R?
 
that theorem which says
 
7:34 AM
My bad. Hm.
 
Yeah. I think a neighborhood of the pt at infinity looks like a Klein bottle coned off
 
when 2 points of an arc form an angel at the center of the circle and at the circumference, the angle at the circumference is half the angle at the center
here the angle made is 180
so half of 180 = 90
\\oo//
am i right?
 
@MartianCactus yes?
 
@MartianCactus yes
 
@BalarkaSen You mean $K \times (0, \infty)$? Sure, that's true.
I guess multiplication by R is the proper version of the suspension. Cute.
 
7:38 AM
Well, that's what a punctured neighborhood should look like. Whereas a punctured neighborhood of the one-point compactification of the solid torus is $T \times (0, \infty)$, which are not topologically the same so they're not homeomorphic
 
@BalarkaSen Those two neighborhoods are not homeomorphic, but it's not obvious that there's not some neighborhood in the latter that is homeomorphic to $K \times (0,\infty)$.
Have to develop a fundamental group at infinity, which is irritating. You could also take local homology at the basepoints, which would show they're not the same. But all of this is kind of more work than I think this should need.
 
@DHMO When I tried generalizing that to the rationals I'm able to show that $f(x)=xf(1)$ for x=0 & $x= \lim_{n \to \infty} 2^{-n}$ and that $f(x)=xf(1)$ being true over the interval [0, $\lim_{n \to \infty} 2^{-n}$] implies that its true for all the rationals, but I'm stuck at proving that it is.
 
(I really like the idea of distinguishing spaces by using proper homotopy invariants, or finding local invariants of spaces by considering the homotopy-type-at-infinity of the complement of the point.)
 
@CompulsiveMathurbator hint: prove that f(m)+f(m)+...+f(m) = f(nm)
 
@MikeMiller Sorry my internet is a bit dumb today. Ah, that is a good point.
 
7:53 AM
@DHMO Ah I see it now (wow I am blind). Thanks.
 
Hey everyone!
 
Hi @Daminark
I honestly don't know how to compute fundamental group at infinity effectively. Sometimes it's just obvious that they're weird (eg, Whitehead manifold, because the end is itself very weird), but I have no idea how to compute it in general.
 
If $X$ is actually the interior of a manifold with boundary $M$, then $\pi_1^{inf}(X) = \pi_1(\partial M)$
 

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