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02:00 - 22:0022:00 - 00:00

22:00
Grothendieck: "An actual prime? You mean like 57?"
also seemingly had no interest in actual examples ...
Say you have a network. Would it be cool to study how the nodes grow as you add edges to the graph in a systematic way?
no, it wouldn't
how about studying evolving networks
22:10
I think you're just saying no
2
but I can't prove it
22:29
Technically he said "nope" the second time instead of "no"
22:44
"Nope" to Demonark :P
if f is not nilpotent, then A_f is not the zero ring, so it has a prime ideal, so A has a prime ideal not meeting f
is this a valid proof of "nilradical is intersection of every prime ideal"
@Ted
Ask Mathein.
that seems fine to me - of course you're still using zorn's lemma here for the existence of a prime ideal
sure, I'm not saying I don't need to use zorn's lemma
yeah i just wanted to make that clear :p since i remember very clearly that you need it to prove intersection of prime = nilrad
22:50
Good morning
hi
@loch might you like to... check a proof for me?
konichiwa
is it long/hard
2
somehow that didn't come out the way i wnated it to
anyway
uh sure
22:51
lmao
now it surely didn't
goodnight, @MikeM.
bonus if you know where it comes from :P
oh is this AM
it's 2018, we're nearly adults, and we're still laughing at dick jokes
bingo
22:52
speak for yourself, a @Balarka
@BalarkaSen speak for urself I'm 23
hides from Fargle
vote yes/no: should I grab this opportunity to learn measure theory? lmao
the end-goal being "post a massive answer expanding on that comment"
yes
you should learn measure theory, regardless of anything
22:57
@Ted I forgot that joke. Did I always say morning?
I like this joke again. Actually, it is my morning right now.
Like this if you like mathematics
Yeah, you did, @MikeM, and I always said good night :P
good joke
I wouldn't call any humor in this room "good."
@BalarkaSen Measure theory isn't exciting but you should use it as an opportunity to learn measure theoretic dynamics
Like your comment
22:58
I don't like measure theory much, but integration theory is great.
basically what I'd do is go through Brin-Stuck along the geodesic that lands me to the proof of what I stated
starting at measure theoretic basics
then write it all down lol
Only write down what isn't tedious; link the important basic lemmas instead of proving them
@LeakyNun looks good to me
thanks
I'm going back to bed for a bit
23:00
Night.
Wait, I already said good night.
Yeah I dunno GMT was a bit fun but if there's anything in that general circle that I think I'd like it'd be ergodic theory
I'd end up writing something like this, this or this
171
Q: Your favourite application of the Baire Category Theorem

Rudy the ReindeerI think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power. Here's the theorem (with proof) and two applications: ...

I think I need to learn this one day
(might as well be today)
My last three mega answers that I remember
23:02
@loch do you have AM?
i have internet
so yes
I assume $\sqrt[3](1 - z^3)$ was supposed to be $\sqrt[3]{1 - z^3}$. Please let me know if I'm wrong. — Ennar 12 hours ago
so what exactly do I need to do for exercise 1.27?
I don't see anything to show
@Leaky: Given a countable subset $S$ of the plane, prove that there's a point $x$ whose distances from the points of $S$ are all distinct.
probably just to understand it
23:03
I actually have no idea what ergodic theory is good for. The proof of the ergodic theorem and the fact that ergodic measures are extermal were both fun but I'm not sure what people really care about it for
@loch interesting
@TedShifrin is this related to BCT?
-1
Q: How to solve $ f\left(\sqrt[3]{1 - z^3}\,\right)^2 = 1 - f(z)^2 $?

mickHow to Find analytic $f(z)$ such that $$ f\left(\sqrt[3]{1 - z^3}\,\right)^2 = 1 - f(z)^2 $$ Koenigs function can not be used here So I am stuck. How does the riemann surface look like ?

what if S is just one point?
woah
I guess one point is fine, anyhow, even though you're being ....
If you have only one number, it is not equal to any others on your list.
oh right
23:06
Finite is no problem still
Right.
"theres a world of difference between diet coke and coke diet" - Akiva Weinberger 2018
4
We don't see too much of DogAteMy these days.
2
In what?
@TedShifrin for any s,t in S, the set of points equidistant to s and t is a xxxxing line, which is nowhere dense.
Oh, well, say hi to him for me, a @Balarka.
23:14
S is countable, so SxS is countable
is that it?
Is a line actually nowhere dense in the plane? I've forgotten the definition.
i mean, a straight line
@TedShifrin It is
Yeah, yeah. That wasn't my quibble.
definition: S is nowhere dense iff S-closure-interior = empty
23:15
I would say empty interior.
Ah, ok.
I'd forgotten.
It seems dense at its own points.
hi @Semiclassic
every space is dense in itself =)
But even thinking of those points in the plane, not just in the line.
every open set can be refined to a smaller open set that avoids the line
@BalarkaSen (is this argument right?)
23:18
not if it's a neighborhood of a point in the line
@LeakyNun This is not my problem broham
What does density at a point mean?
@TedShifrin eh, I mean, a smaller open set, that doesn't need to be a neighbourhood
@Daminark nothing
what is meaningful is that the point is a limit point
Munkres carefully avoids this terminology, I believe. I'm with him.
Among the interesting applications of BCT are the fact that complete metric spaces without isolated points must be uncountable, the fact that a function between complete metric spaces is continuous on a $G_\delta$ subset if you're looking for cool stuff that follows from BCT
23:21
There are also cool things like deciding about the continuity of a pointwise-convergent sequence of continuous functions.
Also a couple of "small lemmas" in functional analysis such as Banach-Steinhaus and the open mapping theorem if you're looking for "more serious" applications
What I've seen as equivalent definitions of nowhere dense is that the closure has empty interior, and that there is no open set in which the set is dense
OK, Demonark, I'll buy that.
Oh regarding BCT there was one problem I had in my analysis final first quarter that involved BCT which I didn't get, I just mumbled something about S-W and was so salty when I realized
Stone-Weierstrass?
23:24
@loch
Yeah
With the BCT you can also construct a perfect subset of $\Bbb R$ containing no rationals, which is completely useless as far as I know, but kinda cool
is there anything BCT can't do
Oh, Rudin has an exercise like that. I think I figured it out eventually with nothing heavy.
disprove BCT
23:25
BCT can't help me decide what to get for dinner
@LeakyNun uh what's your question lmao
@loch, perhaps you haven't asked it nicely.
@loch is that a sheaf morphism i'm looking at
is AM implicitly introducing the spec map
oh
yeah that's secretly $\mathcal{O}_Y \rightarrow \phi_*\mathcal{O}_X$
well only on the global sections, but same idea
@TedShifrin Basically the proof I have in mind reduces to showing that $\bigcup_{q\in\Bbb Q}C+q$, where $C$ is the usual Cantor set doesn't cover the whole of $\Bbb R$, this can be shown with a few basic facts about the Lebesgue measure instead of invoking the BCT
23:31
use BCT sempre
even when non ci hai bisogno
either "non ne hai bisogno" (you don't need it) or "non ce n'è bisogno" (there's no need for it)
ne
can I say non hai bisgono di lei
@AlessandroCodenotti
Okay I can't find the final, I thought someone wrote up the problems sent it on our old group chat but apparently not
In any event it was something like
Like if you have a smooth function on a compact interval such that at any point, one of its derivatives is 0, then show it's a polynomial
I give no warranty on the adjectives
@LeakyNun that's gramatically correct (you don't need her), but "lei" would be understood as referring to a person most likely
46
A: Your favourite application of the Baire Category Theorem

David MitraIf $P$ is an infinitely differentiable function such that for each $x$, there is an $n$ with $P^{(n)}(x)=0$, then $P$ is a polynomial. (Note $n$ depends on $x$.) See the discussion in Math Overflow.

is it BCT on n?
23:42
Discrete spaces are Baire, but I don't think that's how it's being used in that answer
I'll check that MO thread tomorrow though, it's almost 2am here
no, I mean, for each n, take note of the set {x | P^(n)(x) \ne 0}
probably doesn't work though
never mind
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