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12:41 AM
@Adam Isn't there like a free PDF of that one online
 
Does the word metric on a group always induce the discrete topology?
Oh, of course. It's obvious. Just center $1/2$ ball centered at any group element.
 
12:59 AM
@user193319 What's the word metric?
 
1:10 AM
@AkivaWeinberger yeah you might be right, this computer is a mess so wasn't able to find it despite having EKHAD
on a non related note have they managed to catch that serial killer in America? the one that had everyone believe they were suicides convincingly
 
1:47 AM
Can the construction of the integers as pairs of naturals be generalized to the ordinals?
 
2:15 AM
explain what you mean using algebraic definitions
potential group of serial killers*
more than one would certainly explain the high level of professionalism
 
Last night dream: Very weird large cardinal numbers:
Recall that an inaccessible cardinal is a regular cardinal and also it is some $\kappa > 2^{\alpha}$ for any $\alpha < \kappa$
 
are you answering user76284's question? if not it's polite to start with "on an unrelated note"
 
ok sorry about that.
@user76284 Ordinals are actually encoded as a well ordering of the natural numbers, at least until you reach $\omega_1$
e.g. $\omega + 1$ is encoded as 1,2,3,4,...,0 algorithmically speaking. No pairs needed
(continue on unrelated note)
Last night dream introduce a very weird kind of large cardinals which is made from non well founded sets (i.e. sets that contain itself such as X={X}). These weird cardinals, are "large" and inaccessible, in the sense that they are larger than any self containing sets, which are already large because they contain themselves as elements and hence an infinitely decreasing sequence. Thus let a non well founded set be Y={Y}. Then these large cardinals are T such that $Y \subsetneq T$ for any $T > Y$
 
define an encoding with an algebraic lemma, as well as well ordering
 
According to discussions with user21820 on this topic, an encoding is basically a function $f : \Bbb{N}^2 \to \{0,1\}$. Thus e.g. $f(x,y)$ outputs $1$ if $x > y$, otherwise outputs $0$
Hence $f$ is a program that tells you how to order the natural numbers, and hence they encode a given countable ordinal
in Mathworks (Not the main chat!), Oct 24 '17 at 8:56, by user21820
We would thus say that f denotes the ordering 1<2<3<...<0.
for details
 
2:32 AM
$f : \Bbb{N}^2 \to \{0,1\}$ absolutely does not necessarily imply $f(x,y)=1$ if $x \gt y$, if I am correct in my interpretation of the former statement, $f$ is any two variable arithmetic function that is equal to either $1$ or $0$, for any $x \in \mathbb N$ and $y \in \mathbb N$
 
2:50 AM
But I feel as if I should point out, this is not a medium for collaboration between the vast many of us in the sense that our thoughts are going to be respected as having come from us as individuals, it's more a sham set up to benefit a select few, much like all things are really, and providing answers isn't going to lead you anywhere but perpetually being taken for granted
 
3:19 AM
But this is pretty funny, when was $0$ ever considered a natural number?
In the philosophy of mathematics, ultrafinitism (also known as ultraintuitionism, strict formalism, strict finitism, actualism, predicativism, and strong finitism) is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers. == Main ideas == Like other finitists, ultrafinitists deny the existence of the infinite set N of natural numbers. In addition, some ultrafinitists are concerned with...
In the philosophy of mathematics, ultrafinitism (also known as ultraintuitionism, strict formalism, strict finitism, actualism, predicativism, and strong finitism) is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers. == Main ideas == Like other finitists, ultrafinitists deny the existence of the infinite set N of natural numbers. In addition, some ultrafinitists are concerned with...
 
3:43 AM
It's very strange, I mean in that yes I have done quite a bit of work in analytic number theory, I cannot see where they have drawn the conclusion of denying the existence of an infinite natural number line, and yet, they seem to agree that there exists a limit of measurability, which is really the same thing as saying that the set is infinite lol
 
Depending on who you ask, you can have the naturals forming a proper class or a set
 
Depends on how long I can pretend to care about linguistic nonsense
 
The notion of a proper class is only a natural consequence of how Peano axioms work but it does not prove there exists a proper class, only that you can define such in logic
It remains open whether the existence of infinite set can be a theorem
 
sounds classy to read
 
 
2 hours later…
6:03 AM
@user193319 they have coefficients in the prime field $\Bbb F_p$ where $p$ is the characteristic of $E$ and $F$ (which is necessarily the same)
so you can consider them as they same polynomial in $\Bbb F_p[x]$
 
6:56 AM
Can we learn the mean value of an analytic function in the interior of some region by integrating something along the boundary of that region?
 
7:33 AM
Morning
 
@ÍgjøgnumMeg Morning!
 
How's it going? :)
 
Pretty well, thanks.
And yourself?
 
Yeah not bad, had a fun day yesterday and now I'm back at work T_T
just about to stick an i7 into someone's PC
 
if you want to keep up with my series on category theory, I'm at part 4 right now
 
7:44 AM
Oh nice! Will take a look
 
7:59 AM
What's the analogue of en.wikipedia.org/wiki/Stieltjes_constants for the prime zeta function?
 
8:34 AM
who really is 76284, seemed to be piling a bunch of googlable questions
 
user76284 is the unique user between user76283 and user76285
 
that's a technically correct and useless answer
(and then they said technically correct is the best kind of correct)
Looking at his chat history, it seems most of the questions are googlable, except for the anormalous messages related to producing new functions
 
Otherwise, I think I started to understand what Adam is warning about earlier:
6 hours ago, by Adam
But I feel as if I should point out, this is not a medium for collaboration between the vast many of us in the sense that our thoughts are going to be respected as having come from us as individuals, it's more a sham set up to benefit a select few, much like all things are really, and providing answers isn't going to lead you anywhere but perpetually being taken for granted
and providing answers isn't going to lead you anywhere but perpetually being taken for granted
I guess combining with the 76284 data, that kinda justify it
Well at least 76284 is not as JEE machine as gateprep, but it really give me the urge to spam about large cardinals
on another note:
Jun 5 at 0:22, by user76284
What’s the order type of the set of all integer polynomials, where $f \leq g \leftrightarrow \exists x \in \mathbb{Z} : \forall y \in \mathbb{Z} : f(y) \leq g(y)$?
Turns out that RudyRucker book I have been reading in the past weeks have an answer to this question
It's $\omega_1$
 
9:05 AM
@Secret come on, don't go and think that a username composed of any alpha numeric characters awards any kind of legitimate identity, everyone is just a unit of a certain monetary fraction of the website owner's profits
we are all Stack Exchange lol
@RyanUnger ok that's quite a number of rhetoric @BalarkaSen compliments yes he knows he is good too I can only assume, therefore probably doesn't get any satisfaction except for the first one you give if he is impressed by you, the rest id keep to references to specific proofs he has shown you if I were to give social advice there
 
Do you think we should insult @BalarkaSen more so that the compliments are all the more special
 
probably not a good idea unless you want lightning bolt singularities everywhere due to barlaka folding simplices after simplices
in response to something of the nature of that starred comment
 
@Secret I dont believe this
First of all because this order is not a well order, second there's countably many polynomials
 
> One last picture of alef-one. Go back to Figure 42. In this picture we saw how various ordinals can be represented as sequences of functions ordered according to steepness (the <bep ordering). How long a sequence of functions can be found with each function steeper than all the ones before? At least alef-one. That is, if S is a set of functions so that for every g (no matter how steep), there will be an f in S that is steeper, then S must have at least alef-one members.
Ok nvm, I made a read error. They are saying all eventually dominanting functions, not just polynomials
 
I still don't believe the book, the set of functions $x\mapsto nx$ for $n\in\Bbb N$ has this property unless I'm missing something
And regardless having uncountably many members and having order type $\omega_1$ are very different
 
9:21 AM
They are talking about aleph 1 as the first number that cannot be injected into $\omega$, so they are talking about $\omega_1$. However, when I first read this I do felt strange. Problem is I do not have as good an intuition nor experience to handle the set of all increasing functions ordered by eventual dominance, thus I have not tried to (dis)prove myself on that yet
 
@AlessandroCodenotti here's a measure theory question
α β : Type u,
_inst_1 : measurable_space α,
_inst_2 : measurable_space β,
ν : measure β,
s : set (α × β),
hs : is_measurable s
⊢ measurable (λ (b : α), ⇑ν {y : β | (b, y) ∈ s})
is this true or false?
 
Hey @Alessandro @Leaky
 
@Slereah lol no I am just saying it probably only feels rewarding when he impresses himself, if I show my math to my family or my cat and the miracle of them saying something nice occurs, I don't think I would be any happier with the math, so if he is the same in that regard, it only feels good when there is mutual appreciation for something he has done
 
Does anyone know the meaning of "unitary inner product"?
 
9:38 AM
@thomasshelby likely a complex scalar product
 
mbari
 
@LeakyNun translate it in English? I'm not that familiar with lean yet, still learning it
Hi @ÍgjøgnumMeg
Actually let me translate it
Suppose that $\alpha$ and $\beta$ are measurable spacces, $\nu$ is the measure on $\beta$ and $s\subseteq \alpha\times\beta$. Then $s$ measurable implies that for all $b\in\alpha$, $\{y\in\beta\mid (b,y)\in s\}$ is $\nu$-measurable?
(If my translation is correct the theorem is false by the way)
 
9:54 AM
@Adam Are you saying we are all as dumb beasts next to @BalarkaSen's majesty
 
Specifically let $V\subseteq\Bbb R$ be the Vitaly set and consider $V\times\{0\}\subseteq\Bbb R^2$. It is measurable having measure $0$, but obviously not all slices are measurable
The correct version is "let $(X,\mu)$ and $(Y,\nu)$ be spaces with measure. Suppose that $S\subseteq X\times Y$ is $\mu\times\nu$-measurable. let $S_x=\{y\in Y\mid (x,y)\in S\}$ and $S^y=\{x\in X\mid (x,y)\in S\}$. Then $S_x$ is measurable for $\mu$-almost every $x$ and $S^y$ is measurable for $\nu$-almost every $y$."
The converse can fail badly, you can ask all slices to be measurable and yet $S$ can be nonmeasurable, you can even ask all slices to be open while keeping $S$ nonmeasurable
@AlessandroCodenotti Vitali*
 
@AlessandroCodenotti eh
(λ (b : α), ⇑ν {y : β | (b, y) ∈ s}) is a function from alpha to [0,infty], and measurable means that the preimage of measurable sets is measurable
also I don't think the Lebesgue sigma algebra on R^2 is the product of the Lebesgue sigma algberas on R
 
No, it's the completion of the product
 
right, so your counter-example wouldn't fit
because we're using the product sigma algebra
 
Which is why some people define the product algebra as the completion of the product, because you really want to say that every subset of a measure zero set is measurable with measure zero
 
10:06 AM
but we don't have a measure on alpha
 
I don't see how's that relevant, build a counterexample with a measure on alpha and then forget about the measure?
 
well, again, we are not using the completion of the product sigma algebra
so your counter-example wouldn't work
 
Sure then it works just because the inclusion maps $X\to X\times Y$ are measurable
 
eh, what's your counter-example?
 
@AlessandroCodenotti I have none, it is true with the product $\sigma$-algebra
 
10:19 AM
I'm confused
 
If you use the product $\sigma$-algebra on $X\times Y$ then $S\subseteq X\times Y$ measurable implies that $S_x$ and $S^y$ are measurable for all $x,y$
 
but does that mean the measure is measurable
18 mins ago, by Leaky Nun
(λ (b : α), ⇑ν {y : β | (b, y) ∈ s}) is a function from alpha to [0,infty], and measurable means that the preimage of measurable sets is measurable
 
Oh, you mean the function $b\mapsto\nu(S_b)$?
Yes that is also measurable
But it's not as straightforward
Look at @Ryan, summoned by measure theory
 
eh... how should I prove it?
 
Hmm actually maybe you need $\sigma$-finite measure spaces, because this is usually proved as part of Fubini
 
10:26 AM
right, but I don't even have a measure now
and the measure on beta isn't sigma-finite
 
10:43 AM
I'm pretty sure there's counterexamples with sigma infinite stuff
 
@Slereah well no I didn't say that at all, that interpretation indicates it isn't worth me paraphrasing, but no I never implied he was from a royal family either. I don't know what the social environment is like in India, I doubt they have a kings etc, if they do they are probably like ours, totally unimportant super rich reality tv stars
also everyone should look at the way @LeakyNun writes questions this makes it far easier for people not necessarily in that specific field of mathematics to attempt it and or learn from the experience, asking in natural language with a f%$# ton of field specific terminology is just silly
 
11:00 AM
@AlessandroCodenotti I'm pretty sure the sigma-infinite stuff either gives you non-uniqueness of measure or failure of equality in Fubini
rather than non-measurability
 
11:10 AM
In the context of local compactness, what is the definition of a compact neighborhood?
 
@LeakyNun Cohn states your theorem for $\sigma$-finite spaces
I'm trying to figure out whether this is to make the proof easier or to make the proof possible
(fun fact: for non sigma-finite spaces Fubini can fail for all possible product measures)
 
11:25 AM
@user193319 a compact subset that contains an open set containing the point $x$ is a compact neighborhood of $x$
 
@MatheinBoulomenos Ah, I see. Thanks!
 
11:51 AM
odd asf
 
 
2 hours later…
1:24 PM
It's a little strange that some people can have a career and declare themselves to be ultrafinitists. Sure ok academics is in all likelihood like any other industry now, it isn't what you know it's who you know. Actually it was always like that wasn't it
oh well
 
1:36 PM
psychiatry is by far the most corrupt industry by a long shot tho
in my own personal experience I mean, of course there are banks that have stolen billions from people and never spent a single day in a cell, so round that up to all authorities and we have a winner
 
@Adam ultrafinitism is a philosophical view. It's really no different from being religious.
It's not like it makes their research in mathematics any less compelling.
 
I am aware of that but I'm just curious as to how Doron can have a career and still go in opposition to classical view, sure yes his algorithms are very impressive we study very similar areas of number theory, that's the only reason I came across them, but we live in a world where majority rules, popular opinion is equal to correct opinion in an institutionalized mind set, so, I don't make any bold assertions like that, only state what i know to be true, its just interesting that's all
 
1:52 PM
@Adam Why would being an ultrafinitist preclude having a career? I don't follow.
 
Because it's a minority
 
Why does that preclude having a career?
 
I already explained why i think this
 
I don't see why any of that has to do with having a career as a mathematician.
 
I guess already being important might explain it so if i were to guess i would have to go with nepotism. Yes it shouldn't have anything to do with it, but it does. \
 
1:59 PM
It's amusing when people try to reduce the careers and success of others down to unfair advantage.
 
well no I wasn't talking about the success and already stated I hold his work in very high regard, I only implied this for the career of the individual
do you truly believe we are all given equal opportunity @anakhro ? That's very amusing and defensive of you
 
His career is his success.
I never said everyone was given equal opportunity.
But I merely find it amusing when people reduce success down to unfair advantage.
 
no they are not the same thing. So in the hypothetical scenario of what I implied being true, which you have agreed, does happen, you find this amusing?
 
I never said they were the same thing. Imagine it is an inclusion, not an equality.
 
@AlessandroCodenotti that was my phone doing mysterious things while I was asleep
 
2:05 PM
$\text{career}\subseteq\text{success}$
 
@Adam I've known Balarka for years
 
no it's just a subset of factors that contribute to the person's standard of living. success isn't measurable
 
I don't know what it being measurable has to do with the conversation.
 
is success a vitali set?
 
that will be a very bad idea
In fact, I think success is beyond measurable
(Think berstein set but more inconceivable by picking a suitable large cardinal axiom)
on another note
Can we have a $\eta_f$?
 
2:14 PM
@Secret you mean a morphism going between morphisms? Concepts like these exist, and the lead to higher category
 
higher nonsense
 
in a 2-category for example, you have objects, which are 0-morphisms and 1-morphisms going from objects to objects and you have 2-morphisms going between 1-morphisms
 
I hear Jacob Lurie is going to be here soon
Wonder if he'll teach courses
 
well, a career is a good measure of how comfortably the individual lives, and how much appreciation for their work they have received from others. This is not in any sense a measure of success
 
The category of small categories is such a 2-category, with objects being small categories, 1-morphisms being functors, and 2-morphisms being natural transformations
another example is the category of topological spaces with 2-morphisms being homotopies between continuous maps
 
2:17 PM
I see. I just had this thought because it will look more like an interpolation if $\eta$ literally translate the top row of arrows to the bottom row, and then for some reason I started to imagine the whole thing to behave like a homotopy where the top row of the commutative diagram looks like a directed curve to me
snipped
 
that's a good intuition to have
in fact, I have a section on the analogy to homotopies
 
hello.
 
hi.
 
wagwan.
 
2:23 PM
sound
 
Haaaabe d'eeehri
 
hello there
 
General Kenobi
 
behold the masterful princeton landscaping
 
lol 39°C right now
I'm sublimating
 
2:26 PM
f*ck that
 
I've got better puddles at home now that I found a pressure cleaner
 
Italy is supposed to be temperate
 
29 here
 
my mother was in Italy recently actually
 
visiting?
 
2:28 PM
yeah i think it was Venice? I wasn't really listening but yeah her brother's wife is Italian so they have family over there enabling her to stay for over a month which was great
 
nice
I've been to Venice only once.
 
Austria is super hot in the summer :(
 
I live in Tuscany
 
Oh ok yeah I did plan at one stage to do one of those semesters abroad things in Budapest, but I literally don't want to travel unless I can live in that place for an extended period, I just don't see the point in vacations personally
 
@ÍgjøgnumMeg given the name I'd imagined you were icelandic.
icelandian?
Viking!
 
2:32 PM
@Simone the name is Faroese :) But actually I'm English
 
My ancestors are French lol apparently they fled from the French revolution to England and then well by no wrong doings ill assume one generation ended up here in Australia
 
old blighty eh?
 
@RyanUnger Hey you don't happen to know a guy named Clem Brown, do you?
 
@AkivaWeinberger no, what makes you ask
 
Screw the UK
Awful, awful place
 
2:34 PM
I agree
 
He's also in Princeton
 
an undergrad?
 
@RyanUnger Homonymy I presume?
 
@Simone wot
 
2:35 PM
@RyanUnger This
I forgot to tag that
 
I'd like to go to America but not die in a new York apartment owned by one of the Olsen twins. I mean ketamine vacuums
same thing really
 
Let $X$ be some compact Hausdorff space, let $C(X)$ denote all of the continuous complex valued functions on $X$, and let $A \subseteq C(X)$ be a subalgebra. If $f \in A$, does it follow that $|f| \in A$?
$|f| = \max \{f,-f\}$, but I don't think that is helpful...
 
@user193319 that's false
 
this guy would be really interesting to meet wow what an extraordinary life to have lived en.wikipedia.org/wiki/William_Breeze#Gnostic_Catholic_Church
 
@AkivaWeinberger I don't know any undergrads here
@Simone lmao my friend just used this word on the NTY spelling bee
nice one
 
2:47 PM
I feel like there is no finite-dimensional counter-example
 
@LeakyNun ?
 
9 mins ago, by user193319
Let $X$ be some compact Hausdorff space, let $C(X)$ denote all of the continuous complex valued functions on $X$, and let $A \subseteq C(X)$ be a subalgebra. If $f \in A$, does it follow that $|f| \in A$?
 
I was trying to think of a counter-example
 
he said complex valued
and then wrote max
:thonk:
 
2:48 PM
1 min ago, by Leaky Nun
I feel like there is no finite-dimensional counter-example
because $f^n$ are going to be generally linearly independent
unless $f$ consists of roots of unity or zero
in which case the magnitude contains only $1$ and $0$
maybe $(i,0)$ will work
no it won't
no it won't work
because $f^N$ would be $|f|$ for some big $N$
and I'm not going to generate infinite-dimensional counter-example
 
$(-2,0)$?
 
(1,1) and (-2,0) are already linearly independent
I'm interpreting subalgebra as $\Bbb C$-subalgebra
so we want a function $f$ such that there is no polynomial $p$ with $p(f) = |f|$
thinking about it universally... $|z|$ is not a polynomial in $z$ because of complex analysis
so let $X$ be the closed unit disc (so that it's compact and we have the identity theorem)
and let $f(z) = z$
then this should be a counter-example
@user193319 there you go
$|f|$ is not in the $\Bbb C$-subalgebra generated by $f$
 
Ah, very nice! Thanks!
 
note that if the $\Bbb C$-subalgebra is closed under conjugation and topologically closed, then it will satisfy $f \in A \Rightarrow |f| \in A$
 
how?
you only have $|f|^2 \in A$ right
 
2:59 PM
yeah, but you can use some trickery with the taylor series of the square root function
 
hmm
I'm not very convinced
 
The subalgebra generated by $f(z) = z$ is not topologically closed, right?
 
@MatheinBoulomenos the question didn’t say that though, did it?
You do that for the proof of Stone-Weierstrass
 
@RyanUnger no, it didn't, I just thought it was interesting
@RyanUnger exactly, that's where I saw it
 
@user193319 right
the subalgebra is $\Bbb C[z]$
the closure is a subalgebra of $\Bbb C[[z]]$
 
3:09 PM
Topological closure?
 
right
 
I think the topological closure is convergent power serires, not formal ones
 
so I said subalgebra
 
a right
 
3:36 PM
Please recommend me good book to learn undergrad algebra. I took algebra course at uni which covered speedily topics of basic structures, vector fields, transformations, polynomials, characteristic polynomials, residue classes and matrices.
I didn't understand well many things so I wish to re-learn by myself things.
don't really care for uni content, I'm rather interested in learning these
 
4:23 PM
no.
 
4:37 PM
:(
 
the link does not seem to lead to algebra topics
 
I will go so far as to dare claim that this was Adam's intent all along
 
I suspect the same thing
 
5:07 PM
lol how could you have guessed
ok well you can learn about multisets before you learn about vectors I guess
 
it would be nice if senior users were to put up a reading list.
 
definitely and demonstrating your knowledge of said reading list is guaranteed to lead us all to career opportunities
 
5:28 PM
who cares about career opportunities
But would be nice to have such list indeed, like in reddit.
 
 
https://www.reddit.com/r/learnmath/comments/8p922p/list_of_websites_ebooks_downloads_etc_for_mobile/

https://www.reddit.com/r/math/comments/7i9t5y/book_recommendation_thread/

https://www.reddit.com/r/math/wiki/faq#wiki_what_are_some_good_books_on_topic_x.3F
 
5:58 PM
*Research*
My research is on number theory, arithmetic geometry, topology and ways to avoid administrative work.
From Ted Chinburg's homepage
 
I've just discovered that he has a 4 semester course on ANT on youtube
 
oh wow
4 semesters is a lot. We did ANT up to local and global CFT and some stuff beyond that in 3
 
Was just looking at the math reddit that @Flowian posted and it links to an MO post that is essentially "Let's compile a list of all the full courses on various subjects that have been made publicly available"
So thanks a lot indeed, @Flowian
99% of Semester 4 focusses on Kato's Class Field Theory
whatever that is
 
@ÍgjøgnumMeg it's ordinary CFT but the mathematical objects are all personified
 
6:06 PM
@Daminark wtf man
 
I guess that's higher-dimensional CFT
as Kato worked on that
 
I see
 
that's some pretty deep stuff
 
what "dimension" does the higher refer to?
 
for example in the local case, you can define dimension inductively: a 0-dimensional local field is a finite field and then a n+1-dimensional local field is the quotient field of a complete DVR whose residue field is a n-dimensional local field
 
6:08 PM
Ah nice
 
for example $\Bbb F_q((x_1, \dots, x_n))$ (formal Laurent series in n variables) has dimension $n$
I actually don't know the definition in the global case
another example would be $\Bbb Q_p((x))$ which has dimension $2$
ordinary local fields are those of dimension 1
 
Right
So which complete DVR is $\Bbb Q_p$ a quotient of?
 
$\Bbb Z_p$
 
and it's residue field is just $\Bbb F_p$ again so you get a $0$-dimensional local field?
its*
 
$\Bbb Q_p[[X]]$?
he said quotient
 
6:10 PM
oh right
yeah
 
I see
 
oh, quotient field
then it's $\Bbb Z_p$
 
there's also some examples with Witt vectors
 
@Leaky No you got what I meant the first time
 
the $p$-typic Witt vectors of $\Bbb Q_p$ are a complete DVR with quotient $\Bbb Q_p$ and you can take the quotient field of that
basically you have two ways to go up a dimension: either take formal power series or Witt vectors
all higher dimensional local fields are constructed from finite fields like that
 
6:13 PM
Cool :)
 
If $\mu$ is a complex measure on a locally compact space $X$, what does $|\mu|(E)$ denote, where $E$ is some subset of $X$?
 
Fesenko was working in Nottingham (where I almost went for my Masters) and he does a bunch of higher dimensional weird mumbo jumbo
 
@ÍgjøgnumMeg yeah I know, he's another big name in the field
 
that's where the Nottingham group comes from!
 
@ÍgjøgnumMeg Isn' Fesenko the guy who keeps claiming that people who don't understand Mochuzuki's work on IIUT have just not spend enough time on it?
 
6:16 PM
@Tobias yeah something like that
he runs a seminar on IIUT in Nottingham I believe
IUTT
lol
 
sure, whatever
He is also being really aggressive in his defense of the theory
 
I gotta jump in the shower, cy'all later
 
is mochuzuki's work culturally like the crazy shit deligne did in the 70-80s?
the comparison is probably stupid for a million obvious reasons i guess
 
Shinichi Mochizuki's father was Fellow of Center for International Affairs and Center for Middle Eastern Studies at Harvard University interesting stuff
 
everybody's father was somebody, sometimes even somebody else
 
6:27 PM
My Dad was a farmer
 
there is a tongue twister there with further, father, farther, farmer and falter
 
6:57 PM
Hi @Ted
 
Hi @Mathein
 
How are doing? Did you have fun with your friends?
 
Yes, fun and some tremendous food. I took them to my favorite restaurant last night.
 
Sounds nice. I applied for intro algebra TA for next semester and got accepted. I went over my past evaluations for the application (since I was asked for them) and noticed how I improved over time (I've been a TA 3 thrice)
 
Félicitations! ... Most people have a rocky start in teaching and those who care do improve. We've commented before on your improvement in explanations and patience in the chatroom :)
 
7:06 PM
Hello
 
@Daminark happy birthday!
 
Howdy, nerd Demonark. Oh oh, are you a year older and allegedly wiser?
 
Happy birthday @Daminark
assuming that's true
 
7:27 PM
Congrats @Mathein
on the TA position
 
Hey sorry I was out for a bit, thanks guys!
@TedShifrin yeah aging a year in one day is quite a feeling :P
 
@ÍgjøgnumMeg thanks
 
Good thing you've been improving! Does TAing in your case give you the chance to do stuff like problem sessions?
 
cdt
@BalarkaSen Remeber I was telling you of meeting an ISI M Math interview guy who admitted that he had no fucking idea of compactness ? Well now that results are out I have seriously no fucking idea (I remember he telling to someone near me after the interview that he couldn't even prove that if a, b are nilpotent then a+b need not to be nilpotent even after crapload of hints and time) how he got selected in M Math that too with a single digit rank...
 
@Daminark yes, TAing means grading and doing problem sessions
at least here
 
 
1 hour later…
8:47 PM
Is there a Schwartz function $f$ such that $\int_0^\infty \frac{f(x)}{x} dx = 0$ and $\lim_{x \rightarrow 0} f(x) = 1$?
 
8:59 PM
Or more directly, is there a function $f$ such that $\lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty \frac{1}{n} f(n,\varepsilon)$ exists and $\lim_{\varepsilon \rightarrow 0^+} f(n,\varepsilon) = 1$?
Such an $f$ would be a "regulator" for the harmonic series.
 

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