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1:04 AM
Let $(F, +, •) $ be a field and $G\le F*$ is a finite multiplicative subgroup. Then $G$ is cyclic.
Clearly $G$ is abelian.
Let $|G|=n=\Pi_{i=1}{k} p_i^{a_i}$
Then $G$ is isomorphic to direct product of cyclic groups i.e $G\equiv \Pi_{i=1}^{k} \Bbb {Z}_{p_i^{a_i}}$
I need to prove the existence of an element of order $n$.
Each $\Bbb {Z}_{p_i^{a_i}}$ has an element of order $p_i^{a_i}$.
Each $\Bbb {Z}_{p_i^{a_i}}$ has an element of order $p_i^{a_i}$, say $x_i$.
Then $|x|=|(x_1,x_2, \cdots, x_k)|=lcm(p_1^{a_1}, \cdots, p_k^{a_k}) $
1:39 AM
Better to use the alternative decomposition $\osum \Bbb Z/d_i$ where d_1|d_2|\dots$.
Stupid malfunction of site on phone. Of course I mean oplus.
There’s a proof without knowing classification.
2:11 AM
there's the one that's just counting roots of polynomials. it's gotta be on MSE. super short, almost no theory.
Hello, leslie. How are the ducklings?
we were visited by a duck in the pool yesterday, but no ducklings. there is a duck on our neighbor's roof right now.
Still stalkering?
2:41 AM
Is it correct to say that $\sum_{n=1}^\infty \frac{n}{n^2+1}\left(\frac{x-2}{x+2}\right)^n$, with $x \ge 0$, converges uniformly in each $I_a =(0,a]$ with $a>0$ because $\sup_{x \in (0,a]} \left|\frac{x-2}{x+2}\right|^n = \left|\frac{a-2}{a+2}\right|^n$ and the series $\sum_{n=1}^\infty \frac{n}{n^2+1}\left(\frac{a-2}{a+2}\right)^n$ converges absolutely?
i.imgur.com/1a7iW1E.png does anyone know if this proves decidability in this problem? The first paragraph is the standard way to define an extension.
I haven't dealt with decidability yet, so I am concerned.
2:57 AM
@leslietownes Ah, the spy.
he's gone now, finally, but he was there for about half an hour.
i told our neighbor about the duck, he didn't know. he and his friend spent a while taking pictures of it.
@ZaWarudo I am not convinced. As $x\to 0^+$, the quantity approaches $1$, which is surely greater than your $|(a-2)/(a+2)|^n$.
@leslietownes Probably an undercover agent of duckduckgo, spying google users to formulate their ultimate plan.
2 hours later…
4:52 AM
Consider a compact matrix Lie group. In general, if $\exp(\Sigma_i t_i \Lambda_i)$ generates a Lie group, does $\Pi_i \exp(l_i \Lambda_i)$ also generate the Lie group?
e.g. for SO(3) this is the case
1 hour later…
6:18 AM
@Semiclassical I am not getting the relation between the first line and the rest of the comment:/
2 hours later…
7:58 AM
My NBHM interview is over.I guess they will not select me.
I have made many mistakes.
Was it online?
8:14 AM
1 hour later…
9:25 AM
Hi, I need some help from the community.
so what is happening is that i am trying to learn the basics of genetic algorithms
and an example is given to maximize the function f(x, y, z, w) = w^3+x^2-y^2-z^2+2yz+3wz-xy+2 with x, y, z, w being integers from 0 to 15
now i want to write a python code for that
and to check that i need the final answer: the tuple (x, y, z, w) and the maxima that occurs.
can someone help with some software?
1 hour later…
10:53 AM
I tried wolfram alpha but it just picks the polynomial whatever input i give
11:34 AM
Maybe Fulton's algebraic curve is one good introductory textbook on algebraic curves? Any other textbook you people would recommend? (as a supplementary for example)
12:19 PM
No regulation for AI generated content? Nonsense, this'll cause all sorts of problems.
Ted is about to berserk if that happens
Our first post-pandemic/Monika uprising.
How can you show that $\lim\limits_{n\to\infty}n^2 a_n$ d.n.e. if $\lim\limits_{n\to\infty} a_n\neq 0$?
12:59 PM
Sorry, I meant "does not exist". I guess this isn't as standard as I thought.
never heard of anything like this abbreviation
my calculus prof. used that, DNE
I'm not from an English speaking country, and I don't hear it on the internet
LCM and GCF are pretty common three letter abbreviations.
1:06 PM
me neither
anyway, if $a_{n_k}$ is a subsequence convergent to non-zero number or $\pm\infty$, then $n_k^2 a_{n_k}$ converges to $\pm \infty$
Alternatively, if $n^2a_n$ converges to $r$, then $a_n = \frac{1}{n^2}\cdot n^2a_n$ converges to $0\cdot r = 0$
@Jakobian Oh thanks! Honestly I had seen a similar thing with functions last semester but I had a brain boop for some reason. Thanks again.
oh, did I help you in the past
@Jakobian Oh no I meant that I have seen it somewhere else.
1:37 PM
My prof. said that for finite groups, induction and compact induction are the same, i.e., if $H<G$ as a subgroup of finite group $G$ and $R$ is a commutative unital ring, then two functors $RG\otimes_{RH}-$ and $\mathrm{Hom}_{RH}(RG,-)$ from $RH$-module cat. to $RG$-module cat. are isomorphic. Any reason why this is true?
1:54 PM
I have decided to quit.
May be PhD is not for me.
Was anxious during interview? I don't exactly know what is NBHM, just googled
I have failed badly to express myself in every exam.
@noballpointpen I have mental health disorder()
Under pressure, I forgot everything.
2:19 PM
@noballpointpen Interview was no so bad.
But I was unable to give my best effort.
I missed few trivial questions, very trivial.
Live and learn from your mistakes.
Depression :-(
Also learn from your successes.
I have nothing to celebrate.
Build upon what you have learned.
2:28 PM
Maybe you should try the 'fake it till you make it' strategy. You see, there are so many great math teachers on youtube, who are very good at expressing their thoughts. Try to imitate how they are doing their jobs, like how they talk, how they explain things, how they response when someone asks a question, etc. Try to imitate their body languages as well. You have the knowledge I believe, just some lack of confidence is holding you back I guess.
Sure learning from other people's success may work for you.
@SouravGhosh so it must be like 10/10 to pass? I thought you made mistakes to the extent that it would be like 5/10.
Hi :) I just thought I'd advertise this:
Q: Clarification on Nonstandard Analysis: Is $0.\overline{9}=1$, is it not, or is there some subtlety that allows both interpretations?

ShaunThis, I hope, is not a duplicate; I am exercising my critical thinking here and I want to understand what going on, and the available content I have found online on this so far has not helped. I'm getting conflicting information regarding whether $1=0.\overline{9}$ (i.e., "$0$ point $9$ recurring"...

It's fairly niche, so it'd help to give it some exposure.
I hope it's not a duplicate.
Hello, Shaun. How are your studies?
Hi @noballpointpen They're going well, I guess. Thank you for asking. I'm behind, but I'm applying for a retroactive Leave of Absence to make up for it. (I had a lot of time off for my health.)
2:41 PM
Cantor set with subspace top. from R is not discrete.
I should be able to get two months added onto the end of my maximum time.
This is quite surprising as from the construction it 'seems' that it should be discrete.
Yes, you said it me last time. I was asking about the last week.
'seems' can be dangerous thing to say at times.
Ah, now I remember . . . Sorry.
2:43 PM
Cantor set or Cantor dust?
Oh. You removed $\Bbb{R}^2$ .
Is there Cantor dust??
I have handed in a six month review . . . late. My nine month review is soon. Also, I have made progress on an exercise I was stuck on.
Cantor chowmein
I asked this question:
Q: Salvaging Exercise 3.2.10(2) of Springer's, "Linear Algebraic Groups (Second Edition)".

ShaunThis uses the soft-question tag because there might be more than one valid answer, and it's a matter of guesswork to some extent; but there is a right answer (in theory). Thoughts and Motivation: As a follow-up to If a hom. $\phi:G\to H$ of diagonalisable linear algebraic groups is injective, the...

@Koro What is the definition of "discrete"?
2:45 PM
That allowed me to move on in my studies. The error in the statement of the exercise must have been the reason why I spent so long on it.
Every one point subsets are open.
By discrete space, I mean a space with discrete topology (all subsets are open).
yes, in other words one pt. sets are open.
All subsets are open iff all one point subsets are open.
How are you, @noballpointpen?
When a set is open in the subspace topology?
2:48 PM
why are you asking?
To Shaun: Good, but I do not spend enough time on my studies, so my progress is slow these days. I am not in school, it's self-studying. Read few pages of Mendelson's logic yesterday and finished two problems, spent 3 hours overall. Do you have knowledge in logic? Of course, besides stuff that every mathematician knows.
@Koro Cantor set in higher dimensional space.
@Koro Ok. I thought you are asking for a proof.
No, I was not. I am appreciating the fact that Cantor set is NOT discrete.
@noballpointpen Three hours is good going, especially for self-study. I used to. I read some notes by Simpson and I'm interested in topos theory, but I wouldn't say I'm knowledgeable; just an enthusiastic onlooker.
Good to hear this.
2:56 PM
@Koro why would it be discrete
Infact, this also displays another not so obvious fact: that the (infinite)product of discrete spaces need not be discrete.
@Jakobian it is NOT.
I was asking why you are "appreciating" it
@Shaun, I just upvoted your question, so you have a relief after unknown downvoters wave.
Not particularly interested in it, but will read the answer, if it comes.
Thank you, @noballpointpen :)
Cantor set is perfect.
Every points in a discrete space are isolated.
3:08 PM
I have a question on my problem set that asks "Find Taylor series representation of $f(x) = x \ln(2 + x) $ about $x = 2$." I can find the terms by hand and I have found the first 6 terms but I can't see a pattern to these terms so that I can show it with sigma notation. Would it be fine to leave it with ellipsis in the end?
yeah. It's Polish
@Shaun may I ask? Are you insecure -- or how would I say it, not a native speaker, sorry; maybe, "feel guilty" -- about duplicates? It is not the first time I see you mentioning that your question is not a duplicate. And you spend lots of time commenting on other posts about duplicates.
Can you construct an outer measure on the intervals [a,b) such that not all Borel sets are measurable?
@SouravGhosh as in, you associative some value to each such interval and proceed with Lebesgue-measure like construction?
and by Borel sets you mean Borel sets with respect to the standard topology of $\mathbb{R}$?
wikipedia mentions two such methods en.wikipedia.org/wiki/…
3:23 PM
@noballpointpen I'm English. In my culture, people apologise for things a lot. Besides, I don't want to waste other people's time with a question I could have answered by searching harder.
@Jakobian He’s reflecting on the fact that the Cantor set is a countable product of $\{0,1\}$ …
@TedShifrin I know
well, at least it seems that way, the other thing is that Koro mentioned the construction of Cantor set on R
either way I was asking about the subjective feeling of it seeming to be discrete
@Shaun ah, then I understand. Indeed. Sorry, if my question was rude :) I mean, people usually don't mention their question is not a duplicate. If it is, or it is a bad question, these sayings do not help.
@ephe Then you’re not doing it right.
@Jakobian Lower limit topology
3:30 PM
@SouravGhosh they generate the same Borel sets
@noballpointpen It wasn't rude. And you have a point there :)
49 mins ago, by Koro
This is quite surprising as from the construction it 'seems' that it should be discrete.
why does it seem to be discrete
isn't that what I asked?
@Jakobian Yessss
I would like to ask people here. Do you just post links in the chat to this site and they automagically get formatted like this link to a comment above? So it does require high reputation, like image uploading? Occasionally I wanted to post a link to MSE here but I don't know how do you format it, the help\faq pages are useless.
3:36 PM
it contains no open interval, which generate the usual topology on R, that's why it seemed that every c in the set should give an open set {c}.
@Koro Cantor had the opposite mindset..
@Koro $\Bbb{Q}$ is totally disconnected. But an open interval doesn't contain only one rational.
@noballpointpen Click the "share" button on the question/answer you want to share here, then post it as a comment here on its own.
If you want to quote a post from chat . . .
@noballpointpen Put the text in [ ] and the link immediately after in ( ).
@SouravGhosh Cantor wasn't the first to define Cantor set it seems
Ted, it only changes the text of the link.
3:42 PM
@Koro Totally disconnected + locally connected iff discrete.
. . . then click on it and the link option should appear. Click on that, and it will take you to a different screen (if you're on a phone). Copy & paste that link into the chat.
We need a special property " locally connected ".
It doesn't work like this, Shaun. I guess, I need high reputation for the chat to format it for me.
why did Cantor introduce Cantor set in his original paper anyway?
May be to give an example of a perfect set which is n.w dense.
3:47 PM
I think he was more interested in set theory
Also because he was Cantor, who believed asking the right questions is more important than finding answers.
especially since the title of the paper in which he introduced it is about set theory
@user726941 what?
Georg Ferdinand Ludwig Philipp Cantor
To ask the right question is harder than to answer it.
I don't understand, then, @noballpointpen. It doesn't seem to be in the privileges
3:54 PM
@Shaun, just to be sure. Post a plain link to a chat message here now.
Like this?
1 min ago, by noballpointpen
@Shaun, just to be sure. Post a plain link to a chat message here now.
Yes. Weird.
Okay, it worked for me (of course) . . .
It is weird.
You need high rep for an "upload" button to appear near "send" to upload images. This is not in privileges list, too. So there are minor privileges that are not documented... at least easily available.
Did Cantor call Cantor set by Cantor set?
3:58 PM
Ah, the upload is documented. Just clicked on a random chat related privilege.
@Koro Open question
you can verify it if you're willing to look into his paper from 1883
well, I suppose, he may call it Cantor set in later paper, though I doubt it's called so in the first appearance
Q: Proof that every non-countable subset of $\mathbb{R}$ has at least one accumulation point.

user34 First of all, I'm not yet familiar with more general concepts like "compactness" and this is a question from the first chapters of my real analysis course. We can regard the reals as a union of the following sets: $\bigcup ([n,n+1]$$\cap$$\mathbb{R}$) with $n \in \mathbb{Z}$. If each of the set...

We can extend this result upto a Lindelof space.
The proof isn't difficult.
Like the proof of "compact sets are limit point compact".
4:22 PM
@SouravGhosh indiscrete space with uncountable amount of points doesn't have this property
but it's Lindelof
@Jakobian Any example other than indiscrete space?
Slight modification: In a Lindelof space, every proper uncountable set has a limit point.
that's also not true
$X\sqcup \{0\}$ where $0\notin X$ and $X$ is uncountable discrete space is a counter-example
ah wait. My bad
$X' = X$
X uncountable indiscrete space
4:37 PM
Okay yeah it is true after all
Yes. The proof is same as "compact sets are limit point compact"
A \subset X uncountable. Suppose $A$ doesn't have any limit point. Then every points of $A$ are isolated.
I see that as a special case of countably compact spaces being limit point compact, so I was being skeptical
$a\in A$ is isolated iff $\{a\}$ is open in the subspace topology.
@SoumikMukherjee agreed, and in relation to the OP it seems like an entire non sequitor
A is compact and $\{U_a: a\in A\}$ is an open cover where $U_a \cap A=\{a\}$
Finite subcover ---->A finite
Correction: A\subset X infinite
4:43 PM
wdym by A compact
Closed subset of a compact space compact
Closed subset of a Lindelof space is Lindelof
Compact: open cover---->finite subcover
Lindelof: open conver ----->countable subcover
ah, yeah
Converse isn't true.
Every uncountable subset has a limit point ----->Lindelof
Uncountable subset has a limit point => realcompact also isn't true
I thought maybe the properties could be related but seem to be different
(the counterexample given in your question, $\mathbb{R}\times \omega_1$, is not realcompact)
5:00 PM
I don't know 🤷‍♀
this could be written more compactly
If Jakobian wrote JAM, then...
$X$ is Lindelof implies $e(X) = \aleph_0$ where $e(X)$ is the extent of $X$
given that extent is defined as supremum, we can't use it to write the same thing for compact spaces nicely, though
$X$ is compact implies closed discrete subspaces of $X$ are finite
@Jakobian Simple enough :)
A is an infinite subset of X ---->A has a limit point in X ----( A closed) limit point in A ---->A is not discrete
5:53 PM
I have a question: How do you construct a regular holomorphic foliation of $\Bbb C^2$ by the class $S=\Bbb C -\lbrace0 \rbrace $ where all leaves are holomorphic to $S?$ Due to the rigidity of the foliations (because of the holomorphic condition) I am lead to believe there does not exist such a construction. In the smooth setting, it's much different because there's more leeway topologically to deform the foliation achieving the goal
Is there a straightforward argument to invalidate a construction in the holomorphic case?
2 hours later…
7:35 PM
Is the function $\frac{1}{i+x}$ summable over $\Bbb R$? I've been asked this question by a friend of mine, but I've not studied complex analysis
@SineoftheTime wdym?
@Jakobian integrable with finite value
Could we agree to create a mathematics free from division? I really don't like division.
It's a common issue in any community
@MatsGranvik I guess you're commenting on that letter to Stack Overflow?
@Jakobian No I was more thinking about the operation division. Addition, subtraction, and maybe multiplication are fine, but division, I really don't like it.
7:46 PM
oh. Lmao
@SineoftheTime It has a Principal Value. It is not summable in the usual meaning.
If I'm correct, the Principal Value is $-\pi i$.
thank you Rob!
Thanks for the verification
Hi @robjohn you could probably help me with this explanation. I was given a sequence $x_k$ defined recursively as $x_{k+1} = f(x_k)$. Also the function $f$ is a contraction mapping. We used this sequence to prove the contraction mapping principle. I was asked to show $\|x_k - x\| \leq \frac{c^k}{1-c} \|x_1 - x_0\|$. I did all this, but it was commented that this gives an estimate on how fast the sequence converges to the fixed point.
What I'm confused about is what is giving the "speed" of convergence? I know it is the term $ \frac{c^k}{1-c}$ but how should I interpret "speed" in this sort of scenario?
all points are multivariable
what is $x$ (with no subscript)?
7:56 PM
it is the fixed point
Ah, sort of $x_\infty$
yea I guess you could say that. Never seen that notation, but makes sense
$\|x_1-x_0\|\cdot \frac{c^k}{1-c}$ is a pretty explicit expression
Well, $\|x_{k+1}-x_k\|\le c^k\|x_1-x_0\|$ correct?
7:59 PM
Speed of convergence is just an estimate on how fast $\|x_k - x\|$ decreases, as $k \to \infty$.
That's definitional, but also intuitive
and $\|x_k-x\|\le\sum\limits_{j=k}^\infty\|x_{j+1}-x_j\|$, correct?
Imagine you are on a computer and are trying to check the error of how many digits of $x_k$ agree with $x$
so sum the geometric series
Oh...I get that it is a geometric series and such. I'm just trying to comprehend what is meant as "speed" in this scenario. So would I say something like "the sequence is converging at the rate of a geometric series"?
8:02 PM
$\|x_k-x\|\le\sum\limits_{j=k}^\infty c^j\|x_1-x_0\|=\frac{c^k}{1-c}\|x_1-x_0\|$
it converges geometrically
You would say that the rate of convergence is exponentially fast. Because, fixing initial seed $x_0$, $\|x_k - x\| = O(c^k)$ for some fixed constant $c < 1$.
Correct. I worked out the proof already. It is the concept behind it
No big O notation yet Balarka, not at that level yet....lol . So "converges geometrically" would be what is meant.
The larger is the $k$, the more computational power you need to compute $x$
The big O notation is just a way of ignoring the junk an saying that the sequence is proportional to $c^k$.
I would use "exponential convergence", that would be the accurate term
The sum converges geometrically. The sequence (which you wrote as a telescoping sum) converges exponentially.
But whatever, really, it doesn't matter
converging geometrically due to the $c^k$ factor that comes about from the necessary manipulations
8:07 PM
If $\|x-x_k\|$ converges to $0$ rapidly, you can be sure that you don't need to compute $x_k$ for as large $k$, hence saving on computational power
I suppose people use "geometric" and "exponential" interchangeably. I forgot $\{c^k\}$ is called the geometric progression, I had geometric series in mind.
Of course those don't have to be real numbers, but I feel like it's a good intuition
I was calling it exponential because $c^k = \exp(k \log c)$
I meant exponential when I wrote that....but wrote geometric....
OK. Thanks for the explanation fellas
Assume $f:X\rightarrow Y$ is a continuous function between metric spaces such that for all small $r>0$, and all $x\in X$, $B_{f(x)}(r)\subseteq f(B_x(r))$. Assume $K$ is compact and $f|_{K}$ is injective, does it follow that $f|^{-1}{K}$ is Lipschitz?
8:27 PM
Howdy @Balarka! Long time no see.
Hi Ted!
@monoidaltransform Don't you need a reverse inclusion for that?
Oh, no, it's right. $f^{-1}(B_y(r))\subset B_{f^{-1}(y)}(r)$.
Howdy @peek-a-boo. Some shocking lack of understanding on that Riemann/Lebesgue integration post ;)
8:48 PM
on my part?
oh you probably mean in general, especially regarding the countability issue
No, certainly not you, silly.
Let $f : X \to Y$ be a measurable function. I guess we could assume that $X$ and $Y$ are (standard) Borel spaces. How does one prove that $graph(f) := \{(x,f(x)) : x \in X\}$ is a (Borel?) measurable subset of $X \times Y$?
the different notions of size, topological vs measure theoretical vs cardinal seems to be a recurring issue (what is small/negligible for one is not the case for the other:)
Not to mention subtleties with induced measures/volume elements on submanifolds :)
ah yes that question as well
8:57 PM
@user193319 Is the function $\text{id}\times f\colon X\to X\times Y$ measurable?
@peek-a-boo The good news is that the good old tensor iff linear over $C^\infty$ functions is not showing up quite so often :P
I got that as a comment though
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