Mathematics - 2023-06-16

D.C. the III

01:26

if $A = A_1 \times A_2 \times \dots$ and $B = B_1 \times B_2 \times \dots$, what is the relationship between the set $A \cup B$ and $\prod_{i \in \mathbb{Z}_{+}}(A_i \cup B_i)$?

After a little work I said the relationship is $A \cup B \subset \prod_{i \in \mathbb{Z}_{+}}(A_i \cup B_i)$. Is there more to it?

Ted Shifrin

Did you draw some 2D examples?

D.C. the III

I did not I just kind of visualized sets in sets......I'm actually going to try that

Ted Shifrin

(Hint: You are correct, however.)

D.C. the III

I deduced it from the logical ideas, but I've never given thought to trying to draw sets in this sort of scenario. SOmething worth trying

Ted Shifrin

After my years of telling you to draw pictures, reallllly?

Remembering, of course, that a picture a proof does not make.

D.C. the III

01:33

Lol.....maybe I did in a way when I was reasoning it and just thinking of a bunch of "dots" in one ball known as A and another ball known as B and expounding from that. But not in a proper geometric way. Now I'm trying to do it on the cartesian plane to see what happens

Ted Shifrin

Wait til you get to product topology with infinitely many factors …

D.C. the III

ha, literally there is a note about that in the part above the one i'm doing

and of course as you suggested, the picture makes it that much more clearer as I now see. :)

Ted Shifrin

@ZaWarudo Did you ever work out your improper integral question? My hint was on the money.

D.C. the III

01:50

and the intersections are equal......just in case you feel like quizzing me.

Ted Shifrin

So does the lack of parallelism trouble you?

D.C. the III

no. should it? Seems like what tends to happen when unions and intersections of sets happen

Ted Shifrin

Like “or” is too inclusive?

D.C. the III

the union does contain the intersection of the two sets.

Ted Shifrin

Not helpful?

D.C. the III

01:59

ah.....If I remove that intersection then I could say the union containment I claimed above is equal

i think....let me look at my "picture" again

Ted Shifrin

Nope. Far from it.

D.C. the III

then I'm stumped

copper.hat

i sometimes wonder if folks take a moment to look at small examples and stop pushing symbols around. math.stackexchange.com/q/4719595/27978

D.C. the III

is $\|\cdot\|_F$ some special norm?

copper.hat

$\|A\|_F^2 = \sum_{i,j} |A_{ij}|^2$. I think we discussed it recently

D.C. the III

02:09

oh oh oh....we did this two days ago! :)

copper.hat

easy to compute with

for come reason i was slightly irked by the OP's attitude in the comments

thanks for the upvote :-) not necessary, of course

basically the inner product is essentially the Euclidean inner product, so it is nice for examples.

D.C. the III

How are you so sure it was me? Could've been the ghost of Cleo...........

copper.hat

:-) the timing

D.C. the III

jeez my typing and spelling have been off lately.......must be age

copper.hat

I just do a periodic scan for easy convex questions.

D.C. the III

02:13

I know you love them

copper.hat

on my way to get the mythical mse jumpsuit

D.C. the III

jumpuit?

copper.hat

i do like convex stuff

its a running joke with leslie the lawyer

when you hit some rep you get some swag

i am very proud of my mse mug

D.C. the III

That I know. you mentioned a mug before. So you saying jumpsuit

copper.hat

the joke was that when you hit 200k you get an mse orange jumpsuit

D.C. the III

02:15

had me thinking what level of rep you have to reach

copper.hat

i don;t think there is any swag at 200k anynmore

D.C. the III

probably would be better if it was an all white jumpsuit with extra baggy sleeves that can get tied together...

and your own personal padded room

copper.hat

:-) i suspect folks might think a bomb shelter would be more appropriate

for me, that is

D.C. the III

man...I know you have some stories from those times.

copper.hat

my earlier gripe was premature, the OP was subsequently gracious

yep, a reminder of some of those days recently on hearing that kaczynski died

D.C. the III

02:19

is there relation beyond the fact that both events included bomb in them?

copper.hat

yep, long story, short version he left a pipebomb beside my computer terminal and my colleague (poor fellow) opened it up and took the blast

D.C. the III

Well..............................bloody hell.

did your colleague make it?

copper.hat

he did, but basically ended his fighter pilot career. incredible grit.

D.C. the III

blimey

copper.hat

i had come from ireland, where the possibility of bombs was a daily thing at the time. was not expecting anything like that in berkeley

only 33k to go for my jump suit

@D.C.theIII have you studied complex analysis?

purely curious.

D.C. the III

02:30

Not yet. Still haven't done Real Analysis yet. Plan on doing it in the fall and if I could squeeze complex in I will too.

WOn't be using Alfors for the first go round of it though. ...lol

copper.hat

real analysis and complex analysis, despite the similar name, are two completely different beasts.

D.C. the III

So I have read....which is why I will probably stick to the plan of real analysis first then attempt complex

copper.hat

i like real analysis, but its a bit of a vortex into which one can easily slip.

i think it is best learned in a few passes.

don't make complete understanding be a limit to progress

i always want to capitalise Boolean, but apparently i should not

D.C. the III

Yea that is my plan. I was going to do it in the way my school does it. They give an introductory to it using Pugh for a semester. Then the next set of semesters they use Folland. The Folland level of it is cross listed between undergrad and grad level

copper.hat

my alma mater was where Boole taught

i like pugh the person.

D.C. the III

02:36

@copper.hat I fall into this trap too much

The author of the book or another Pugh?

copper.hat

he once wrote out a proof for me

i am guessing it must be the same hippy fellow

D.C. the III

well that probably answers the question...lol

copper.hat

he was very human in a nice way, which i needed at the time. he was on one of my committees.

was->is. i jut have not seen him in decades

Ted Shifrin

@D.C.theIII wait ‘til you’re half my age!

@D.C.theIII Complex analysis is more beautiful …

D.C. the III

well considering I'm in the vicinity of Leslie's age I probably am already

copper.hat

02:40

complex analysis is like competition ice skating. real analysis is like technical mountain climbing.

Ted Shifrin

No one is as old as Leslie.

copper.hat

i can't poke at him unless he is present

D.C. the III

before I can become graceful I need to learn how to skate

Ted Shifrin

You poke at me regardless!

D.C. the III

Leslie is always present....watch him just pop up to retort

copper.hat

02:41

only when you are present :-)

i have mellowed in my age, but i am fairly direct

Ted Shifrin

Or unfairly …

robjohn

02:55

@copper.hat Mine is one of the two mugs I use.

copper.hat

@robjohn what is the other?

@TedShifrin true, sometimes directness is a bit selfish. truth is not always necessary.

Ted Shifrin

Well, I am ignoring an increasing number of chatters …

robjohn

@copper.hat Just a 16 oz mug. I use that for a large chai in the morning with breakfast. I use my MSE mug for tea at other times.

@TedShifrin like me?

Ted Shifrin

You’re not so lucky, @robjohn!

I have no mug. Copper has hijacked mine.

D.C. the III

@robjohn tea bags or loose leaf tea?

robjohn

03:01

Darn, I'm being watched, I knew it.

@D.C.theIII I have some loose tea, but the convenience of bags wins

D.C. the III

2 min difference in prep time?

robjohn

and the prep and clean up is cleaner

copper.hat

i was wondering why my mug had the initials TS followed by an infinite list of twin primes on the bottom.

i have given up on tea prep. the water just needs to darken slightly and that is sufficient. no more heating the pot, letting it draw and all that mess

bags are good.

D.C. the III

All this blasphemy......

copper.hat

one step closer to my jump suit.

my kids cringe when i produce my mse mug, especially when i use to to serve a drink to one of their friends.

i know they know i know that it is THE mug.

robjohn

03:06

@copper.hat I often make sun tea in the summer. It's been too cloudy here to make it this year, for the most part.

copper.hat

they just hope that by not acknowledging it that the matter won't come up.

@robjohn i had to google that, not my cup of tea so to speak

robjohn

@copper.hat It is pretty good, and a lot of the acidity is muted. Sugar is not needed.

Ted Shifrin

Sugar is despicable!

copper.hat

@robjohn i think tea is more of a habitual thing for me rather than taste

i agree with Ted's sugar perspective, but it took me many years

D.C. the III

Been gradually cutting it out. Definitely is hard thing to do

Ted Shifrin

03:11

Ted is, of course, bitter and sour.

copper.hat

i like limes

robjohn

@TedShifrin I usually use honey

@copper.hat in tea?

copper.hat

@robjohn ohh nooooooo, just skim milk.

the limes was a response to Ted's comment

@D.C.theIII if you want to get a feeling for the true an delightful evilness of real analysis i would suggest perusing "Counterexamples in Analysis" by Gelbaum & Olmsted.

D.C. the III

will add it to the list

copper.hat

wow, i ramped up 70 rep today. truly am a rep hound.

the closer gang seem to have disappeared

because some of my recents are definite closing fodder

that was a bear i was happy to poke

robjohn

03:18

@copper.hat Trader Joe's chai mix has milk in it. Otherwise, I don't use milk in tea.

copper.hat

i need my calcium :-) but i have always had tea with my milk

robjohn

@copper.hat I'm on calcium supplements to counteract another med I'm on.

copper.hat

osteoporosis seems to be popular in my family

peek-a-boo

@copper.hat wow I took a brief preview of that text, and looks like with that book in hand, first year students can answer a good chunk of their assignments lol

copper.hat

@peek-a-boo it is not for the faint hearted, but yes indeed, many read to write answrs

Thomas Finley

05:44

2

Q: Possible Absurdity while proving $1-\frac 12-\frac14+\frac 13-\frac 16-\frac 18+...$ converges to $\frac 12\log 2.$

Prove that the series $$1-\frac 12-\frac14+\frac 13-\frac 16-\frac 18+\frac 15-\frac {1}{10}-\frac {1}{12}+....$$ converges to $\frac 12\log 2.$ I tried solving the problem as follows: The series given is, $1-\frac 12-\frac14+\frac 13-\frac 16-\frac 18+\frac 15-\frac {1}{10}-\frac {1}{12}+...$. L...

0

Q: Problem while proving the series $1+\frac 13-\frac 14+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+...$ converges to $\frac 32\log 2.$

Prove that the series $$1+\frac 13-\frac 14+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots$$ converges to $\frac 32\log 2.$ I tried solving the problem as follows: The series given is $$1+\frac 13-\frac 14+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots.$$ We writ...

real-analysis sequences-and-series

Does these two questions have the same mistake ? I gazed upon them for a long while now, but failed to find any mistake with the 2nd one.

So, yeah, I need a bit help with the 2nd post above.

@copper.hat which book ?

copper.hat

"Counterexamples in Analysis" by Gelbaum & Olmsted.

@ThomasFinley You cannot rearrange terms in a conditionally convergent series.

and expect to get the same answer.

Thomas Finley

@copper.hat I rearranged the terms in the partial sum,not the series. So, isn't that allowed?

@copper.hat oh, thanks!

Sourav Ghosh

Q1: Sequential space+( ? ) \implies first countable

Sourav Ghosh

08:16

A topological space $X$ is disconnected $\implies \exits U, V$ two non empty open sets such that $Cl(U) \cap Cl(V) =\emptyset$

Contrapositive: A topological space $X$ be such that $Cl(U) \cap Cl(V) \noteq \emptyset$ for any two open sets $U, V$ , then $X$ is connected.

Thomas Finley

I found the issue, with my second post i.e

2

Q: Problem when proving the series $1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+...$ converges to $\frac 32\log 2.$

Prove that the series $$1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots$$ converges to $\frac 32\log 2.$ I tried solving the problem as follows: The series given is $$1+\frac 13-\frac 12+\frac 15+\frac17-\frac 14 +\frac 19+\frac 1{11}-\frac 16+\cdots.$$ We writ...

real-analysis sequences-and-series

I had not evaluated the limit correctly

But then, a new issue arises about which I wrote in the post. I am writing it hereby as well, to make things simpler.

## A New Issue

The evaluation of the limit .i.e $\lim S=0$ is incorrect. It should be, as follows:

>$$S=\frac1{6n+2}+\frac{1}{6n+4}+\cdots+\frac{1}{12n}\implies 2S=\frac1{3n+1}+\frac1{3n+2}+\dots+\frac1{6n}\\=\left(1+\frac12+\dots+\frac1{6n}\right)-\left(1+\frac12+\dots+\frac1{3n}\right)\\=\big(\gamma_{6n}+\log(6n)\big)-\big(\gamma_{3n}+\log(3n)\big).$$ This means, $\lim 2S=\log 2,$ or $\lim S=\frac 12 \log 2.$

But here's the problem, what went wrong with this reasoning, i.e the reason by which, I concluded $\lim S$ is $0$ which is:

Sourav Ghosh

[Contd.] Call such topological spaces "strongly connected spaces."

Now consider the topological $(\Bbb N, \tau_{\text{Golomb}}) $.

Given any two basic open sets $B_1=B(a_1, d_1) =\{a_1 +k d_1 :n\in\Bbb{N}_0\}$ and $B_2=B(a_2, d_2) $

$d_3=d_1d_2$ is a limit point of both $B_1$ and $B_2$

So Golomb space is strongly connected.

Golomb space is a strongly connected $T_2$ space.

I am not able to find any other strongly connected space with at least two points that satisfy a stronger separation axiom.

Alessandro Codenotti

08:44

@SouravGhosh because there is none

Sourav Ghosh

09:03

@AlessandroCodenotti Correct✅. There is no such completely $T_2$ space.

Ok. Such spaces are at most $T_2$

Thomas Finley

09:20

Hey guys! Any idea how to show that the series: 1+1/3-1/2+1/4+1/5-1/6+... is divergent ?

冥王 Hades

09:35

Wonder what the average IQ of this room is

Thomas Finley

16 mins ago, by Thomas Finley

Hey guys! Any idea how to show that the series: 1+1/3-1/2+1/4+1/5-1/6+... is divergent ?

Comparison Test, Limit Form Test, D'Alambert's Test, Cauchy's Root Test, Cauchy's Condensation Test, Raabe 's Test don't work

This is because, they all are defined for series of positive real numbers

Neither does Gauss Test and Logarithmic Test works, because of the same reason.

Leibniz's Test isn't working either.

The series is not an absolutely convergent series either.

Not conditionally convergent as well.

Now, I am out of ideas after seeing that the general root rest and ratio test not working.

Koro

65

A: Is the cofinite topology on an uncountable set first countable?

You are correct: it is not first countable. However, this is not because each point of $X$ has uncountably many nbhds: each point of $\Bbb R$ also has uncountably many nbhds, but $\Bbb R$, being a metric space, is certainly first countable. To prove that $X$ is not first countable, you must show...

I understand this.
But I don't think I could have ever come up with this on my own without taking a look at the solution.

and this is discouraging.

Koro

10:08

@ThomasFinley do you know that given $H_n= 1+1/2+1/3+...+1/n, \lim_{n\to \infty}(H_n-\log n)\to \gamma $?

you can try to generalise this: start with m 'odd' terms then one even term and keep repeating like this. in your current case, you have m=2. (2 'odd' terms followed by one even and so on.)

Thomas Finley

@Koro yes, I am aware of the result.

@Koro no idea, how?

Koro

write the sum of first 3n terms $s_{3n}$

find $\lim s_{3n}$

do the same with $s_{3n-1}, s_{3n-2}$.

b_jonas

10:27

@ThomasFinley I don't understand how that series is supposed to be defined. Is there a post?

oh, it's math.stackexchange.com/q/4719788/24908

Sourav Ghosh

@Koro Assume the contrary that $X$ is first countable.Fix $x\in X$ and a countable local base $(B_n)$ at $x$. Let $y\neq x$.Then $X\setminus \{y\}$ is an open set containing $x$ , hence $\exists m\in\Bbb{N}$ such that $y\not \in B_m$.Then $B=\cap_{n} B_n=\{x\}$. Hence $X\setminus \{x\}=\cup_{n} X\setminus B_n$

$|X\setminus B_n|<\aleph_{0}$ implies $|X\setminus \{x}|=\aleph_{0}|$

Koro

@SouravGhosh did you read why I posted this question here?

Sourav Ghosh

10:43

You can prove it yourself with a few hints,no need to see the full solution.

Koro

@ThomasFinley: so to take care of these type of questions, try to prove the following: $1+1/3+1/5+...+1/(2a-1)- (1/2+1/4+...+ 1/(2b)) + 1/(2a+1)+...= \log \left(2\sqrt{\frac ab}\right)$. This is the sum of a 'odd' nos. followed by b 'even' terms.

Sourav Ghosh

First countable implies sequential space. Not sequential implies not first countable.

But cofinite topology is sequential 🙃

s.harp

11:08

How do you say something like "Independent of the above characterisation, these things are objects of interest and have been studied before" in a better way? The formulations "Independent" and "objects of interest" are very clumsy

Koro

@ThomasFinley I answered your question.

Balarka Sen

11:38

@copper.hat laughed audibly a little

user1176409

11:50

Hello

Koro

2 hours ago, by Koro

I understand this.
But I don't think I could have ever come up with this on my own without taking a look at the solution.

2 hours ago, by Koro

and this is discouraging.

@Jakobian please advise.

:(

I want to show first that cofinite R is not 2nd countable. So it should suffice to show that given any countable collection of open sets {O_n}, there is an open set U such that not even the intersection lies completely inside U.

It is clear that if $A= \cap_n O_n$, then $A^c= \bigcup_n O_n^c=$ a countable union of finite sets so $A^c $ is countable. $\implies A$ is non empty.

not sure how to construct $U$ now.

infact, A is uncountable.

Jakobian

12:11

@SouravGhosh Urysohn is a weaker axiom

Koro

U= A^c\cup (R- {a}) works, where a is an element of A.

Jakobian

@SouravGhosh could you explain this?

s.harp

12:28

nounification of holomorphic, do you take "holomorphy" or "holomorphicity"?

Koro

similarly, using the same idea, it follows that cofinite R is not 1st countable.

$\ddot\smile$

Monty

13:19

does anyone know how to construct a function orthogonal to the gradient? For instance if I have a function $f:\mathbb{R}^d\to\mathbb{R}$ and I consider the vector field $J\nabla f$ where $J$ is an antisymmetric matrix, then $\nabla f(x)$ and $J\nabla f(x)$ are orthogonal. Are there any other ways to construct orthogonal vectors?

copper.hat

13:55

@BalarkaSen at what?

Thomas Finley

14:27

I had a strange thought bugging me. Are these two statements same:" If P, then Q" and "Q is true, if P is true."

Monty

@ThomasFinley i think so

sunny

> A set $F\in\mathbb{R}$ is closed $\iff$ the limit of every convergent sequence in $F$ belongs to $F$.

I am reading a proof of this proposition and am stuck on the $\Longleftarrow$ direction.

> $\Longleftarrow$ Suppose that the limit of every convergent sequence of points in $F$ belongs to $F$. Let $x\in F^c$. Then $x$ must have a neighborhood $U\subset F^c$; otherwise for every $n\in\mathbb{N}$ there exists $x_n\in F$ such that $x_n\in (x-1/n,x+1/n)$, so $x=\lim x_n$, and $x$ is the limit of a sequence in $F$. Thus, $F^c$ is open and $F$ is closed.

I am getting so hung up on the word "otherwise". Second, I do not understand the point the author is trying to make with $x_n\in (x-1/n,x+1/n)$. What does this tell us?

robjohn

16:13

@ThomasFinley yes

also "P is true only if Q is true"

Ted Shifrin

16:34

@ThomasFinley The pattern is not clear from this. But it looks like ++-++- …. Except in the first set you switched two.

Ted Shifrin

16:53

@sunny You mean $F\subset\Bbb R$, of course. If $x$ has no neighborhood contained in $F^c$, then every neighborhood, e.g., $(x-1/n,x+1/n)$, must contain a point of $F$.

So he's creating a sequence $\{x_n\}$ of points in $F$ that converges to a point $x\in F^c$.

Jakobian

17:14

If we take a neighbourhood $B(d, a)$ of $d$, $d+k_0a = a_1+k_1d_1$, $d = d_1d_2$, $k_0, k_1\in\mathbb{N}$, $\gcd(a_1, d_1) = \gcd(d, a) = 1$

the question is if such $k_1, k_2$ must exist

everything else is given

robjohn

@ThomasFinley Can you say what the general term of the series is? from what you've given, it is hard to tell what we are summing.

Jakobian

I guess, this is $k_0a = a_1 \pmod{d_1}$ and then $k_0 = a^{-1}a_1\pmod{d_1}$ since $\gcd(a, d_1) = 1$

robjohn

@ThomasFinley Is the series supposed to be $1+\frac12-\frac13+\frac14+\frac15-\frac16+\dots$?

the switching of $\frac12$ and $\frac13$ is confusing

Ted Shifrin

@robjohn Precisely what I mentioned. We are of a single mind.

robjohn

Unfortunately, Thomas seems to have popped in for a bit, and is now gone.

Same thing happened yesterday

Jakobian

17:25

okay now I see why Golomb topology has the property that closures of any two non-empty open sets intersect

Ted Shifrin

@robjohn A lot of nerve.

robjohn

It certainly makes things take longer.

Jakobian

it's an interesting concept, reminds me of hyperconnectedness, but it's a weaker property, maybe it could be called "anti-Urysohn" just like people call hyperconnected spaces "anti-Hausdorff"

or maybe "weakly hyperconnected" or "weakly irreducible"

if it has any relation to algebraic geometry then we could try borrowing a name from there

Thomas Finley

@robjohn,@TedShifrin Oh! A lot happened since I was gone for a bit. Yes, the general term of the series $1+\frac12-\frac13+\frac14+\frac15-\frac16+\frac 17+\frac 18-\frac 19+\cdots$

It seems it was asked here, before.

Ted Shifrin

Oh, now it’s corrected andceasy.

robjohn

17:37

Show that is greater than $1+\frac14+\frac17+\dots+\frac1{3n+1}+\dots$

Ted Shifrin

Oh, now it’s corrected and easy .

Thomas Finley

4

Q: Abbott Understanding Analysis 2nd ed Exercise 2.7.2

I am trying to decide whether the series converge or diverge and am struggling with two different series. I am able to use the comparison test, the absolute convergence test, the alternating series test, and the ratio test. The first series is: $$1 + \frac12 - \frac13 + \frac14 + \frac15 - \frac...

real-analysis

Ted Shifrin

Again @robjohn and I are of a single mind.

Use no test at all. Just thought.

Thomas Finley

@TedShifrin thought made me conclude, the series was divergent. But I needed some rigorization.

Jakobian

I found arxiv.org/pdf/1509.01420.pdf mentioning anti-Urysohn spaces, but they demand that intersection of any two non-empty regular closed sets to be non-empty

Ted Shifrin

17:40

Group the 2nd and 3rd terms in each group.

Thomas Finley

@robjohn I think I know what you're talking abt. But yes, I think, rigorizations are often hard to come by.

Ted Shifrin

Looks like ChatGPT has struck again.

@ThomasFinley Not in the least. Compare partial sums $s_{3n}$ to those of an obvious series.

Thomas Finley

@TedShifrin Maybe, grouping 3 terms makes things simpler. It made things magical and the answer in that link seems to be following that.

Ted Shifrin

That’s what thinking leads you to. :)

Thomas Finley

@TedShifrin That's the trick, but say, if one is unaware of this way of approach, things will surely go messy in his/her life!

Jakobian

17:43

closure of any non-empty open set is a non-empty regular closed set, and conversely

Thomas Finley

@TedShifrin Yes, everything logical is achieved by thinking, imho.

Ted Shifrin

Math takes experience to gain insight.

5

Mittens

@ThomasFinley: you original problem is about putting paranthesis in a series. the general term of your series is $a(3n)=-1/(2n)$, $a(3n-1)=1/(2(2n)-1)$ and $a(3n-2)=1/(2(2n)-3)$. Adding parenthesisin this case means that instead of considering $\sum_na_n$ one considers $\sum_nb_n$ were $b_1=a(1)+a(2)+a(3)$, $b(2)=a(4)+a(5)+a(3)$, etc.

Jakobian

okay this is precisely the same condition then. So I found a paper mentioning those "strongly connected" spaces as anti-Urysohn in literature

Thomas Finley

@TedShifrin yes, that's the thing.

Ted Shifrin

17:44

I meant thinking as opposed to algebraic contortions.

I often put this very question on my exams in the Spivak course. Pretty much all the students got it.

Thomas Finley

@Mittens maybe, but now I have to run off for a bit again.

@TedShifrin good students?

I should say, insightful guys

Jakobian

@SouravGhosh check out arxiv.org/pdf/1509.01420.pdf , those spaces are called anti-Urysohn

ugh

Thomas Finley

Anyways, bye, running late....

Oh, @robjohn and @TedShifrin thank you for ur time. Not to forget @Mittens, as well

robjohn

18:05

@ThomasFinley just to complete the thought $1+\overbrace{\left(\frac12-\frac13\right)}^{\ge0}+\frac14+\overbrace{\left(\frac15-\frac16\right)}^{\ge0}+\dots\ge1+\frac14+\dots$

missed them again

Mittens

@ThomasFinley: It is a basic result in that if the $n$-th term of a series $\sum_na_n$ converges to $0$, i.e., $a_n\xrightarrow{n\rightarrow\infty}0$ and parenthesis of bounded length are introduced(i.e. given strictly increasing function $p:\mathbb{N}\rightarrow \mathbb{N}$ with $p(n+1)-p(n)$ bounded), then $\sum_na_n$ converges iff $\sum_nb_n$ converges, and in such case the sum is the same.

Mittens

18:41

@ThomasFinley: the more general series that you wrote $1+\frac13+\ldots+\frac1{2a-1}-\frac12-\ldots-\frac{1}{2b}+\frac{1}{2a+1}+\ldots +\frac{1}{4a-1}-\frac{1}{2(b+1)}+\ldots \frac{1}{4b}+\ldots$ is handle also by introducing parenthesis. In this case we introduce parenthesis of length $(a+b)$

@ThomasFinley: The form of the $n$-th term of the series $\sum_n\alpha_n$ is a little more complicated, but the term of the series with parenthesis is easy:$ b_n=\frac{1}{2(n-1)a+1}+\ldots \frac{1}{2na+1}-\big(\frac{1}{2((n-1)b+1))}+\ldots+\frac{1}{2nb}\big)$. To obtain the sum you use the scaling property of the harmonic series: $H_n:=\sum^n_{k=1}\frac1k = \log n +\gamma +O(1/n)$ where $\gamma$ is a well known constant.

冥王 Hades

Heh, speak of the devil

sunny

18:57

@TedShifrin thanks for the reply! which is a contradiction, right? Because our assumption was that the limit of every convergent sequence of points in $F$ belongs to $F$. The proof is by the way from here, page 92.

My question regarding this proof was asked on the main site before. As I have commented to the answer, the contradiction seems still somewhat unclear, but it has to be what you said...that we have a sequence of points in $F$ converging to a point in $F^c$...contradiction! The answer in the link is in-line with this.

Ted Shifrin

19:46

@sunny Yes, of course.

Balarka Sen

hi Ted

sunny

Cool, thanks!

Ted Shifrin

19:57

Howdy, a Balarka. That is a long and weighty tome. I'll skim through it, but it'll take a while!!

Balarka Sen

20:16

Thanks, @Ted! No hurries, or obligation.

冥王 Hades

^My honest reaction

Sha Vuklia

21:00

Let $S=\bigoplus_{d\geq 0}S_d$ be a graded ring. For $n\in\mathbb N$ let $S(n)$ be the graded $S$-module defined by $S(n)_d=S_{n+d}$. I've seen the claim that the canonical map
$$
S(n)\otimes_S S(m)\to S(n+m).
$$
is an isomorphism. Surjectivity for me is clear. Injectivity not. Since the tensor product commutes with direct sums, we have
$$
S(n)\otimes_S S(m)=\bigoplus_{i,j}S_{n+i}\otimes_S S_{m+j}
$$
while
$$
S(n+m)=\bigoplus_d S_{n+m+d}.
$$
Let's assume $n=m$. If we take $i\neq j$, then we have the direct summand

冥王 Hades

Just realized how awfully old the chat room looks

Straight out of 2007

SE (the company) doesn’t seem to be interested in improving the experience, just as they aren’t interested in what the community says

copper.hat

what is wrong with old?

冥王 Hades

What isn’t? It’s clunky and just looks bad overall.

copper.hat

it does the job

冥王 Hades

Some buttons can also be relocated and changed in size and much more

It’ll be far more convenient

Sha Vuklia

21:07

oh, hm

9

Q: If $M_*$ and $N_*$ are graded modules over the *graded* ring $R_*$, what is the definition of $M_* \otimes_{R_*} N_*$?

Quick question (hopefully): What is the correct definition of a tensor product of two graded $R_*$-modules and/or graded $R_*$-algebras $M_*$ and $N_*$ over the graded ring $R_*$? $M_* \otimes_{R_*} N_* = ?$ If R is not graded I know how to do this, but when $R_*$ is graded the usual construct...

algebraic-geometry commutative-algebra graded-rings graded-modules

apparently there is a different definition of tensor product in the case of graded modules

hm, no, it does seem to be the ordinary tensor product

Ted Shifrin

@冥王Hades Who the h*** cares?!

冥王 Hades

@TedShifrin the guy you pinged does

Ted Shifrin

Irrelevant ...

Everyone's complaining about the so-called improvements on the main site (vote buttons, etc.) ...

冥王 Hades

They’re right to complain. The updates are horrible

Ted Shifrin

So of course updating the chat site would be horrible, too.

冥王 Hades

21:17

For a network that prides itself in being a collection of high quality questions and answers they sure suck at using all that knowledge

Ted Shifrin

I think you're just full of hot air.

冥王 Hades

I just ask for a better experience.

Ted Shifrin

Why don't you worry more about contributing good mathematics to the site?

冥王 Hades

Are they mutually exclusive?

Ted Shifrin

I haven't seen much of the latter.

冥王 Hades

21:25

I post questions/answers that I find fun, if you do not, I cannot and will not change your mind.

s.harp

I don't know why, but visual updates tend to make every corner round, every button shiny/transparent

leslie townes

21:45

it makes the user experience less hazardous. reduces the chances that you will scratch or cut yourself on a pixel

ted: munchkin may finally have gotten covid. a kid who was in her class on wednesday tested positive yesterday, and now she has a fever and sore throat. on the plus side, she's home, and can assist me in working from home.

sunny

> Complete [metric] spaces are ones in which the converse is true.

What does this mean? I've tried Google, but without success. Maybe an example would help.

Sha Vuklia

my question regarding the tensor product has been resolved on main btw

copper.hat

i always get anxious around tensors

who chose the name anyways

Ted Shifrin

22:08

@leslietownes Sounds like it. I hope she doesn’t suffer too much and become more insufferable!

@sunny The converse of what? The statement is that every convergent sequence is Cauchy?

shintuku

have we always been in such a frenetic élan of technological progress or is it just me that feels like this year is packed

sunny

@TedShifrin Yes, exactly. I misread something :) My bad, sorry, but thanks! (so in a complete metric space we also have that a Cauchy sequence converges)

Ted Shifrin

That’s actually the definition of complete.

sunny

Yes.

Ted Shifrin

So no questions, then.

sunny

22:17

No :)

leslie townes

@TedShifrin she's mostly napping, which is kind of sad, because it means she really is sick.

Ted Shifrin

When I got it a few months ago, that’s what I did for a few days.

冥王 Hades

I really gotta learn to put the phone down

Wonder how much of my deteriorating eyesight is because of this habit

Transcript for

$Mathematics$ Mathematics