Mathematics

3:13 AM

Isom. thms for rings hold for nonunital rings right?

I looked up the proof but I think it doesn't matter if the ring is unital or not.

4:34 AM

hard to say for sure without more specificity as to what you mean by "isomorphism theorems" and how you phrase them, but broadly speaking yes, it shouldn't make too big a difference, maybe no difference as long as you phrase the results the right way. "old enough" algebra books often include some results of this type.

shintuku

i have never encountered a ring without unity in my 2 weeks of ring theory

christ this profile pic is terrible

why would they do this

what happened to the old profile pictures

leslie townes

did yours change? the automatically generated ones sometimes do. koro's changed recently.

shintuku

yes it changed

it's no longer anthropomorphic and it's way greener

leslie townes

oh, you can switch back. you just need a copy of the image, or the hash that was used to generate it. gravatar.com/avatar/… in your case.

shintuku

WOAH!!!

thanks leslie!

leslie townes

4:44 AM

i noticed koro's because it was accompanied by a change in color even more dramatic than your green-getting-greener.

shintuku

yeah he went to like grey

how did you find the hash btw?

test123

leslie townes

the internet archive keeps occasional copies of old profiles. web.archive.org/web/20220816162640/https://…

shintuku

ahh very cool

leslie townes

it might take a minute for the servers to catch up to your "new"/old image.

i used a gravatar for sufficiently long that i grew attached to it and kept a copy of it, but i like my current image.

shintuku

there was not enough water in the pool

or the pool was not adequately sized

D.C. the III

4:58 AM

WHat have they done to my avatar???

Reading up a few lines I see I'm not the only on eperturbed

shintuku

find your profile page on internet archive

and save your pic

D.C. the III

And we shall return to our "old" tempered green in a few

Sourav Ghosh

5:20 AM

1) Can a discontinuous function on a connected domain have connected graph?

2) Can a discontinuous function on a path connected domain have path connected graph?

I know 1) is true.

For an example, we can choose f:I\to\Bbb{R} such that f is discontinuous only at a single point which also an end point the interval and the corresponding point in the graph is a limit point of the rest of the graph.

f:[0, 1]\to\Bbb{R} defined by f(x) =\sin(\frac 1 x) ;x\in(0, 1] and 0 otherwise

Alessandro Codenotti

@SouravGhosh what is a domain here?

leslie townes

if you look at a portion of a helix such as {(cos(t), sin(t), t): t in [0, 2pi)} as the graph of a real-valued function on the unit circle, you get a nice example for both (1) and (2)

and yes what is a "domain" here, is my example even an example

the things that are often taken as branches of the complex logarithm on C \ {0} would also work, they will have discontinuities on whatever path you cut along, but this won't disconnect or path-disconnect the graph

Sourav Ghosh

6:15 AM

@AlessandroCodenotti Any path connected space

Alessandro Codenotti

7:13 AM

Then Leslie's examples work

leslie townes

you can change the answer to (2) if you impose more conditions on the domain. e.g. math.stackexchange.com/questions/2577156/… shows that if 'domain' means 'subset of R' the answer to (2) changes. it also links to examples like the one you gave for (1).

leslie townes

7:31 AM

it makes sense that R would be somewhat special here. it is pretty restrictive in terms of how you can "get to" one point from another. not a lot of ways of sneaking in a discontinuity in a way that doesn't also disconnect a graph.

Sourav Ghosh

7:50 AM

@leslietownes Thanks :)

user1176240

Can anyone cite the source, page # for the scan in math.stackexchange.com/q/4003576? Thanks!

PM 2Ring

For people wondering why their gravatar changed recently:

4

62

Q: We're regenerating all identicons

We've recently discovered a vulnerability in our identicon generation process. To remedy it, we have changed how we approach generating them and regenerated all identicons. We do not have any indication that any personally identifiable information (PII) was leaked as part of this vulnerability. S...

^ @shintuku @D.C.theIII @Semiclassical

user 85795

Thanks @PM2Ring 🙏

PM 2Ring

No worries.

leslie townes

8:06 AM

@user1176240 it appears to be from some edition of steven lay's "analysis with introduction to proof." maybe the fifth edition. page 280 in the version excerpted here: math.colorado.edu/~walter/Math3001/section7.1pp275to283.pdf

PM 2Ring

I made an animated interactive version of Prince Rupert's cube in Sage.

sagecell.sagemath.org/…

user 85795

\o @copper.hat how's it going, pal?

Are you over the jet-lag yet.

Sourav Ghosh

I am going to show that every metric space is normal.

Let A, B be two non empty disjoint closed sets. f:X\to \Bbb{R} defined by f(x) =d(x, A) /{(d(x, A) +d(x, B) }

Continuous and well defined.

Koro

@shintuku you got it back :).

Sourav Ghosh

f(A) ={0}

f(B) =d(A, B) /d(A, B) +d(B, B)

A, B disjoint closed sets. Does this implies d(A, B) >0 ?

Koro

8:18 AM

yes

Sourav Ghosh

Gr(exp) and x-axis in R^2

Alessandro Codenotti

@SouravGhosh no

Koro

ohh 😮😮

Sourav Ghosh

I guess so. One closed set must be compact.

Sourav Ghosh

8:35 AM

f(x) =0 implies d(x, A) =0 implies x\in A

f(x) =1 implies d(x, B) =0 implies x\in B

Ignore what I have done above 😪. f(A) ={d(x, A) /(d(x, A) +d(x, B) : x\in A}

d(x, A) +d(x, B) >0 for all x\in X

Done!

f^{-1}{0}=A and f^{-1}{1}=B

We can choose two disjoint open sets containing 0 and 1 respectively (thanks to Hausdorff :).

Their pre images separates A and B

Hmm

10:50 AM

Hey guys, I'd love to chat with a transdisciplinary, meta / holistic mathematician about a problem I'm trying to solve.

It has to do with sweeping the entirety of an N dimensional state space with a single variable, or finding a mapping from an N dimensional space to an X dimensional space (without any data, not at all talking about dimensionality reduction techniques like PCA or t-SNE or UMAP that require data)

But I want to do this hierarchically (like an autoencoder, or how I imagine a renormalization group works even though I barely know anything about it), and (hopefully) in some loc…

So imagine a second (44.1K samples / numbers) and we want to hear every possible second. This problem is a bit like using a hilbert space filling curve, but I've already tried implementing that and running into some problems programmatically. That solves my "many to one" dimension problem but not my "many to many"

or "many to fewer"

It's also a bit like a poincare disk, or any fractal

Any ideas, or concepts, or books you think I should look at would also be really helpful!

one potato two potato

12:12 PM

For UFD $R$, if $a = bq$ for some $a,b,q\in R\setminus\{0\}$ then $R/(b)\simeq (q)/(a)$ as an $R$-submodule of $M$. Does this imply there exists a unique maximal ideal $M$ of $R$ with $M\simeq R/M$ as an $R$-module?

writing $a$ as a product of irreducibles,,, but it's PID that irred element generates maximal ideal

冥王 Hades

Hamburger

one potato two potato

Correction: there exists a unique maximal ideal $\mathfrak{m}$ of $R$ such that $M\simeq R/\mathfrak{m}$ as an $R$-module

Well actually, $(q)/(a) = \{r+(a)\in R/(a)\mid sr\in (a)\quad\forall s\in(b)\}$.

one potato two potato

12:52 PM

what is the initial object of the full subcategory of unital rings with $1+\cdots+1$ invertible (in the ring)?

Thorgott

1:29 PM

how many additions is that supposed to be

one potato two potato

Fixed $n$. Sorry

Thorgott

2:03 PM

$\mathbb{Z}\left[\frac{1}{n}\right]$ then, no?

one potato two potato

ah

thanks. it factor through fraction ring

Koro

2:26 PM

3

A: Existence of a set $A\subset\mathbb{R}$ such that $|G\setminus A|=\infty$ for every open set $G\supset A$

You are indeed right that your set $A$ need be non-measurable and uncountable (after all any countable set is measurable). Let $S \subseteq [0,1]$ be a non-measurable set. By the criteria that you stated there must exist $\epsilon_0 >0$ such that $|U \setminus S| > \epsilon_0$ for any open set $U...

why is $|U\setminus (S+n)|\ge \epsilon_0$?

Koro

3:21 PM

I understand this now.

But why is $|U \setminus (S \cup (S+3))| \ge |(U\cap(-1,2)) \setminus S|+|(U \cap (2,5)) \setminus (S+3)|$?

Sourav Ghosh

3:45 PM

"Möbius transformation is always a bijective holomorphic function from the Riemann sphere to the Riemann sphere"

Tz=(az+b) /(cz+d)

for c=0 , it's a polynomial. So infinity is a pole.

How we define holomorphicity at infinity?

Semiclassical

4

Q: Holomorphic at infinity (definition)

I struggle quite a bit with the usage of $\infty$ in complex analysis. In some cases, I can translate a definition involving infinity to equivalent statements using limits, or in the case of continuity I just make use of the topology defined on the Riemann sphere. However, what I don't understan...

complex-analysis

Sourav Ghosh

I haven't understood the answer.

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$Mathematics$ Mathematics