2:05 AM
Balarka!

Hi @anak!

I came across your papers on stratified spaces the other day and thought of you.

Oh lol glad you remembered me

I was trying to find something about symplectic reduction and it turns out you did a stratified symplectic reduction.
Hope you have been well and having fun with math still.

Sort of. Our version of stratified symplectic spaces do not quite gel well with symplectic reduction, but we state a relationship.
I was very tired after thinking about stratified spaces for a long time, so I switched gears and now I am trying to learn contact topology.

2:14 AM
Contact topology is thrilling.
I was going to try to do some stuff with contact geometry, but I don't think anything will come out of it. I have been doing mostly just Riemannian stuff recently, so I strayed from my beginnings.
What sort of stuff are you looking at in contact topology?`

Oh cool. Maybe we can read together again, finally.
@anak My MSc thesis, which I am currently writing, is about loose Legendrians. I am thinking about Honda-Huang a lot lately.

Have you been in contact with Emmy at all?

Nope, but that'd be fantastic lol
Maybe I'll have an occasion someday
I'm a big fan of her work

Yeah, you were really into the Eliashberg-Murphy paper.

What are you doing in Riemannian geometry?
(In the sense of, what the hell are you doing over there, chief?)

2:22 AM
flopping around like a fish trying to find an interesting enough result for a thesis and failing
the usual

Pretty much. Found any good music lately?

I think I went through a couple of musical phases since we last spoke.

A couple?

In the middle I was listening to Burial a lot. Future garage/UK garage/dubstep.
The umbrella term for what I was listening to at that time is "hauntology", you can look that up.

2:28 AM
british deconstruction movement, intriguing

yeah lol

UK garage is also british...I am sensing a theme

is figuring out whether $I_? = \{h/h' \in K[x,y]_{(x,y)}[1/y]: h \notin \langle y \rangle \frac{K[x,y]}{(x,y)} \}$ is an ideal of $K[x,y]_{(x,y)}[1/y]$ messy or am i doing it wrong

the notation is sadistic in the least

@anak indeed

2:36 AM
I think the only british house artist i know well is Fatboy Slim
but he's like big beat/acid

$\langle y \rangle$ powers of $y$; $(x,y)$ ideal at which we localize, $[1/y]$ is what we're extending by.

Any current rec.s for what you are listening to right now?

I really enjoyed "Engine of Hell" by Emma Ruth Rundle, which I heard a couple weeks ago.

but she isn't from the UK!!!

Yeah, I got over the UK phase :P

2:39 AM
American post folk now? Or this is a one off?

One off, do not have any coherent themes ongoing.

Sometimes it's good to not have a musical commitment.

True

I like those seasons where I can jam to pretty much any tune.

2:59 AM
Fill tool is a bit strange in Inkscape but this will do I think

3:13 AM
Does there exists any set $A$ containing the Cantor set $C$ such that $D(A) =A-A$ doesn't contain any non empty open interval?

i used a frac instead of a set minus
we call that optimal notation

Q.2) Let $A_n\subset \Bbb{R}$ be such that $D(A_n)$ doesn't contain any interval. Can $\cup_{n\in\Bbb N} A_n$ contain an interval?
@shintuku This is not setminus.It is Minkowski difference or algebraic difference.
We call the "difference set"

Doesn't $A - A$ contain $C - C$? Isn't $C - C$ an interval?

@BalarkaSen Done!
Ignore :)
Both problems are done:)
So $S=\{A\subset \Bbb R : D(A) contains an interval }$ is not a sigma ideal.
Countable sets, measure 0 sets, meager sets, sets having the property of Baire are sigma ideals.
I am trying to find more interesting sigma ideals in \Bbb R
*Correction: $D(A)$ doesn't contain any interval.

3:41 AM
What are the various ways/strategies to prove that a set G (G\subset R) is dense in R ?

choose a point in R and make an $\epsilon$ neighborhood, and find at least one element of $G$ in that neighborhood

Now, I know the elementary defn that a set G is dense in R, iff for every x,y in real set, there exists a g in G, s.t x<g<y
@shintuku is that a definition ?

it's equivalent @ThomasFinley

improved

3:43 AM
@ThomasFinley cl(G) =R
Any non empty open set intersect G

64

I am having trouble proving the statement: Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that for every $\epsilon > 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty.

Say for eg in this post
The user suggests a point in OP to prove the set is dense
It says for every epsilon>0, (0,epsilon)\cap S is non-empty
Is that a criteria for being dense in R ?
I mean an equiv defn or something?

Side note: Any additive subgroup of $\Bbb R$ is either discrete or dense.

I am not quite sure about it?

G\subset R is dense iff every basic open set ( i.e open interval) intersect G

@SouravGhosh yes, but is :"for every epsilon>0, (0,epsilon)\cap S is non-empty" a debn for being dense in R ?

3:47 AM
No. One way is true.

Because this criteria/defn takes only sets (0,\epsilon) into consideration
@SouravGhosh So what about the link of the post I posted here ?

If $S$ is dense then $S\cap (0, \epsilon)$ non empty for all \epsilon >0

@SouravGhosh Oh, but then in the post, the user is trying to prove, S is dense by the other way

For an example take [0, \infty)

That's erroneous, isn't it?
Anyways, I am trying to prove, that the set $\{m+n\sqrt 2:m,n \in\Bbb Z\}$ is dense in R. I dont know where to start ?

3:51 AM
@ThomasFinley The additive structure will be crucial here.

My knowledge about dense sets is that: a set G is dense in R, iff for every x,y in real set, there exists a g in G, s.t x<g<y or equivalently, a set G is dense in R, iff, every point of R, is a limit point of G.
@SouravGhosh You mean that the set is a group under addition?

$S\cap (0, \epsilon)$ non empty implies $S\cap (-\epsilon, 0)$ also non empty for every $\epsilon>0$

@SouravGhosh but why will it be useful here ? Has this something to do with a set being dense in R ?

4:19 AM
I just got the ability to chat here :)

@SouravGhosh this is indeed helpful but here in this case, S is an additive subgroup, so what if it's discrete ?

4:40 AM
@ThomasFinley Forget m for a moment and consider only the fractional part of $n\sqrt 2$.
then show that it is dense in [0,1).
@ThomasFinley think of discrete as without any limit points.
if you know groups then you can use the fact that additive subgps of R are either dense in R or cyclic.
but that's not necessary really. You can prove this using pigeon hole principle (which is roughly - if you have n>m objects, m placeholders for those objects, then atleast one of placeholders will have more than 1 object in them).
closure of totally bounded set is totally bounded?
cl A= \cup G, where G is open in cl A. Take any e>0. Take a p in cl A.
I want to show that B(p, e) sits inside some G.
B(p,e)= {a\in cl A: d(a,p)<e}
if p is in A \subset cl A, then it does sit inside some G by total boundedness of A so let's suppose that p is in cl A\A.
A\subset \cup G. Given any e>0, for every a\in A, there exists a G such that B(a, 2e)\subset G. closed ball B(a,e)\subset B(a,2e)\subset G for some G.
So B(a,e) (this ball is in cl A) should also be in G.
(first attempt, it should probably work.)

It looks fine to me.

thanks.
2

4:58 AM
$\forall \epsilon >0$ there exists $a_1, a_2, \cdots, a_n\in A$ such that $A\subset \cup_{k=1}^{n} B(a_k, \frac{\epsilon}{2})$
Then $cl A \subset \cup_{k=1}^{n} B[a_k, \frac{\epsilon}{2}] \subset \cup_{k=1}^{n} B(a_k, \epsilon)$

I think this is what I should have done.
my answer is not correct. I confused between Lebesgue covering and total boundedness somehow.
I started with cl A = union of some open family
and then tried to find e>0 such for every x in cl A, B(x,e) is in some member of the open family.
this was not required at all.

@Koro can you please elaborate the meaning of discrete, i.e some definition? I only know that intutively?

@SouravGhosh indeed, this is how one would do it. thanks.
@ThomasFinley $A\subset R$ is said to be discrete if $A$ has no limit points in R. e.g. the set of all integers Z

@ThomasFinley Derived set of A i.e $A'=\emptyset$

5:17 AM
@Koro I tried to prove that $n\sqrt 2$ is dense in $[0,1)$, using the fact that irrationals are dense in $\Bbb R$ and so in [0,1) hence, for any x,y in [0,1) we have, an irrational i s.t x<i<y ...but then I am stuck?

An interesting fact: $Z, β2Z$ both are discrete but their algebraic sum/Minkowski sum Z[β2] is dense.
@ThomasFinley $\{ β2 n : n\in\Bbb Z\}=β2Z$ . Isn't β2Z discrete? Does it have any limit point?

5:52 AM
@ThomasFinley I meant fractional part of $n\sqrt 2$ not $\sqrt 2 n$. Anyways, here is one simpler way (it doesn't use showing fp of $\sqrt 2 n$ dense in [0,1) first): step 1: show that $0$ is a limit point of {m+ n\sqrt 2: m,n integers}=T. step 2: take any interval (a,b), a<b and show that this interval has an element of T. This uses step 1 and also completeness of R.

$A \subseteq B \subseteq C$ with $A,C$ local rings and $B$ a ring does not imply $B$ local
this is a sad moment for $K[x,y]_{(x,y)}[1/y]$

@ThomasFinley Denoting $\sqrt 2$ by $b$, note that for any integers $n\ne m$, $\{nb \}\ne \{mb\}$ (here {,} denotes fractional part) because if $\{nb \}= \{mb\}$, then $b=\frac{[nb]-[mb]}{n-m}$, which is rational but b is irrational so this is not possible.
Step 1 from the last message: Take any $n\in N$. Divide [0,1) into $n$ parts: [0,1/n), [1/n, 2/n),... Consider n+1 numbers {b},{2b},...,{(n+1)b}. The subintervals are only n but the numbers are n+1 (distinct by the remarks in the first part of this message) so some subinterval must contain at least two of the numbers-say {i b} and {jb}.
now, you try step 2.

@Thomas Dirichlet devised the pigeon-hole principle for precisely this stuff. See math.stackexchange.com/a/56054/207316
It's a beautiful result. And it's probably a good idea to play with a few simple examples until it "clicks".
On a (somewhat) related note, take a look at Beatty sequences. en.wikipedia.org/wiki/Beatty_sequence

6:39 AM
@TedShifrin: good evening.

Hi, Mr. sheriff!

it was a beautiful day, here. Mid 70s, sunny. There is a chance of rain on Sunday.

6:53 AM
@robjohn Well, maybe here, too.

We have a bunny living in our backyard. We just hope it's successful at avoiding the coyotes that come through the neighborhood.

7:09 AM
Consider the first Reidemeister move, but take an axis and spin it to get a "Reidemeister move" for 2-knots in R^4 (in terms of knot diagrams, immersed surfaces in R^3).
Iterate to get it for $n$-knots in $\Bbb R^{n+2}$. What should you call this thing? "Torus symmetric Reidemeister move"?
There's different ways to generalize the first Reidemeister moves, hence the question.
Ignore the fact that there are no knotted $S^n$'s in $\Bbb R^{n+2}$ for $n \geq 3$. I am working with knots with more structure, so this is no longer true in my setup.
Torus-symmetric... is a lot of words to say. Toric would be better, if not for that word already being taken. Toral sounds pretentious.
Toroidal? lol

7:49 AM
is the strike till going on

8:11 AM
@HopefulWhitepiller Yes. Over 1100 people have signed the letter, and the strike announcement post has over 74k views & +813/-10 votes.

8:24 AM
@ε₯ηHades Do you remember Hashirama Senju once said-always conflicts, no matter what the era?

8:40 AM
Has there been a Fields Medal for numerical analysis?

9:12 AM
@PM2Ring I wonder who those 10 downvoters are... (spoiler: probably the staff)

@PlaceReporter99 Probably not. There are some huge fans of ChatGPT who are happy to vote against anything that seems anti-ChatGPT to them.

@PM2Ring like the staff...

:) No. The staff / company actually don't want the sites flooded with ChatGPT stuff.

@PM2Ring they should step down then and let the community run themselves.

But they are freaking out about the drastic reduction in traffic on the network, especially SO, since the release of ChatGPT.

9:19 AM
@PM2Ring It's because most people are asking ChatGPT now.

That's certainly an effect.
Numbers have been dropping for years, for a variety of reasons, ChatGPT has just accelerated that.
One of those reasons is that SE Inc seems to be persuing quantity at the expense of quality. That has frustrated and discouraged expert answerers, disappointed in the increasing percentage of low quality questions & answers that have been accumulating, and has contributed to an atmosphere where new members (especially question askers) feel hostility.
So for several years, we've had inflated numbers of people making low-quality contributions. And a lot of those peopke who were asking low quality questions on SE can now ask ChatGPT their crummy questions. ;)

Canβt we just go against the command of SE staff?

@PlaceReporter99 Not exactly, but there are ways... See how Physics.SE are dealing with this issue. The Physics mods don't care about ChatGPT, they just use their usual policies on people who persist in posting low quality content.
144

We, the moderators of physics.SE, are deeply concerned about the recently announced "network policy" regarding computer-generated content on the SE network, which effectively attempts to forbid moderators from issuing suspensions or deleting posts for the undisclosed usage of computer-generated c...

That works on Physics.SE, because all the mods are very knowledgeable, so it's not too hard for them to recognise when a post is plausible-looking rubbish. It doesn't work so well on SO, with its much larger volume of traffic covering a broad spectrum of technologies that no single person can have expertise in.

9:57 AM
logged into an image ai website and asked it to remove some high tension wires from the foreground of a random image
after many failed attempts, it added more wires

Current AI can do some impressive stuff, if you steer it correctly, and are aware of its limitations. But it can easily go off the rails. It's scary that so many people think it's good.
ChatGPT is pretty good with language structures. OTOH, it can't even balance parentheses. See writings.stephenwolfram.com/2023/02/…

do i really want to read an article on ChatGPT by the author of A New Kind of Science

10:19 AM
@BalarkaSen It's actually a pretty good article. And he (mostly) restrains himself from reminding you how awesome Stephen Wolfram is. ;)
He wants to train a LLM on his writings. He's got decades of material in his archives...

lol
apt
from what i can see the greatest contribution of Stephen Wolfram on this planet is helping a couple thousands of precalculus students cheat on their homeworks
which is a feat
and now chat gpt is taking his place, so a collab will be one for the decades

From Scott Aaronson's blog, scottaaronson.blog/?p=7188#comment-1948579 > If youβd asked me 20 years ago, βhow will Wolfram react to one of the great AI developments of the 21st century?β β I wouldβve said, βpresumably, by figuring out some way to make it about him.β He doesnβt disappoint! π
They developed a plugin to let GPT-4 interact with Wolfram alpha. It still requires manual supervision to correct hallucinations, but it is pretty impressive. writings.stephenwolfram.com/2023/01/…

11:25 AM
Why is real part of the function $f(z) = z^{1/2}$ positive?

CheatGPT

11:48 AM
@BalarkaSen he's a great expositor and has a knack for pointing out beautiful things of a certain kind

I have no idea why people are so obsessed with ChatGPT as if it is the solution to every problem for humanity.
@Koro I guess you are on your own in your math journey. Good luck o7

@PrithuBiswas chapt gpt is something special from a technology / society perspective. This is a new technology that basically overnight became accessible to everybody. And its really impressive. Its natural to think "if this is just the beginning, what will its successors be able to do?". Remember how chess players thought that computer programs will not be able to beat humans in their lifetimes?

12:07 PM
@s.harp maybe. i personally would much rather watch a good kung fu movie than read stephen wolfram
im bored with science and tech expositors. they should all jump the AI blackhole and make AIs ghostwrite their books
especially when the expositors are CEOs of multi-million dollar business sharks

@BalarkaSen I would much rather watch a bad movie than read wolfram

i think when you compare any movie with wolframs writings, it automatically gets the adjective "good" added in front of it

I think that these characterisations are coming from a dishonest place. Sure Wolfram is a jerk, a crank, and completely full of himself. So you don't want to give him any credit. But I don't think you can honestly deny that he can write compelling stuff

yeah i can
i just did

Its the "honestly" that I'm contesting

12:14 PM
im being absolutely honest. in fact im so honest that i could never get through 5 pages of anything wolfram has ever written
compelling is in the eye of the beholder, which is fine, really
i think hes boring as shit lol

Maybe its my physics background that influences my idea of beautiy then :-)

id probably hazard a guess that its just you and not physics
but sure

Do you think feed-forward networks are boring?

i know epsilon about neural networks
never felt the need to know more than "its just gradient descent on steroids"

Yeah, for gradient descent you need to have a candidate for the gradient. With a feed-forward network you get this via is a rule called "backpropagation" which is literally just the chain rule (you have a bunch of layers and the signal goes from one layer to the next, the transform depends on some parameters, if you want to find the gradient you use the chain rule)
from the perspective of mathemnatics everything is barely more htan linear algebra, just a tiny bit of basic analysis
theres a paper from the 80's that is now cited like 20.000 times (number made up) by Hinton where he literally just explains what the chain rule is
still, I think its beautiful, and its not just me. These are the kind of things Wolfram likes as well, if you dont care for them I can understand why you dont enjoy his writings

12:23 PM
oh im sure its interesting to many people, i was just suggesting your interests may not be completely decomposed into your physics background
i like physics, a chunk of the math i think about has physical motivations behind it. i admit i dont find neural networks at all interesting

You are probably right, this is more computer science aesthetics. But I dont have a computer science background so I cant make a one-liner about it

i think interests are complicated emergent social phenomena and thats cool
as long as stephen wolfram doesnt write his complexity theory explanation of interests

By the way, does anybody know about this thing mathstodon.xyz ? I stumbled upon it today and have absolutely no idea how to interact with it. Where are the threads / chats?

1 hour later…
1:32 PM
The proof goes on to show that $I(y)=H(y)+C$ by switching the order of integration of $(2)$. Anyway, I am confused about the notation in $(2)$. The partial derivative is taken with respect to $y$, but clearly $f$ depends on $x$ and $t$. On second thought, however, I don't know how I'd write it differently, but it is confusing.

correct proof is using DCT

Ok.
It's like writing $\int_a^x f(x) dx$...
The author tries to avoid it, it seems.

1:56 PM
If you are referencing papers in the middle of a beamer slide show what information do you include? (first author) surname et al, publication year ? Of course full references will be given at the end of the presentation

1 hour later…
3:01 PM
Depends on the topic, usually I put [AuthorLastNames, year]
If there are too many authors, et al. is acceptable

2 hours later…
5:02 PM
@TedShifrin In the end, it was good. I made mango salsa (tomatoes, mango, a couple of habaneros, cilantro, lime, salt, red onion), fried up some cod, and made some corn tortillas (I need to get more masa next time I go to the store).

5:17 PM
Sounds yummy, Xander.

It turned out well. Though I am not sure if my brother liked it. I wish that he were more communicative. :/

I thought lawyers were plenty communicative when they wanted to be.

Though, to be fair, I think that he would eat nothing but roast beef and spaghetti carbonara all day, every day if he had the choice.
@TedShifrin Other brother.

Oh, other brother. Sorry.

He has Down Syndrome. I am one of his legal guardians, and he lives with me.

5:18 PM
Right, I only remembered after you corrected me.

So I have to find food for him to eat. And he doesn't really like anything green.

At least he doesn't ask for spaghetti-os every day.

@TedShifrin I mean, I think if he really had his druthers, he would eat nothing but hot pockets (since, at least, he knows how to use the microwave). But those things are both (a) expensive and (b) so so so salty.

Yes, terribly unhealthy.
I was shocked a few years ago to discover that a can low sodium Progresso soup, which has almost no taste, has something like 1/3 the dayβs sodium allotment in one serving.

@TedShifrin Yeah. It is frightening.

5:41 PM
Let $f:E\to\mathbb{R}$. I know that $f^{-1}(\mathbb{R})=\{x \in E \ | \ f(x) \in \mathbb{R}\}$, is it the same as $f^{-1}(\mathbb{R})=\{x \in E \ | \ \exists a,b \in \mathbb{R} \ \text{s.t.} \ a \le f(x) \le b\}$?

Yes
both are just the set $E$

I donβt see the point of it.
What are you trying to do?

Thanks. I want to prove that for $f:E\to\mathbb{R} \cup \{+\infty\} \cup \{-\infty\}$ the set $f^{-1}(\mathbb{R})$ is a measurable set given that the sets $E$, $\{x \in E \ | \ f(x)=+\infty\}$ and $\{x \in E \ | \ f(x)=-\infty\}$ are measurable.
We defined $f(x)=+\infty$ if for each $t\in\mathbb{R},f(x)>t$ and $f(x)=-\infty$ if for each $t\in\mathbb{R},f(x)<t$. So I wanted to say that $f^{-1}(\mathbb{R})=(\{x \in E \ | \ f(x)=+\infty\} \cup \{x \in E \ | \ f(x)=-\infty\})^c$.
But I was not sure on how to write $f^{-1}(\mathbb{R})$ with logical symbols, since I never written it explicitly before.
And so I concluded that it should be measurable, because complement of the union of two measurable sets.

5:58 PM
Why not write it as the complement of $\{\pm\infty\}$?
Oh, you said that.
No different from taking the complement of two finite values?

No different indeed, sometimes I get scared by new stuff :D

Yeah, weβre not used to functions having actual value $\infty$.

6:27 PM
Let $I_K:L^2\to L^2$ and $I_Kf(x)=\int_X K(x,y)f(y)\,dy$ and assume all the required conditions on $X,K$. I want to show that $I_K$ is compact. I know the proof given in Axler's book but I am trying to do in a different manner. Let $(f_n)$ be bounded and $L^2$ is reflexive banach space so it has a weakly convergent subsequence $(f_{n_k})$. Now, $I_K$ can be shown to be continuous so it is weakly continuous. So, $I_K(f_{n_k})$ is a weakly convergent subsequence
Any idea? Is it possible to proceed from here?

7:18 PM
ooohhh, i got a +4 rep bump because a user was removed...

Imagine if they removed me!

Last month I got -28 because a question that I answered got removed.
What do they mean by "removed"? I don't understand.

I guess your answer had not been accepted? I think once there's an accepted answer, a question cannot be deleted. Except by mods.

7:40 PM
Sanity-check: Say $X, Y \subset Z$ are submanifolds. Saying that $N^*_Z (X \cap Y) \cap T^*Y = N^*_Y (X \cap Y)$ is exactly saying that $X$ and $Y$ are transverse.
I think what I wrote doesn't make sense, but something related does.
I hate thinking dually

Transversality says the conormal bundle of the intersection is the sum of the conormal bundles.

$N^*_Z(X \cap Y) = N^*_X(X \cap Y) + N^*_Y(X \cap Y)$?
So just like tangent bundles

No, I meant conormal of $X$ in $Z$ plus β¦.

but I think what you have is isomorphic ,
Think concretely in terms of $df$ and $dg$.

7:55 PM
Shouldn't take long to figure out, but my brain is protesting so I will try to find a different way to say what I want to say.
Yeah, exactly, thinking in terms of functions now
I have an embedding $f : P \to M$, which is transverse to a submanifold $S \subset M$. If $d\phi \in T^*M$ is conormal to $S$ (this just means it vanishes on the tangent spaces to $S$), then $f^* d\phi \in T^*P$ is conormal to $f^{-1}(S)$.

Yes.

8:54 PM
Hi, I apologize if my question is too vague or not well formed. I'm trying to figure out what subject to look at more specifically, or ideally some resources.

I'm interested in the case where I have a differentiable, real valued function f, parametrized by a large set of weights (a good example could be a standard neural network).

I want to consider operators T that act on the function to produce another function of the same type: f' = T(f). I want to use the fixed point theorem to show that some operator is a contraction and therefore applying it many times must lead to a fixed point.

9:52 PM
@PNDas The question was unacceptable because it was a PSQ. It was closed and since it was not improved, it got deleted.

10:30 PM
skymonkey: at that level of generality the goal makes sense. but also at that level of generality, it may be no easier to identify which changes in weights produce a contraction than it is to prove that a fixed point exists via some other route.
e.g. if the expectation is that the contraction mapping theorem will make it 'easier' to find/guarantee fixed points, at least in some circumstances, the reality might be, it isn't any easier. it's just translating a hard problem about fixed points to an equally hard problem about what parts of a parameter space yield a contraction.
maybe this would change if you have a lot more structure/hypotheses on the operators involved and the weights.

10:56 PM
π­π­π­π­π­π­π­π­..........I wanted to graduate next year, but being realistic it isn't feasible.......life is so heartbreaking sometimes........π­π­π­π­

matter of perspective

Very true and I have been in that frame of mind all day adjusting the perspective
it is coming to terms with it and reconciling it which has been frustrating. Part of the process of things though...
and most importantly being at peace with it

@leslietownes if $U\subseteq X$ is a proper non-empty path-connected open subspace of a path-connected metric space $X$, does there always exist $x\in \partial U$ such that $U\cup \{x\}$ is path-connected?
i.e. I want to find some path contained in $U$ which would connected to the boundary
for $X = \mathbb{R}^n$ for example, the answer is yes
and in general, can have $x\in \partial U$ with $U\cup \{x\}$ not path-connected

@D.C.theIII I often ran into math majors who were taking loads impossible for them to be successful with (like 4 upper-level courses a semester). They insisted they had to do it in order to graduate. I tried to suggest that taking a lighter load and succeeding would be better than failing courses their way.
4

@TedShifrin That's exactly the reconciling I am coming to grips with........sure I "could" finish and "get the piece of paper", but at the expense of burning out and not learning anything and scraping by with low marks?.............That is exactly how I ended up in the predicament I'm in now.

11:06 PM
Well, pragmatism has different aspects. :)

In truth the only real "downside" is I will be finishing a year later and I'll be a year older......even if I'm not a complete spring chicken anymore nothing else truly changes. I could still do everything I would like to do....

Sometimes one needs to modify goals β¦

it's true.
I'll get through my bout of existential crisis. Just needed to do some digital venting

@leslietownes thank you, I'll think about it

-1

Let $0<n$ be a positive integer. Let $p_n$ be the $n$ th odd prime. $$p_1 = 3,p_2=5,p_3=7,p_4=11,...$$ Let $$f(n) = \exp(1 + \sum_{i=1}^{n}\frac{1}{p_i-1})$$ Let $$g(n) = \prod_{i=1}^{n}(1+\frac{1}{p_i-1})$$ Let $$t(n) = \frac{f(n)}{g(n)}$$ Then this limit exists  \lim_{n \to +\infty} t(n) = \f...

strange
i guess the downvoter did not believe it ?
OR thinks it is one of mertens theorems...

11:22 PM
Maybe because it looks like "no context, homework question".

yeah but this definitely NOT homework
i like privacy but i do not like anonymous :p