00:00 - 23:0023:00 - 00:00

12:04 AM
@Jakobian Thought: For a path $f : [0, 1] \to [0, 1]^2$, let $V(f)$ be the set whose elements are a collection of sub-intervals $[a_i, b_i] \subset [0, 1]$ whose interiors don't intersect, such that $f(a_i) = f(b_i)$. Then $V(f)$ is partially ordered: just say a collection is less than the other if each interval in the former is contained in some interval of the latter.
Apply Zorn's lemma on $V(f)$ to extract a maximal such collection. Then crush these.

well, Zorn's lemma might not be applicable as constant function shows
If your family is $[0, 1/3]$ and $[1/3, 1]$ then maximal such set containing it doesn't exist

ah...

I think of browsing those

@Jakobian Wait, $\{[0, 1/3], [1/3, 1]\} \leq \{[0, 1]\}$
So it's fine, yes?

Oh I didn't realize you were using a different ordering than subsets

12:14 AM
Subset is bad, you want to be able to forget nesting

What about a function like $f(x) = x\sin(1/x)$ then, and intervals $[z_n, z_1]$ where $z_1, ...$ are zeros ordered from the right?
well that doesn't work but maybe something similar?
no it won't work we will converge always

I think this works, because to verify increasing chains have an upper bound you're basically going to end up using that if $[a_i, b_i] \uparrow [a, b]$ and $f(a_i) = f(b_i)$ then $f(a) = f(b)$
By Hausdorffness???
Lol

I suppose this is where we use it

I don't see an immediate problem, but it's almost morning here

There is probably a subtelty in the way we crush it

12:23 AM
It's a collection of closed intervals in $[0, 1]$ that we crush, so that always gives $[0, 1]$
By Bing shrinking if you like
It's a cellular set

My ice cream scoop has a hole in it. :(

someone ate the cherry

No cherries. Just a hole in the scoop.
Though I guess that, after 25 years, I can get a new one. :/

Did the mice gnaw on it?

Possibly. Though I think that it is more a problem of corrosion.

12:36 AM
consider
getting a new one
and it's on me

@BalarkaSen but if $A_\alpha$ is a chain, and $t\in\bigcup\bigcup_\alpha A_\alpha$ then picking an interval $t\in [x_t, y_t]$ with $f(x_t) = f(y_t)$ won't you have to consider the problem of intervals $[x_t, y_t]$ and $[x_s, y_s]$ overlapping?

None of the intervals in $A_\alpha$ overlap, and something in $A_\alpha$ overlaps with something in $A_\beta$ only if one is contained in the other. So how can overlaps form in the "limit"?
The worst overlap is sharing of an endpoint, which I allow.

yeah. I was too focused on formalism and didn't see this obvious point

A Balarka, have you broken your sleep schedule again?

Given this works, I still feel Zorn's lemma is a kind of cop-out here and there is a more systematic way of doing this by tracing the curve and restarting whenever a loop is encountered.
Which is what true loop erasure is
@TedShifrin You mean I fixed it! :)
Yeah I have to go to sleep at some point soon... I might miss the student seminar today...

12:52 AM
I assumed that it was fixed during your absence 🤷‍♂️🤷‍♂️

I fixed it for a brief week during the annual department colloquium, because it happened early in the morning (and by that I mean like 10 AM...)

1:07 AM
Okay, I believe this will induce a function $\tilde f:[0, 1]\to X$ (most of the time)
but I'm still not certain about injectivity
What if $\tilde f(x) = \tilde f(y)$ but there is $x < t < y$ that we identified to a point?
Won't we need to iterate this process transfinitely?

That can't occur, draw a picture
Remember $[0, 1]$ is our new domain after collapsing stuff. If $\widetilde{f}(x) = \widetilde{f}(y)$ in this new domain, then $[x, y]$ must necessarily contain a bunch of collapsed intervals in the old domain
Pictorially your situation is: you have a loop and the curve, after exiting the loop, comes back and hits the loop... but you had already collapsed the loop!
So the curve doesn't hit itself after all

@BalarkaSen thank you for your answer. Actually this is part of a paper on Teichmuller polynomial (by McMullen). Here, $\lambda$ is an $n$-dimensional lamination. I'm not sure that universal covering argument is what he meant.

Is $\lambda$ the leaf-space?

Ah OK it is literally the union of leaves
Why not, then? $\lambda$ is a perfectly fine space, you can take its universal cover and the lamination lifts upstairs also

1:18 AM
well $\lambda$ can be disconnected

You can take the universal cover of all the components in question
Universal cover should "open up all the loops"

so it's a union of the universal cover of each component?

Ought to be
Q: Why do people care about the non-fibered faces of Thurston norm?
What does it tell me about the manifold

Well just a "natural question"?

I care about the Thurston norm because of Gabai's theorem; norm-minimizing surfaces give rise to taut foliations. Is there no natural reason to care for the non-fibered faces other than curiosity?
At the end of the day the goal of a topologist is to understand manifolds, yes?

1:30 AM
of course. But I think people try to understand stretch factor of pA map using non-fibered faces (purely my guess I haven't read anything related to non-fibered faces)

Do you know what the stretch factor tells about the topology/geometry of the pA's mapping torus?
Does it have to do with the size of the systole, or volume, or something?

as far as I know volume is related

The only trivial observation I can make is that the stretch factor of $\phi^n$ is $n$ times that of $\phi$, and the same is true of the volume of $M(\phi^n)$ (it's a cyclic $n$-sheeted cover of $M(\phi)$)

there's a big conjecture on the stretch factor of pA, I can't remember explicitly but it's about realizing problem of pA map of the given (possible) number. But if your motivation is based on manifolds, then don't care!
arxiv.org/abs/2210.13418 some questions related to stretch factors are written here btw

1:44 AM
Thanks, @onepotatotwopotato. I'll take a look

someday I will use the phrase calibrated associative submanifold in a way I understand
10 year goal

The main purpose of Teichmuller polynomial is for all possible stretch factors of pA monodromy of given fibered 3-manifold. It's defined on fibered faces. Veering/taut polynomial is an analog but is also defined on non-fibered faces.
I think their primary goal is not about geometry or manifolds. So... maybe not appealing to you!

@onepotatotwopotato Or maybe we don't understand it simply enough to see what it's saying about the manifold

that's the hope

1:53 AM
How do you compute the stretch factor? Suppose I write down a pseudo-Anosov braid for you. Do you compute how the weights in the invariant train track scales?

Use the following idea to prove The \textbf{Fundamental Theorem of Algebra}:

Let $p(z) = z^d + a_{d-1}z^{d-1} + \ldots + a_0$ be a complex polynomial of degree $d \geq 1$.
For $0 \leq r$ let $\gamma_r$ be a closed curve given by $\gamma_r(t) = p(re^{it})$, $t \in [0, 2\pi]$. For $R$ large enough, prove that the closed curve $\gamma_R(t) = p(Re^{it})$, $t \in [0, 2\pi]$ is homotopic in $\mathbb{C} \setminus \{0\}$ to the curve $C_{R, d}(t) = Re^{idt}$, $t \in [0, 2\pi]$. Conclude
$$\text{Ind}_{\gamma_R}(0) = \text{Ind}_{C_{R, d}}(0) = d\, .$$ Assuming, towards contradiction, that $p(z)$ has

@BalarkaSen yeah. I don't know, I guess this works. I'm a bit upset that it does

Well, its the most naive thing one could think of

@userunkown You have to start by specifying how large $R$ has to be.

@TedShifrin I have no idea. I am taking complex analysis, but have never studied topology before.

2:16 AM
@Ted: good evening. How are you doing?

This has nothing to do with topology other than what you’ve studied. Think about the homotopy $(1-t)f+tg$ between $f$ and $g$. What do you need yo be sure of?
Hi @robjohn. I can see great to drive at night again :)

Wonderful! I know someone else who's had similar results as well

Getting old and bionic isn’t all bad!

2:40 AM
is it possible to integrate $\int_0^L\frac{kx dx}{4\pi\epsilon_0 \sqrt{(X-x)^2+Y^2}}$ for some coordinate $(X,Y)$
$k,\epsilon_0$ some constants

2:54 AM
Yes. Work on it.

do I do a sub like $u = X-x$ so $du = -dx$ and then do polar coordinates?

@TedShifrin that's the hope by year 5 of my 10 year goal ;)
I briefly watched a talk (2023 symposium) by Sebastian Goette about calibrations just to be fascinated by it and not understand much
many many arrows in those diagrams lol
I watched that because he commented on my post yesterday

3:40 AM
You need to be super-solid on differential forms. Then look at the original paper.
@Obliv This is a single integral, not a double integral. So what do you mean by polar coordinates? But a trig substitution should work.

3 hours later…
6:36 AM
I think these two papers by Brock shows some connection between volume of pA mapping tori and Weil Petersson translation length. The first one mentions a reference to a connetion between Teichmuller translation length and volume. @BalarkaSen
Brock wrote many many interesting papers btw

7:23 AM
My prof has completed all but law of large numbers and central limit theorem
But it's barely been half a semester
Is this acceptable behaviour

7:40 AM
it is impossible to assess reasonableness from a syllabus or list of topics alone, one might also want at least some idea of what (if anything) the prof expected to assess, and how, and what background/prerequisites were assumed
having said that, that does not seem like an unreasonable pace for a university level class devoted to those subjects
i'm also glad that i'm not in that class

@leslietownes analysis on R was assumed, and multivariable calculus (apostol vol-2) level was assumed. My problem is that, the prof has an option to, you know, use the entire 40 hours. She covered all this is slightly less than 20 hours.
I don't know what's to be done in the rest of the semester

what do they plan on doing with the remaining time?

I have no idea

well, it's definitely bad that they haven't told you this. is the instructor new to teaching? or new to teaching this class?
this is kind of a different issue, haha. i still think it could be a reasonable pace in the abstract, but the facts on the ground seem to be an instructor way out of sync with their own announced plan for the course (i just now noticed the third column of the above image)
which is another kind of bad
i assume the above was the disclosed timeline for the entire class, i.e. there aren't rows of that chart other than the ones in the image

@leslietownes I presume she's new, she's only 32, and is an assistant prof.
@leslietownes yes, this is the whole semester, 40 hours worth

7:54 AM
going "too fast" is a really common problem for new instructors, although, this seems to be an extreme case of absolutely speedrunning a syllabus. i'd guess that at a minimum they haven't taught this class before, and maybe also that they haven't taught a lot, period

I don't really have a problem with fast paced classes, but this is a bit too fast for me. I need stuff to sink in, and I need time to work out through the book. My ideal method is to work through the book primarily, and try to keep up or stay ahead of the class. This has been next to impossible with this class

which isn't to say that speed problems don't always develop, but if you have more experience teaching you notice when it's happening and don't let things get this far out of sync
the most common variant of 'teaches too fast' is more localized, i.e., 'prepared what they thought would last for one lecture and used only a third of the time to get through it'
maybe there's some master plan to revisit the subjects again in sequence on a higher level, and maybe even to repeat that cycle, i somehow doubt it, but it would be one way to use the time
other than just "i guess it's just study hall for the rest of the semester"
it would be unconventional to teach an intro stat class this way but i've definitely seen "cover the whole class in a day, then again in a week, then again at normal speed" work
its actually not a bad way of structuring a closer to research level class
like a topics course on something

Our Galois theory professor covered the whole semester in day 1 as an overview, then continued in normal pace. That gave us extra motivation.

1 hour later…
9:33 AM
Can we say that the above graph is a tree with a root 1? I mean if the root is 1, is it a rooted tree?

the answer (as well as the reason for why the answer is what it is) depends on your definitions, of course, but i would imagine so

Thanks a lot @leslietownes The definition for a tree?

maybe it's drawn in kind of a funny way for purposes of visualizing it, but if you shifted 11 down so it's roughly on the same horizontal line as 17 and 18, and moved 18 and 6 below that, and 2 and 7 further down
yes, your definition of "tree" and "rooted tree"

Hmm okay

sorry, 16 and 6 below that

9:39 AM
A tree is a connected acyclic undirected graph

if you made those modifications and then flipped it vertically it would even look kind of like a real life tree :)

Yes. :)
There, so we anyway have to look at in the modified way right?
Else, it doesn't have any meaning in the visualized levels it shows? Like 14 and 16 are in the same level and 11 above it doesn't mean that 11 is a parent right?

"have to" for what purpose? i think it looks better that way, but under most definitions, how you depict the thing wouldn't affect whether or not it is a tree (which would be determined only by the vertex set and edge structure)

Have to means, like to see it more clearly that it indeed is a tree
@leslietownes Hmm okay :)

well if you gave me an enormous mess of points and edge data, depending on how you gave it to me, i guess i'm not sure if i could tell at a glance whether it was connected or not, or acyclic or not
and maybe depicting it graphically more like a real life tree helps you 'see' that

9:43 AM
Okay :)
Then the above one can be called a tree right?

yes, "can be called" could even be "is," under the definition you gave above

Can also say this tree has a sub tree with root 2, if I name 2 as a root for a subtree?
@leslietownes Okay :)

your new question now puts a focus on what "subtree" and "root" are defined to be

Hmm :)
A rooted tree is a tree where one of the nodes is designated as the root node.

offhand i don't see how singling out any particular vertex would automatically single out some subset of the graph as "the subtree" associated with that vertex
you could think about rewriting this picture as to center more or less any vertex at the top or bottom of a redrawn picture of a real life tree

9:48 AM
A subgraph H of a graph G is another graph formed from a subset of the vertices and edges of G if all endpoints of the edges of H are in the vertex set of H.
Then subtree is a subgraph of a tree.
@leslietownes Okay

well if you wanted say to talk about some subtree like "the points connected to 2 via a path in the above picture that only goes upward, and the edges that make those paths," this singles out a proper subset of vertices and edges up above (that would include the vertex 17 for example but not the vertex 10)
but that's something i'm doing with reference to the entirety of the above picture
that isn't coming just out of the edges and the adjacency data
so i wouldn't think of that particular subtree (or any other one) as "belonging" just to my choice of 2 as a "root"
it comes from me using the above diagram to make a verbally complicated reference to a set of edges and vertices that i could also just specify by listing the vertices and edges i want to include
which is generally what you'd need to do to specify a subgraph or subtree
the tree structure alone wouldn't just tell you "oh, that's the subset that goes with 2"
@Hasini as an example, pick vertex 11. draw it on a new piece of paper. on a line above it, draw vertices new 14, new 16, and new 6, connected to your new 11 by edges. on a line above those, draw vertices new 2, new 10, new 17, and new 18, connected to the ones on the line below it as to reflect the adjacnecy relations in the above picture. continue
you could do something similar with any other vertex instead of vertex 11
and when you do this you are drawing, not different trees, but the same tree
same vertices, same adjacency relations, just drawn differently with "11" at the bottom instead of somewhere else
a sufficiently abstract textbook might discuss this somewhere in connection with the notion of graph isomorphism

10:04 AM
Okayy Thank you very very much @leslietownes :) :)
I get a clear idea now :)
This was present in a question which says as follows:
(May be this questions is not very nicely formulated)
For the answer someone has written as "This is not a tree as the node 14 can not have two parents"
But he must be mistaken I think acording to what you explained earlier :)

okay, someone is using a different definition of graph than what you mentioned above
it is really common for people to use different definitions
like, unbelievably common, for something that goes as underappreciated as it does

Nope the definitions were the ones I gave
From his reference

well, isn't very enlightening to be in partial dialogue with something written by someone who isn't in the chat right now, and who apparently isn't using the definitions above

What I've shared as defintions are the exact quotes from the definitions given
But this is a great help for me if I can get help to understand this
Hmm okay

you haven't defined "parents" yet
if you use the definitions of en.wikipedia.org/wiki/Tree_(graph_theory) which more or less coincide with what you've said above about "tree" and "root", and which also defines "parent" for vertices in a rooted tree, the picture above does specify a tree in which it is possible to label a root node 1, and in this rooted tree, 14 does not have two parents under that definition
it has one parent, 10
node 11 (for example) is not a parent of node 14 under these definitions (in fact, node 14 is the parent of node 11)

10:18 AM
@leslietownes Okayy @leslietownes
That is a great help!
Many many thanks!!!! :)

the problem here is that "below" in that image is being made with reference to a particular drawing of a tree
in which "is below" is exactly the same concept as "is further away from the root, via a path in the tree"

Hmm, then what is the way to think about it?
@leslietownes Okay

in the picture you've drawn previously above, "visually below" no longer has this meaning and so does not let you "see" the parent/child relation as easily

I'm not exactly clear

there's a fundamental confusion here, about properties of trees, vs. properties of drawings of trees
if you "re-drew" your first picture to look more like an "actual real life tree" with 1 at the root, then you'd visually see the parent-child relationship as a vertical one
in general, however, the parent relationship is defined for arbitrary trees as it is on wikipedia, which is with reference to the adjacency structure of the tree only, and not a visual representation
this is a different visual depiction of the first graph
it is possible to more easily see that 14 does not have two "parents" in this picture, because the graph has been drawn to make distance from the root correspond more closely to horizontal drops

10:29 AM
Hmm okay
Still it is then possible to see that it it a tree :)
Thank you very very much @leslietownes
Thanks a million!!!
Have a nice day!

you too
generally when one draws a picture of a graph, visual concepts like "above," "below," "to the right of," etc. are not a part of the adjacency data, and so are not properties of the graph, but of the picture only
with some extra structure (like a tree and a choice of root) it is sometimes possible to come up with pictures where these visual notions do reflect things with graph theoretic meaning (like the "is a parent of" relation), but this is not an automatic consequence of visually drawing a graph

Okayy I will use it to explain others

there's a pretty cool open source utility, "graphviz," for automatically generating pictures from adjacency data, which is pretty configurable and can do things like "take this adjacency data, which describes a graph theoretic tree, and draw it as a real-life-looking tree with a chosen vertex as the root"
and all kinds of other fun stuff
if you are ever in a position where you need to generate a picture from raw adjacency data, its worth checking out

Okay, that's cool :)
I will check it :) :)

here's an example of what you get if you feed it the adjacency data for US states and ask to plot it as a planar graph
its more interesting, of course, with data sets where you don't already have a visual picture :)

10:43 AM
Yeah this is very interesting :) :)
I'm most grateful to you @leslietownes
Many many thanks again!!! :) :) :)

3 hours later…
1:28 PM
@user85795 - i did! thanks for asking.

hey, folks. I've another question for you. trigonometry. beginner stuff. chapter about sine and cosine rules. i wanna confirm that my diagram is correct, as I've tried several times, but I get gibberish.

the problem: "A and B are two points 1500 metres apart on a road running due west. A soldier at A observes that the bearings of an enemy’s battery is 295.8 degrees, and at B, 301.5 degrees. The range of the guns in the battery is 5 km. How far can the soldier go along the road before he is within range, and what length of the road is within range?"
is my interpretation correct?

2 hours later…
3:03 PM

### CURED

For feedback/discussion/requests of Close/Undelete/Reopen/Edit...
That question is hilarious

4:02 PM
@nevvermind Your picture doesn't look right to me, but I don't have all the context. Usually, "bearings" are measured on a compass. 0° = 360° is due north, 90° is east, and so on. So if the bearing from the soldier to A is 295.8°, then I would imagine that the soldier is south of the road, and that A is NNW of the soldier (approximately). The bearing from the soldier to B is 301.5°, which puts B a little to the east of A.
Your picture looks backwards to me.
But maybe I am interpreting "bearing" wrong?
Otherwise, the picture seems to be fine.

4:19 PM
@nickbros123 isn't that a good thing she skipped all the boring topics so now she can talk about the good ones?
:P

4:35 PM
The space of real analytic functions, say on the line, doesn't have a natural Fréchet structure--the appropriate topology thereon is notoriously more complicated. The case of real analytic functions on a compact interval (which is simpler) leads to a Silva space and these are never Baire except in trivial situations

4:48 PM
in Helpful Commentary, 20 secs ago, by Shaun
Can I get some feedback on this question, please? It's getting a few downvotes . . .

hey, @XanderHenderson. thanks for pitching in. i can confirm "bearing" in this case means "absolute bearing" (as described here: https://en.wikipedia.org/wiki/Bearing_(angle)#Absolute).

however, I think you misread the problem a lil' bit. it says "a soldier at A", which is different from the bearing "from a soldier to A". i read it as "the bearing from A to P" and "the bearing from B to P".

5:14 PM
@Shaun if someone wanted to comment they'd ask there, no
it would make more sense if, I don't know, you wanted to people notice the channel more or something
@JohnZimmerman interesting. What's a Silva space?

@Jakobian I don't know what you mean.

How to construct a Cantor like set all of whose points are irrational?

@Jakobian OP gravatar is disturbing :(

@SoumikMukherjee Why don't you consider irrational numbers as the Baire space $\omega^\omega$ and then the subspace $\{0, 1\}^\omega$?
The map $f:\mathbb{Z}\times \omega^\omega \to \mathbb{R}\setminus\mathbb{Q}$ given by $$f((a_0, a_1, ...)) = [a_0; a_1, ...] = a_0+\cfrac{1}{a_1 +...}$$ is a homeomorphism
So here the Cantor set will be those irrational numbers with continued fraction consisting of ones and zeros

5:31 PM
I have a basic doubt. Parseval's formula. Can it be applied even when the Fourier series does not converge pointwise to the function? I hope the question makes sense.

also see this great answer by Martin Sleziak
@SineoftheTime I think the OP watches simpsons a lot. Great show

@psie as far as I understand, the identity has little/nothing to do with pointwise convergence

@psie that's right. It doesn't have

👍

@Jakobian Thanks

5:42 PM
@psie all this formula requires is that $f$ is in $L^2$
@TedShifrin did you see my comment about $\text{RSpec}$ being the "real spectrum"?

@Pizza how are you doing?

21

Let $C$ be the middle-thirds Cantor set. If $x\in\Bbb R$ and $q\in\Bbb Q\cap(x+C)$, then there is a $y\in C$ such that $q=x+y$, and hence $x=q-y\in q-C$. If $\Bbb Q\cap(x+C)\ne\varnothing$ for every $x\in\Bbb R$, then $$\Bbb R=\bigcup_{q\in\Bbb Q}(q-C)\;.$$ Verify that each of the sets $q-C$ fo...

@SoumikMukherjee pretty sure I saw this before, cool question. You can argue topologically or measure theoretically
you didn't specify in what way you want Cantor set to be in irrationals so I've assumed you want it there as a topological space

6:00 PM
I actually didn't think about any way of specifying.

Just trying to follow along with the cases here, and I follow up until the third case ($n \in E$). How do we know $E \neq \emptyset$? Did one of the previous cases exclude it?

@Jakobian Yeah. I still don't get it.

Now obviously if $E = \emptyset$ then it's equivalent to the natural number 0 which is in every $n$
But Halmos didn't treat that case separately so I wondered if somehow it "fell" under one of the earlier two cases

@EE18 What's his definition of proper subset?

The usual I think, $A \subset B$ and $A \neq B$

6:07 PM
Then the empty set comes under that category.

What if $n = 0$ though?

Ah.
For me, the empty set is not a proper subset.
2

@TedShifrin in algebraic geometry you want to study the affine space, if $R = k[t_1, ..., t_n]$ where $k$ is an algebraically closed field then spectrum works great. Here its similar but $k$ will be a real-closed field. Apparently in this case real spectrum is better, and this is what you do in real algebraic geometry and semi-algebraic geometry and results in something nice. In words of Pete Clark, the space of orderings of a field contains algebraic, geometric and cohomological information.
you should check section 7 of his article

I just would never have thought orderings on a real field would have anything to do with the ideal structures.

@SoumikMukherjee fun fact: you can even get a Cantor set in $\mathbb R$ made of points that are algebraically independent over $\Bbb Q$

6:41 PM
Sorry, another question about Halmos :(
In particular, I don't follow the sentence beginning $If n \notin E$, then the restriction..."
Doesn't this presume $E \neq n$?
In which case there is seemingly no contradiction, as we've just demonstrated that $n$ is equivalent to $E$, i.e. to itself ($E = n$)

6:54 PM
@TedShifrin whaaat
I never saw anybody saying that $\emptyset$ is not a proper subset of a non-empty set

@Jakobian Well, now you have.
Unfortunately, I am forced to agree with Ted. A proper subset is a nonempty subset which isn't the whole shebang.
The emptyset is not proper.

I confess I don't use the language much, but my impression was that the empty set and the entire set are "trivial" subsets and should be discarded.
@Xander But Munkres agrees with them.

Well, what does he know?!
Silly Munkres.

Well, in my algebra book in the appendix on basics on sets, I agreed with them, too.

@TedShifrin I'd call those trivial subsets, restricting the meaning of proper to the relation $\subseteq$

6:58 PM
I guess I have to agree.
This is a surprising question. I don't see an easy answer.

well, its a PSQ

Regardless, I don't see how to do it.
I suspect there is no short-cut.

The equation gives you some relation between second and first derivative, but I also doubt its enough to say anything at all
seems like an XY problem

Global max (in $(x,t)$) is another issue. That occurs on the boundary, of course.

7:14 PM
@AlessandroCodenotti There was a way to do this problem using some geometric group theory which I do not remember anymore. $\mathrm{PSL}_2(\Bbb Z)$ acts on $\Bbb H^2$, and the orbit of $0$ in $\partial \Bbb H^2$ is $\Lambda := \Bbb Q \cup \{\infty\}$. You'd like to pick out an appropriate subgroup of $\mathrm{PSL}_2(\Bbb Z)$ whose limit set misses $\Lambda$. An infinite-index, infinite, virtually non-cyclic Fuchsian group has limit set a Cantor set, so if you could do this you'd be done.
Can you fill the details? I tried for a bit but failed.

7:48 PM
@Jakobian my limited understanding is that a Silva space is related to to the space of real analytic functions on a compact interval. It seems fascinating but related to banach spaces and functional analysis which I don't know anything about

@BalarkaSen I've never seen this approach before
The proof I know is just pretty much an immediate corollary of Mycielski's theorem

@Jakobian I really want to understand a much more simplified version - not considering the entire space of real analytic functions but just one class that forms a monoid structure
analytic monoid lool

8:08 PM
see section 3
real analytic functions are interpreted as germs of holomorphic mappings, and thus they admit subspace topology from the space defined in 3.3

8:54 PM
all you know is $au+bv=6$ for some integers $a,b,u,v$ what could $(a,b)$ be? I put it has to be divisible by 6, is that enough or should i say more
or is $(a,b) = 6$ :o
i think by some theorem $d = au+bv$ so yeah I guess the gcd is 6
as for whether u,v could be relatively prime, it depends on if $a,b$ are divisible by $6$?
jk it doesn't imply $d = (a,b)$
i guess $(a,b)$ could be anything.
let $p$ be a positive prime integer. Isn't it trivial to show $\sqrt[3]{p}$ is irrational?
like, by the fundamental thm. of arithmetic $p = p\cdot 1$ has no other factors but itself.

Gcd = 6 is plain wrong.

yeah I caught it

gcd has to divide 6.
Remember that the gcd of $a,b$ is the smallest positive integer that can be written that way.

what if $6 \mid u$ and $6 \mid v$ then we have $ag+bh = 1$

What?

9:06 PM
it's not explicitly stated that $au+bv=6$ is in reduced form
just that "that's all you know"

I told you the answer. Take the time to understand it.

wait yeah $1 \mid 6$ oops
how would I find the solutions if any of $2000x\equiv 4\pmod{19875}$?
I can reduce it to $500x\equiv 1\pmod{19875}$
i know it doesn't have solutions
i just don't know how to explain why
I guess showing the class of $2^2\cdot 5^3$ is not invertible $\pmod{19875}$

9:21 PM
You’re right, but be careful. Does $2x\equiv 4\pmod6$ have a solution?

1 hour later…
10:27 PM
@Tedshifrin yes
$x = 2n+1$ for $n\geq 0$
why did I write that

So what’s different about the two questions? … Yeah, that solution is wrong.

uh for $x \in \{2,5,8,11,...\}$

GUYS HELP
Hiiiii btw
Ted you gotta know some thermodynamics right?

im taking that this semester even though it's pain and suffering
maybe i can help @allie

im so happy they put calculus in chemistry
so i had an exam today and the quesiton confused the fuck out of me. I think one of the pieces of info like didn't add up with the rest of the problem
So just a quick question: for a reversible adiabatic expansion, can the pressure increase?
like increase in pressure but do expansive work
in the same process

10:40 PM
oh i gotta get my notes brb

reversible expansion for an ideal gas?

yeas

so $pV^{\gamma}$ is constant if I recall correctly

mhm
and gamma > 1

the pressure can increase or decrease i think
since $\Delta(V^{\gamma}P) = 0$ you can have $\Delta P > 0$ I guess

10:45 PM
if you look at the graph of the adiabatic expansion in the Clapeyron plane, $p$ is in general no constant

im looking at $xy^\gamma = k$ on desmos right now, and all of them show p decrease as v in crease
for the specific gamma i have
and for all values of k i think?

exactly
that oen shows p decrease as v increase

for $V_f^{\gamma}P_f = V_i^{\gamma}P_i$ with $\Delta V >0$ we need $\Delta P <0$ I think

yup

10:48 PM
can't be the case where $V$ is increasing and $P$ is increasing as well unless we add heat

hmm

but when $V$ decreases we have to have $P$ increase vice versa
maybe I'm thinking of isotherms actually
$\Delta Q = 0$ for adiabatic processes means $\frac{dS}{T} = 0$ means no change in entropy. if $V$ increases, that increases the entropy so $T$ must drop and so $PV \propto T \implies$ P drops as well

Let $A,B$ be nonempty subsets of some poset $X$ which are bounded above and suppose $\sup (A \cup B), \sup A, \sup B$ all exist. Am I correct that it is still not in general true that $\sup \{\sup A, \sup B\}$ exists and that a necessary and sufficient for said existence is that $\{\sup A, \sup B\}$ have some upper bound?
My textbook doesn't explicitly say this, it's just something I'm playing around with in my head

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