It's just gone 1am here, so @ me if you have an answer, please. Night :)

@Shaun as well-orders

@leslietownes I once avoided old typewriter font textbooks but after I realized many famous books are written in that way, I just read and don't care any longer.

some notes are really handwriting not even typed

but still, I only read books or papers in English.

oh, i had a similar thing. i avoided typewriter stuff, unless it was the least worst option. and when a book's typesetting is bad i do dislike it. i try not to be superficial about it, but i always notice.

there's actually the third type of book, incomplete book. I recently found some possibly good book or preprint of a book that contains quite a lot of important concepts I'm interested in but (quite) incomplete and latex is not well rendered so lots of ??? are written (for example Lemma ???, or see [??]). But I won't send email to the author to ask some better version, maybe it's rude.

haha, yeah. sometimes the material is even there, they just didn't run latex twice to make the cross references work.

you sometimes also see stuff like that on book sites/torrents/etc. where, someone has added it or uploaded it to some archive without noticing that it isn't a final version. i imagine i'd find it especially rude if someone contacted me about something i'd written like that when i don't even maintain it on my own website anymore. :)

This is something I didn't expect.

i took a class from a prof who had nearly a complete book written, and only circulated it in hard copy form. i thought he was being unreasonable at the time (and lost my own copy and wish i still had it), but that form of limiting distribution has a lot to recommend it.

Is there a fast way to convert from Polar form to Cartesian Form?

"fast" meaning what exactly (e.g. do you have implementation in software in mind, and if so, what kind of constraints are there on that)?

No, as in on pen and paper.

Like in an exam setting.

I tend to draw out the picture and then go from there

but that usually takes too much time.

i'm not sure if i know what you mean. if you mean remembering the sines and cosines of "nice" angles without a device, i don't know of a good way of speeding that up

in my own mind i think in terms of a hazy mental graph of the sine function with a few remembered points on it, i guess i use that when i'm dealing with e.g. multiples of pi/3

for anything more complicated than that it would just take me time

@Jakobian Thank you :)

Hello Everyone...

@leslietownes okay... pi/4 ;-p

When do you use community wiki to answer?

I usually only see them used for two reasons: (1) For the wiki aspect, when you want your post to be wiki-fied and edited by many users over time, and (2) When you want to include an answer but not receive any benefits from answering, often used when you are repeating someone else's solution.

I have the following definition of what it means for a function $f: I\to\mathbb R$ to be strictly increasing in an interval $I\subseteq\mathbb R$: $$f(x_1)<f(x_2)\quad\text{if }x_1,x_2\in I\text{ and } x_1<x_2.$$ I'm looking for the negation of this statement. I assume the definition has a hidden "for all" quantifier in it, so the negation would maybe be $$f(x_1)\geq f(x_2)\quad\text{for some }x_1,x_2\in I\text{ and } x_1<x_2.$$

Is this correct? Bonus question; does this imply the function can not be one-to-one on $I$?

I guess to write this correctly in logic I would have to be more formal: $$\forall x_1\forall x_2(x_1<x_2\implies f(x_1)<f(x_2)).$$ So the negation would be: $$\exists x_1\exists x_2(x_1<x_2 \wedge f(x_1)\geq f(x_2)).$$

I don't like the way that the definition is phrased. I think it would be more clear if you wrote $$x_1,x_2\in I \text{ and } x_1 < x_2 \implies f(x_1) < f(x_2). $$

Yeah, I agree.

the negation implies a single pair where the function is not strictly increasing

the negation is an existence statement

@shintuku but there could be more than a pair, right?

yep

@sunny Yes, but there needn't be.

right

nope

gimme a sec

$\exists x_1, x_2: (x_1, x_2 \in I \land x_1 < x_2) \land f(x_1) \geq f(x_2)$

parenthesis for readability

ok, so can we draw any conclusions whether or not the function can be one-to-one on $I$? If it is strictly increasing for some $x$ and not strictly increasing for other $x$, that doesn't sound very injective to me...

@sunny What?

That doesn't make sense.

Monotonicity is not a local property. It is defined on an interval, not at a point.

@sunny you have to make more explicit your reasoning here, i don't understand the link you're making with injectivity

If $f$ is strictly increasing on some interval $I$, then it is injective (one-to-one) on $I$. But the inverse does not hold: there are functions which are injective, but not strictly increasing (nor decreasing). For example, consider the function on $[0,1]$ which maps $x$ to itself for all $x \ne 1/2$, and which maps $1/2$ to $47$. This function is not strictly increasing, but it is injective.

@XanderHenderson alright, I was just about write something similar. If the converse would hold of "strictly increasing $\implies$ injective", we could conclude that the contrapositive of the converse, i.e. "non-strictly increasing $\implies$ non-injective" would hold, but I guess it doesn't then

"The contrapositive of the converse"?! O_o Why say this in such a complicated way? It is the inverse.

Why is CA (cone of the inclusion A-->X) contractible in $X \cup CA$?

I don't see it from the pictures: what point does CA contract to?

it can't be the cone point.

Of course it's the cone point.

You can write down the homotopy explicitly.

I understand that X is the base, base of the cone is attached to the face, the cone point is in the air.

@Koro "...in the air"?!

What does that mean?

@XanderHenderson I meant to convey that the image looks like this

I know what a cone looks like... I don't understand your objection.

so I don't understand how could it possibly contract to the cone point. In order to do that the cone has to detach itself from the base.

The cone is $A \times [0,1] / \sim$, where $(x,y) \sim (x',y')$ when $y=y' = 1$, right?

Hint: Write $t$ for the $[0,1]$ variable, not $y$.

@TedShifrin Much better hint than what I was writing.

yes, in my case this can be considered as upside down cone...

I made only the mildest of suggestions :P

it's not intuitively clear to me how a mapping cone can contract to a point.

What is the mapping cone of the inclusion map?

it's just $X\cup CA$, where CA = $A\times I/ A\times 1$.

to me [0,1] is from bottom to top, i.e., I'm collapsing the top of A \times I.

So how is CA contractible in this mapping cone?

That's not literally correct. The $X\cup$ part has an identification. But I see your concern. What hypothesis do you have about $A\subset X$?

(X,A) is a CW pair.

Ah, so isn't there a retraction $X\to A$?

I haven't yet come across this fact.

What's the definition of a CW pair?

But I don't see why there won't be a retraction.

Do you see why your writing the disjoint union is incorrect?

You're forgetting the definition of the mapping cone.

with identifications, I should have written $X\bigsqcup_{in} CA$.

in means inclusion map here.

(thinking geometrically, it seemed to me that it won't matter here as $A\subset X$ but I should have stated it this way)

Which attaches the cone to $X$ along $A$. No, of course it matters. Disjoint union is NOT at all what you mean.

@TedShifrin $A\subset X$ is 1) closed in X, 2) union of some cells of X.

ah okay.

for the record, we are discussing example 0.13 of Hatcher's

So the cone itself is obviously contractible to the cone point.

agreed

Your issue is that you must first get $X$ mapped to $A$.

you mean using 'retraction'?

Let me also write out one equality/homotopic equivalence: $X/A=X\cup CA/CA \simeq X\cup CA$

For the homotopic equivalence, Hatcher says 'since CA is contractible subcomplex of X\cup CA.'

I understand the first equality.

He means the literal cone, not the mapping cone. $X\cup CA$ is the mapping cone.

ah, I see.

supporting this fact is example 0.8 wherein he considers an arc to be contractible.

it makes sense now. many thanks.

Sure thing.

A subset of a complete metric space is completely metrizable iff it is a $G_{\delta}$ set.

How to prove the forward implication?

What if you take the intersection of all $1/n$ neighborhoods of the subset?

In analysis on R, when we declare sets (to prove theorems, like monotone convergence), such as set $ S=\{x_{n}: n \in \mathbb{N}\}$ , and for the purpose of this example, say $x_{n}$ is a sequence that goes like $\{1,2,3,1,0.5,0.333..,0.25.....\}$. When speaking about this set, can I treat the set S like it is $\{1,2,3,1,0.5,..\}$ or should I treat it like $\{1,2,3,0.5,...\}$

@nickbros123 you need to distinguish sets that are sequences from sets that are unordered

@nickbros123 if $A = \{1,2,1\}$ and $B = \{1,2\}$ are unordered sets, then $A=B$

i.e. any repeated element in the presentation of an unordered set is only a single element in the set

@shintuku is it correct to apply arguments that make use of the completeness axiom of R then, when we speak about sets that are sequences (or as you said, ordered)

@nickbros123 A set is a collection of objects. A sequence is a function from $\mathbb{N}$ to some set---this is usually written as an ordered set of objects.

@nickbros123 completeness is usually used in the context of real numbers and more specifically intervals of real numbers, sequences don't tend to involve intervals of real numbers

However, from the point of view of analysis, you really shouldn't think of a sequence as a set. Sure, at a foundational level, everything in mathematics is a set, but this isn't a helpful way of thinking about sequences and is likely to be confusing.

Yeah, sequences are really functions (from $\Bbb N$ to your set).

Oops. Xander already said this.

$\mathbb{N}$ or $\mathbb{Z}$?

either of them works

but the natural way to think of it is $\mathbb N$

The rational way to think of it is $\mathbb{Q}$

in monotone convergence theorem, the completeness property is used, as in, a set $S$ is declared based on the sequence $x_n$ , like: $S=\{x_n: n\in \mathbb{N}\}$ and there is an assertion that, as this set is bounded, it has supremum. So I'm confused as to how we view this set S

$S = \{f(1), f(2), f(3), \dots\}$, where $f$ tells you what the $n$th element of the sequence is

Though I understand this is the range of the sequence which is the function from $\mathbb{N} \to \mathbb{R}$

I did not get an answer on mathoverflow so I'm going to ask on mathstack exchange

@shintuku if suppose $f(1)=f(2)$, do we only write it once or we dont change the occurrences?

Do NOT write sets.

using the uno reverse card

Sets

The sequence is the function.

@冥王Hades Any well-ordered countable set.

@nickbros123 we write it twice since we want to convey an ordering, i.e, not $\{f(2), f(141), f(5), \dots\}$. but as ted states the sequence is more specifically the function itself

@TedShifrin ok, so if my sequence is bounded, the set (which is the range of the function) has a supremum, and I can apply all the rules and theorems pertaining to the completeness property of R?

Yes.

For that purpose, the order of the terms in the sequence is not relevant (yet), and so looking at the set of values is all you need.

@shintuku I understand. I'm in the middle of trying to prove a theorem and this shakes my understanding of proofs based on completeness property, mainly because intuitively I've been building a picture of a set, which looks like the sequence itself, with ordering and all that.

@TedShifrin yeah this clears it

yeah the sequence itself is the arrow that points 1 to something, 2 to something, 3 to something ,etc.

and it's often represented as an ordered set where the first element means what 1 points to, the second element what 2 points to, etc.

Yeah, got it, thanks. If anyone's wondering, I'm trying to prove this theorem:

@AlessandroCodenotti is there a connection of some kind between Suslin lines and Dowker spaces of certain kinds?

The way I've defined it I have a closed form volume for each of the 4 'zonoid' components individually but I don't have a closed form volume for the enclosed volume of their intersection

very similar in spirit mathoverflow.net/q/430504/123449

the volume of each of the 4 components is a sum of determinants naturally (because it's a zonoid i.e. hausdorf limit of sequence of zonotopes). I'm tempted to say the enclosed volume is another sum of determinants but can't quite get the precise result

@Jakobian not that I know of

I'm asking because I was watching a lecture by Marie E. Rudin, and she hypothesized that such connection might exist

I was wondering if anything is known about this

I'll just assume that it's $\pi/4$ and start deriving results based off of that

@AlessandroCodenotti how did you learn model theory? Did you learn mathematical logic first?

what kind of model theory is being used in topology?

I was thinking about this - maybe I should try to learn it when I need it for anything

Model theory is part of mathematical logic. So is recursion theory.

@nickbros123 That's an excellent exercise.

I never had a model theory course, I learned it by reading what I needed and by osmosis since I'm officially part of the model theory group here and I'm surrounded by model theorists

Model theory and topology don't interact very well though. There are some nice results on model theory of complete metric spaces, but they all work in continuous logic (of which the usual first order logic is a special case, corresponding to discrete spaces)

Now I'm reading some lecture notes about NIP theories and Dp-rank since I need to understand a bit more about those

Well, nonstandard analysis is a consequence of the compactness theorem in model theory. So there's that.

(Yes, I remember that from the one graduate course I took in logic as an undergraduate ... and dropped in week 10 or something after 3 infernal weeks of boring Turing machines.)

@TedShifrin What were they boring into?

My brain, apparently.

Hi :) I thought I'd advertise a new question of mine.

I think of them as a topological space together with a ring of continuous/smooth/holomorphic functions on it (and on its various open subspaces). This is the structure sheaf. What else is there to do?

I'm sort of curious about why you rolled back my edits with (a) improved readability, (b) conformed to typical typesetting and referencing standards, and (c) removed the unnecessary (and whiney) "please help" (which is among phrases on the SE network which are typically removed when other edits are made, see also "thank you in advance").

Well: (a) That's debatable. (b) The (A) and (B) are as in the text. (c) I'm just being polite.

@Shaun You don't have to slavishly copy the text. Numbered lists are a natural part of markdown, and improve readability.

Regarding "just being polite": math.meta.stackexchange.com/q/31750

Hello

hope everyone is having a good day.

And your [ . . . ] breaks across lines awkwardly---the standard typsetting convention is to simply use an ellipsis to indicated elided material.

(oh, and "second edition" is not part of the title of the text, does not need to be capitalized, and should not be italicized; titles of texts should be italicized, but not quoted)

I mean, at the end of the day, it is your question, but, like, why insist on writing it in a way that is harder to read and does not conform to convention?

Thank you, @XanderHenderson. I didn't know that. I'll be more careful in future.

it would seem that a lot of users who upvote me get their accounts deleted.

@Shaun Just to reiterate. Whining/begging is not, to my mind, being polite.

Everyone on this site wants/expects to be helped. I personally am inclined to help people who've made efforts and I can build off what they have done right or wrong. Entreaties to "please help" just push me away.

I don't like being ordered, even with a "please."

@copper.hat SE rolled out a new set of tools for tracking down suspicious voting. Lot's of sock puppets have gone down in flames recently. The smart socks tend to spread their votes around to try to hid what they are doing in the noise, so a lot of users are losing XP.

black market of mse reputation points

so i should sell my rep while i have the chance...

@copper.hat Yup. I'm pretty sure you can get a reasonable amount of lesliecoin for an upvote.

but then the nsa will track me

Well, yeah. Get over it.

They're already tracking you.

my idyllic life is soo disturbed now

I still think leslie has pocketed the loot he promised me a year ago.

@copper @Xander This is an intriguingly simple question. I keep wanting to say there can't be such a function, but ...

Any time the function is $1$ on a (unit) vector, it must be $0$ on the orthogonal great circle. So take the union of all such great circles, and the function must be $1$ on the complement. Is that possible?

Why would sock puppets bother upvoting or downvoting on this site?

@TedShifrin Nice, aggravating question.

@TedShifrin Ugh... I've been totally nerdsniped by that question.

It's a cool question.

(Even if I simultaneously believe that it should be closed for lack of context.)

I keep thinking I have reasonable approaches, but then they don't work... :/

I commented on the approach the OP was trying (that he put in comments). It won't work. I feel like there must be a BCT result with covering the sphere with great circles.

Sometimes "context" gets overblown here. I think it's an interesting "continuous" pigeon-hole question.

Oh, @robjohn is here! Maybe he'll swoop in and make an obvious comment on it.

Oh, hi @robjohn :)

@TedShifrin Hey, there. That seems like one of those non-measurable functions, but I can't think of anything right off

Yeah, that was what I was thinking at the outset.

But now I'm thinking the OP's idea might work.

If you take a vector more than $\pi/4$ from the north pole or south pole, its orthogonal great circle will pass through that $\pi/4$ cone. So I think he wins.

You mean take a cap around each pole, and a band around the middle? I was thinking of a band with latitude between $\pm\arcsin(2/3)$, but I haven't given that enough thought and I am going to PT soon.

However, the $\pi/4$ option does sound good.

Do well at PT, @robjohn. Mine is tomorrow.

PT: as in physical therapy? Or perverse theology, or part time, or...

Whatever the case, sending my well-wishes, @Ted, @robjohn o/

I only care about discrete techniques now

and semi-definite programming

metric geometry

matrices

I need to re-write someone's definition as a semi-definite program

perverse theology? What does that even mean

Can an atheist participate in perverse theology?

What is perverse theology?

An offshoot of perverse sheaves.

what should I know to solve these (#2, #3 & #5)?

I don't know much category theory.

also what is HP^n?

Quaternionic projective space.

what in the world is that?

thanks, I'll look it up.

@Koro It is like a real or complex projective space, just more quaternionic.

I wonder how S^3 acts on S^\infty

@Koro Magically.

that makes sense

these 5 questions seem inexpugnable to me

I'm studying a continuous function which is neither strictly increasing or decreasing in a neighborhood around $0$ and it is claimed the function therefor has no local inverse there. I do not understand that claim. What proposition/theorem is this claim based on?

Try IVT

"local inverse" presumably has some hypothesis requiring at least continuity?

oh, i see, f itself is required to be continuous

Ok, I will try. I was thinking that every continuous bijection from $\mathbb R$ to $\mathbb R$ is strictly monotone, and hence a not strictly monotone function can not be a continuous bijection, however, it could still be a discontinuous bijection.

use the continuity of f. if f is not assumed continuous there is no relation between invertibility (local or not) and preserving or reversing order

@sunny How do you prove that?

@TedShifrin hmm, maybe with MVT 🤷‍♂️ ??

I

V

T

MVT requires derivatives, no?

Mr Shifrin sir....I have a question of 2 dimensional manifold importance

I am trying to convert your example from the video lecture of $F(\bf{x}) = [ \|x\|^2 - 4, x_1x_3 + x_2 x_4] = [0,0]$ into the parametric form you wrote out

which was the torus

@D.C.theIII DR Shifrin.

True

If you are going to use honorifics, use the right ones.

shrug

How would I conceptualize this? Doing other examples and going from parameterization to level set or level set to graph I could get, but this one has me miffed....maybe if it wasn't trig functions I could get it possibly

So what’s the question?

so you challenged us to get the parameterization form which was: $g(u,v) = (\cos u + \cos v, \sin u - \sin v, \cos u - \cos v, \sin u + \sin v)$...I'm embarking on said challenge

I’m never touching trigonometry again

it contains sin

Note that $u=(x_1,x_2)$ and $v=(x_3,x_4)$ are orthogonal vectors. The way I see it is to make a change of coordinates … what is $(u+v)\cdot (u-v)$?

Sorry, didn’t mean to reuse the same letters.

What is $\|u\pm v\|^2$?

$\|u\cdot u + v\cdot v||$?

whys Chatjax freaking out on my PC

working out $\|u + v\|^2 = x_1^2 + x_2^2+x_3^2+x_4^2 + 2x_1x_3+2x_2x_4$ which almost brings together the two equations from the level set form.

Is it unusual to have ODEs, numerical methods and introduction to partial differentials at a high school level?

hades: depends on what you mean by these terms. it might not be unusual to see stuff that arguably fits those descriptors in a high school class, at an appropriate level. i wouldn't expect a hs treatment to cover what a university-level course with a similar description would

e.g. some ODE are covered (maybe not by name but by subject matter) in many calc classes. if you study euler's method for numerical solutions (calc books often have a section on this, even if instructors choose not to cover it), that's a numerical method.

@leslietownes Probably not as much depth as a college level course but if I remember correctly, ODE chapter went quite in-depth covering most usual techniques seen in college courses.

similarly in textbooks, partial derivatives and partial differentials are often introduced at or near the same time, again, whether or not an instructor chooses to cover that, or in depth.

a standard calc sequence will often include separable ODE and linear ODE with constant coefficients (maybe arbitrary order) or nice enough coefficients (maybe first, second order). and a standard move with the "AP" program in the US is to stuff as much of that into high school as possible, so it sounds like the AP program is delivering a high quality consumable.

the high quality package

when i taught classes that required that stuff in college, the people who had 'already taken it' had to be reminded of everything.

the deluxe bundle

the premium wrapper

shintuku has picked up on my slight sarcasm here. a lot of this "having ___ at a high school level" is just trying to get the gee-whiz factor of "wow, little johnny is 14 and he's already mastered PARTIAL DIFFERENTIAL EQUATIONS" (parents have no conception of what this means but wow, it sounds great)

i don't mean to single out the AP curriculum with that. it's definitely more general than that.

"imagine when he gets to entire differential equations!"

High school linear algebra …. rolls $5^e+\pi^3$ eyes

it's a little funny, if you think about how the stuff that makes up even a university-level curriculum in all of that stuff is not going to get you beyond, i don't know, the mid 1800s, in terms of known techniques. and a person whose education stopped at the mid 1800s in a lot of other subjects would be regarded as pretty ignorant.

@leslietownes you just called me out. Though I wasn’t doing PDEs at 14, just integrals

we had differential equations in calc 2 but they were just separation of variables and like one other technique

and that's basically what you can do at that level. there isn't much more than that.

linear odes, yeah that was it

and in a college class it might be crammed into two halves of a single lecture. i don't know why the brand names for this stuff are so attractive at the high school level.

it has the prepackaged formula for solving it

I'm stuck.

it makes more sense if you think of math not as a subject in its own right, but as an abstract signifier of substantive education. at one time, instruction in latin or greek might have checked this box, these days in a lot of places it is only math.

@leslietownes my history education is barely getting past the 18th century

formation of the state and right-left division

interesting stuff

absolutism, arrival of absolute monarchs

concentration of power not seen since romans

important stuff

Hardly known to me

@D.C.theIII $u+v$ and $u-v$ each live in a circle, DC. Solve for $u$ and $v$ in terms of those.

initiation of industrial revolution

arrival of legal techniques to deal with abstract financial instruments

allows concentration of capital into ships

then TRAINS

Ok Ted. I have to turn off the tag sound always scares me when I am in a bout of concentration.

structure of industry ravages domicile-based economies

formation of a proletariat to power the industrial machine

the exodus from the country side

end of 18th century might not be there yet

but everything is announcing it

so much stuff if you restrict yourself to the english empire alone

but it's all a system, you have to get history to tell you the formal concepts of the machinery of politics and state, which unlike other physical objects has a schematics which is essentially historical

@TedShifrin Thanks, I had left at 1:00. I have more on Thursday.

@amWhy Thanks.

Hades-1 COVID-0

An insignificant pathogen can’t stop The God of Underworld