Mathematics - 2021-01-20 (page 3 of 3)

Ted Shifrin

10:00 PM

So I'm fine with your saying $\forall a,b\in\Bbb R$ with $a<b$, we define ...

Thorgott

also important: what is $(a,a)$?

polite proofs

@Thorgott Also the empty set

Thorgott

right, so that covers all cases

Ted Shifrin

However, it becomes important when you teach/think about the integral to allow $\int_a^b$ even when $a\ge b$. So in that situation, you don't really want the empty set for the interval.

So for many reasons I don't want to do what Thor and Astyx want.

Moving on ...

Fargle

Aww, I missed interval discourse

Ted Shifrin

10:04 PM

Your life hangs by your answer to this question, @Fargle. Are you on my side or are you on the obnoxious Europeans' side? :D

Fargle

I'll play centrist here---depends on the context.

Thorgott

Well, there's no issue, $\int_a^b$ for $a>b$ is not $\int_{(a,b)}$. the latter is a Lebesgue integral and zero as it should be, the former is an oriented integral over the oriented interval $(a,b)$, but with negative the usual orientation.

Astyx

So ... in the Atlantic?

Ted Shifrin

And feel free to help out polite with his quandry.

Obviously, the geometer in the room thinks it's important to understand the two orientations on $[a,b]$. I stand by my objections.

Fargle

I can see why for some definitions you'd want $(a,b)$ to be empty if $a \geq b$, particularly in cases where you're not gonna be concerned with the interval as having orientation.

polite proofs

10:06 PM

Jeez, I feel like I accidentally caused a ruckus. Sorry.

Ted Shifrin

LOL

Astyx

is it clearer polite?

Ted Shifrin

All polite wanted to discuss was whether to incorporate the quantifiers and, if so, how.

I made the big deal about $a<b$.

Thorgott

I also think orientations matter, but the issue is here is that notation makes it easy to mix up oriented and unoriented integrals, not the interval notation. You wouldn't write $(b,a)$ for the interval $(a,b)$, but negatively oriented, would you?

Ted Shifrin

Many times in mathematics there is an understood "for all," and he wants to be concrete and include it.

If you want to state $[a,c]+[c,b] = [a,b]$ for integration, @Thor, then it is sometimes convenient to understand it with my point.

Otherwise you have to write zillions of cases, and it honestly is just not worth it.

But no student of mine has ever raised this as a crisis, tbh.

Fargle

10:11 PM

My milquetoast answer was because I was concerned that orientation might make certain basic topological facts on $\Bbb R$ harder to state, but I suppose either definition can be WLOG'd down to the case where $a < b$ if you're working in cases where orientation does not matter.

Ted Shifrin

I just preface everything with $a<b$ in that situation, @Fargle.

Anyhow, answer polite's question re quantifiers.

Fargle

That also works.

@politeproofs $a$ and $b$ don't need to be quantified because they're already bound by you stating $(a,b)$.

polite proofs

Well, I feel like now we are back to square one, but thank you.

(Ted wrote that same idea a while back)

Ted Shifrin

LOL. You see — Fargle is savvy: He sides with Ted.

It is never wrong to include the implied quantification, @polite. It's only wrong with the quantifier does not belong there.

copper.hat

Why is it never Square Zero?

Ted Shifrin

10:16 PM

Does a zero look very square to you?

Fargle

I mean, I suppose you might want to say in your definition, "For any $a,b \in \Bbb R$, we define $(a,b) := \{x \in \Bbb R\;|\;a < x < b\}$", but the quantifier would be in that sentence rather than in the actual defining equality.

$0 = 0^2$, yes

copper.hat

No, but it is closer than a skinny one.

polite proofs

Right, well anyway, I think that saying: Let $a, b \in \mathbb{R},$ such that $a<b$. Then define $(a,b) = \{ x \in \mathbb{R} : a < x < b \}$

Ted Shifrin

LOL. Score one for Fargle.

Oh, no, score 0.

polite proofs

Oh well Fargie typed the same thing I suppose

Ted Shifrin

10:17 PM

Move on to something more interesting now, @polite :)

Fargle

Well, I didn't include the extra condition, but ye

copper.hat

i feel as if a burden has been removed today, cannot quite figure out what it is...

omg

the goal is communication not notation.

Thorgott

rip astyx

Ted Shifrin

Yes, copper, you and I are more similar in our lack of interest in formalism, etc.

Astyx

It was going to be so beautiful :(

Fargle

10:19 PM

:56814936 If you really want Ted's definition, you need $\{x \in \Bbb R | \min(a,b) < x < \max(a,b)\}$, otherwise you end up with every backwards interval just being $(\emptyset, -1)$.

There is always a higher degree of pedantry, just as there is always a bigger fish :D

Ted Shifrin

No, that's not Ted's definition. I just explained why I wanted $a<b$ in the original definition.

copper.hat

this is true is most endeavours...

Astyx

Right good point

$$]\cdot,\cdot[ : \begin{cases} \Bbb R^2\to \mathcal P(\Bbb R)\times\{\pm 1\}\\ (a,b) \mapsto (\{x\in \Bbb R \mid \min(a,b)<x<\max(a,b)\}, {(b-a)\over|a-b|})\end{cases}$$

Ted Shifrin

ROFL

copper.hat

some things cannot be unseen

polite proofs

10:20 PM

Sure, let me do a quick limit proof (or maybe that's even less interesting?): Let $f$ and $g$ be functions with domain $\mathbb{R}$. Suppose $\lim_{x \rightarrow a} f(x) = b$ and $\lim_{x \rightarrow b} g(x) = c$. Prove or disprove that $\lim_{x \rightarrow a} (g \circ f)(x) = c$.

Since the first two limits exist (too lazy to LaTeX), we know that $\forall \epsilon > 0, \exists \delta_a > 0, |x - a| < \delta_a \implies |f(x) - b| < \epsilon,$ and $|x-b| < \delta_b \implies |g(x) - c| < \epsilon.$ Since $f,g$ are (assumed to be) real valued functions, we can say that $\forall x \in \mathbb{R…

Fargle

@TedShifrin I meant for your oriented interval.

polite proofs

Though I just typed it out here in this chatbox, so maybe my proof is bogus or has wrong steps

Ted Shifrin

I also wanted it just for closed intervals. :)

Fargle

As I said, there is always...

Astyx

$$[\cdot,\cdot] : \begin{cases} \Bbb R^2\to \mathcal P(\Bbb R)\times\{\pm 1\}\\ (a,b) \mapsto (\{x\in \Bbb R \mid \min(a,b)\leq x\leq\max(a,b)\}, {(b-a)\over|a-b|})\end{cases}$$

Enough Latex for a week :)

copper.hat

10:21 PM

what stops $f(x)$ being $b$ for $x$ close to $a$?

Ted Shifrin

I would actually use different letters in the domains of $f$ and $g$, @politeproofs.

But copper is giving you a counterexample.

copper.hat

why not $2^\mathbb{R}$ altogether?

Ted Shifrin

Your limit definitions are incorrect. You are missing the $0<$ in both.

polite proofs

But I wrote that $\delta_a,\delta_b > 0$

Ted Shifrin

No, no. What's the definition of limit?

polite proofs

10:23 PM

Oh I didn't write that about $\delta_b$ (in fact I didn't write anything about it)

Ted Shifrin

Right, you omitted to specify $\delta_b$, but there's a more serious issue here.

Fargle

Ted's point is that you need $|x - a|$ to be nonzero, because a limit doesn't have to equal the function value.

polite proofs

$\lim_{x \to a} f(x) = L$ means that $\forall \epsilon > 0, \exists \delta > 0, |x - a| < \delta \implies |f(x) - L| < \epsilon$

Oh ok

Fargle

So you need $0 < |x - a| < \delta$.

Ted Shifrin

NO.

polite proofs

10:24 PM

Oh good point about the zero remark

Ted Shifrin

You specifically never require that $x=a$ is in the domain of $f$.

polite proofs

Indeed

Ted Shifrin

So you should question whether your proof is valid.

Astyx

(Bourbaki would like to differ)

Ted Shifrin

Yes, I know, @Astyx.

Shaddup, Frenchie.

Astyx

10:25 PM

grumble

Ted Shifrin

@politeproofs And, aside from that, I honestly think it's easier for you and for your reader to talk about $g(y)$ and $f(x)$, using different letters. Because, ultimately, you need to let $y=f(x)$ in the composed function.

And you are less likely to write nonsense or make mistakes if you do that.

polite proofs

Argh! Can't edit my post to get all the LaTeX.

Ted Shifrin

Yeah, only five seconds for editing.

Astyx

reload the page and copy pasted the unformatted latex

Thorgott

five minutes, not seconds

Ted Shifrin

10:27 PM

Whenever I run out of time, it feels more like five seconds.

BigSocks

I think it is only 2 minutes for the proletariat

btw I got the physical copy of the Lorenzini book @TedShifrin also we worked on understanding a lemma here earlier, Astyx and Thorgott and I

very nice book, makes me happy to read it

Astyx

What is it about?

BigSocks

An invitation to Arithmetic Geometry

Astyx

oooh

I should read that

BigSocks

yeh- so that thing you don't like but are pretty good at now

lol but I thought you said you wanted to go back to analysis?

Astyx

10:29 PM

lol I failed the AG exam hard today

Ted Shifrin

I'm glad you appreciated my rude suggestion, @BigSocks.

Astyx

I'm not pretty good at anything

BigSocks

damn... sorry to hear

Astyx

Meh, it's life

Ted Shifrin

Astyx, I would disagree with that statement, based on years of discussions in here.

BigSocks

10:30 PM

Some of the best suggestions come in the worst forms @TedShifrin

polite proofs

Since $\lim_{x \to a} f(x) = b$ and $\lim_{x \to b} g(x) = c$, by definition, $\forall \epsilon > 0, \exists \delta_a > 0, \exists \delta_b > 0, 0 < |x - a| < \delta_a \implies |f(x) - b| < \epsilon,$ and $0 < |x - b| < \delta_b \implies |g(x) - c| < \epsilon.$ Then since we assume $f,g$ to be real valued functions, let $f(x) = y$ for some real number $y$.

So $y \in \text{dom}(g)$, so $|x - a| < \delta_a \implies |f(x) - b| < \epsilon \implies |g(y) - b| < \delta_b \implies |g(y) - c| < \epsilon$

Hmm, this is not quite as good

Ted Shifrin

Actually, it was telling you Hartshorne was inappropriate for you that was more rude. But I'm glad you like Dino's book.

BigSocks

right, I figured

I just respect your experience

Ted Shifrin

You need $g(y)$ from the beginning, @polite.

BigSocks

and eek! an $\epsilon - \delta$ proof. excuse me while I get something to drink

Ted Shifrin

10:31 PM

You have errors in there.

And what is the flaw in your logic?

I'm going to start martinis an hour early today to celebrate Inauguration Day.

polite proofs

I suppose the implication isn't clear with $f(x) = y$

BigSocks

nice!

Ted Shifrin

How do you know $f(x)$ satisfies $0<|y-b|<\delta_b$?

But your very first sentence with $g$ should have $y$'s there ... You don't just switch at the end. The confusion has already occurred.

polite proofs

Well the problem is that I don't know if the image of $f$ contains all of $\mathbb{R}$

Ted Shifrin

Huh?

polite proofs

10:36 PM

Did I write nonsense?

Well, I am saying that $f(x) = y,$ for some real number $y$.

Hmm

Ted Shifrin

No, you are taking $y$ in the definition of limit for $g$ (which you didn't put there, as I said), and you are substituting $y=f(x)$ for some $x$ with $0<|x-a|<\delta_a$.

So we need to know that the image of $f$ is in the domain of $g$, at least for $x$ near $a$.

polite proofs

Yes, but we don't know that

Ted Shifrin

You need to have assumed it to write $g(f(x))$ at the very beginning.

But look at the particular question I asked above that I just starred.

I'm going to unstar it as soon as you answer me.

Fargle

If your $f$ is a constant function $f(x) = c$, and $g$ has a limit of $L$ at $y = c$ but isn't continuous there (say $g(c) = M \neq L$), what happens?

Ted Shifrin

I messed up, @Fargle. He was doing $g\circ f$. My fault.

polite proofs

10:40 PM

Well, $0 < |y - b| < \delta_b$ because $\lim_{y \to b} g(y) = c$

Ted Shifrin

No.

Remember, we just took $y=f(x)$.

polite proofs

But we can't do y - b

argh

Fargle

You only know $|f(x) - b| < \delta_b$ by using $\delta_b$ as your "epsilon" for $\lim_{x \to a} f(x) = b$. You don't know that $|f(x) - b| > 0$---it might equal zero. The limit definition doesn't assure that it doesn't equal zero

Ted Shifrin

You know that $|f(x)-b|<\delta_b$. You never did relate $\epsilon$ and $\delta_b$. One of the things I commented that was wrong earlier (but didn't make the list).

polite proofs

Well I kind of did that tacitly, since I set $\delta_b = \epsilon$

Fargle

10:46 PM

(That's why my leading question above points in the direction of a counterexample.)

Ted Shifrin

No, you need to set $\epsilon = \delta_b$. You need to make that explicit.

polite proofs

Well, equality does work both ways

Ted Shifrin

But @Fargle is telling you the real mistake in your argument, which both copper.hat and I were telling you earlier.

No. This is important.

polite proofs

(I seem to start a lot of sentence with well)

Ted Shifrin

You have a for all $\epsilon>0$ in your definition. You need to apply it with a particular choice, namely, $\epsilon=\delta_b$. This is super important.

Saying equality is symmetric shows your mind is muddled.

Similarly, we're setting $y=f(x)$ for our choice of $x$. This is assigning the value of $y$. Not vice versa. These are the important issues of logic you do need to work hard to understand.

Astyx

10:49 PM

That's weird to be honest. I don't get why you wouldn't want a definition that lets you do composition of limits like this

polite proofs

@TedShifrin I don't know how to work on them.

Ted Shifrin

@Astyx: Then talk about continuous functions only.

Bourbaki's definition is of continuity.

Astyx

Yes right

Ted Shifrin

This is a very hard proof, tbh, @polite. Don't you have some easier ones on limits to start with?

Or maybe they just wanted a counterexample.

Have you done some simple, concrete $\delta$-$\epsilon$ proofs with specific functions?

polite proofs

Well, it does indeed say to prove or disprove

Yes

Ted Shifrin

10:53 PM

And you understand the logic in those?

A lot of students write garbage for those, too.

polite proofs

I would like to say so

You can ask me an example if you would like

I can prove it in this chatbox

Ted Shifrin

If you email me, I can email you a brief set of notes I wrote with some examples. And there are lots more exercises there.

polite proofs

Oh, sure, but I meant that I think I can do simple examples rather easily

Ted Shifrin

Yes, so we have told you here that you need to give a counterexample. Finish it up by doing that :)

Well, getting the language and logic precise is not necessarily so easy.

I guess I can just post that pdf here, actually. I think I've done that before.

polite proofs

I actually was thinking of it now. How about $f(x) = 0$ if $x \le 0$, and $f(x) = 1$ if $x > 0$. Then if $a$ is like... $5$, then indeed the limit holds true.

Ted Shifrin

10:56 PM

I assume that can be clicked on and enlarged, saved.

Huh? What is $g$?

polite proofs

I haven't defined $g$ yet, I was just thinking about $f$ for now. We can say that $g(x) = 1$ if $x \le 0$ and $g(x) = 0$ if $x > 0$ perhaps?

Ted Shifrin

And you're looking at $\lim_{x\to 5}g(f(x))$?

Astyx

Sanity check: given K a finite field, there's a unique extension of K of degree n for every n

Ted Shifrin

raises hand for insanity

Astyx

Because $K = \Bbb F_q$, and the extension of degree n is $\Bbb F_{q^n}$

Ted Shifrin

11:01 PM

Yeah, I would have reduced to the case $q=p$ first, but yes.

Astyx

With $q = p^k$ for a prime p and a positive integer $k$

Thorgott

unique up to isomorphism, of course, yes

Astyx

Ok, cheers

polite proofs

I haven't defined $g$ yet, I was just thinking about $f$ for now. We can say that $g(x) = 1$ if $x \le 0$ and $g(x) = 0$ if $x > 0$ perhaps. Then $f(x) = 1$ as $x \to 5$, and $g(x) = 1$ as $x \to -5$, but $g(f(x)) = 0$ as $x \to 5$

Thorgott

it's the splitting field of $x^{q^n}-x$ in an algebraic closure of $\mathbb{F}_p$, whence exists and is unique up to iso

Ted Shifrin

11:02 PM

I don't follow what you typed, @polite.

polite proofs

Sorry, let me do it one message.

Ted Shifrin

Isn't $g(y) = 0$ whenever $y$ is near $1$?

Astyx

That implies there's a unique non-ramified extension of degree n of a given local field

Anyway, bedtime for me

good day/night to you

Ted Shifrin

Bonne nuit.

polite proofs

Oh wow I think I have it finally. Let $f(x) = -x, g(x) = -1$ for $x < 0$, $g(x) = 0$ for $x \ge 0$. Then $g(x) = 0$ as $x \to 5$, but $f(x) = -5$ as $x \to 5$. Therefore, $g(f(x)) = -1$ as $x \to 5$

Thorgott

11:08 PM

gn

Balarka Sen

Hi all

Ted Shifrin

@Polite: You have confused yourself for exactly the reason I wanted you to use different letters. Write $g$ with $y$, as I've said 20 times. We don't care about $g(x)$ for $x$ near $5$.

Hi, a @Balarka.

copper.hat

morafterevening.

polite proofs

Wait, why don't we?

Fargle

Yeah, you need $y \to b$---i.e. $y \to \lim_{x \to a} f(x)$.

Ted Shifrin

11:10 PM

I think it's middle of Balarka's night/morning.

Yes, precisely. And that is $-5$, not $5$.

polite proofs

So isn't that a counter-example?

Ted Shifrin

No, most definitely not.

Go off and write things on paper and think carefully.

Fargle

No, because $g(y) \to -1$ as $y \to -5 = \lim_{x \to 5} f(x)$.

copper.hat

it might help to draw a little diagram @polite. formalising understanding is easier than understanding formalism.

obviously missing my daily commute

Ted Shifrin

Draw graphs. Draw an $xy$-graph for $f$, and then a graph for $g$ with its domain labeled $y$.

Fargle

11:13 PM

i.e. if you let $f$'s variable approach $5$, you need to let $g$'s variable approach $f$'s limit at $5$.

Ted Shifrin

Just go ride your bike in circles (or ellipses will do), copper.

Fargle

I was going to say one can easily bike on any conic section, but I guess that's a bit hyperbolic.

copper.hat

good idea :-). will go for my run approximation.

Ted Shifrin

Ah, somebody flipped Fargle's switch.

copper.hat

mine will be more like a hypobolic.

polite proofs

11:17 PM

sketchtoy.com/69470222

Not to scale

The piecewise function is $g$

And the appearing to be continuous function is $f$

Ted Shifrin

Yeah, this is not a counterexample.

Graph $g(f(x))$.

Balarka Sen

It might be because I just woke up but I'm getting very confused by something

Ted Shifrin

Are you going to confuzle me?

polite proofs

$f(x) = \lfloor x \rfloor $, $g(x) = \lceil x \rceil $, then $g(f(x))$ has the graph of $f$

Ted Shifrin

I haven't checked that yet. What is $a$ and what is $b$?

Yep

11:26 PM

I would like to answer questions on this site but alot of the time topics posted are well beyond me and i dont know if it helps to just find old easy questions to answer, does

well not easy but things i can do, does that help?

Ted Shifrin

Answering old questions that have good answers isn't a good idea. If they haven't got an accepted answer and you can do better than what's there, it's OK.

Yep

@TedShifrin right ill be sure to get on it :)

polite proofs

I've never been this frustrated

Balarka Sen

@TedShifrin Maybe I can unconfuzzle me in the process of explaining what I want to do. I have a manifold with boundary $(M, \partial M)$ and a smooth function $f : M \to \Bbb R$ such that $\ker df_x$ is transverse to $T_x \partial M$ for all $x \in \partial M$

Ted Shifrin

This isn't a question of $\delta$s and $\epsilon$s. @polite Here's a huge hint. Suppose $f(x)=b$ is a constant function.

OK, so $\partial M$ is nowhere tangent to a level set of $f$.

Balarka Sen

11:34 PM

Right. What I want to do is "tug" $\partial M$ inwards along the level sets of $f$

Ted Shifrin

OK, that makes sense to me.

Not unique unless the level sets are curves.

Balarka Sen

Agree, not unique. I just need to choose a vector field in a small collar around $\partial M$ which is tangential to the level sets of $f$ and points in the $I$-direction of the collar, inwards.

Namely, a collar is of the form $\partial M \times [0, 1)$, consider projection to the $[0, 1)$ factor. Restrict to each $f^{-1}(c)$, and take gradient.

Ted Shifrin

Why take gradient?

Oh, you want a vector field.

Balarka Sen

Yeah, to flow along that realizes the tug

Ted Shifrin

Maybe that's actually smooth.

Balarka Sen

11:37 PM

should be!

Hm, I don't mean, gradient. I mean $f|_{f^{-1}(c)} : f^{-1}(c) \to [0, 1)$ is also a submersion, so you can lift $\partial/\partial t$ along this

Ted Shifrin

By compactness, you can choose $1$ small enough to make sure the projection is never dying.

Balarka Sen

Yeah our pictures are matching

It's insane how fast topologists/geometers communicate! Imagine telling an algebraist "Choose $1$ small enough"

Ted Shifrin

Right, you want to project the normal vector to the level surfaces.

Well, I did that for fun. :)

So you are doing an intrinsic gradient on the level surfaces.

Balarka Sen

Yeah.

We can cook up some appropriate Riemannian metric that respects the foliation and write $X$ down as such an intrinsic gradient. It's doable, for sure.

Ted Shifrin

I'm fine with projection of $\partial/\partial t$ onto the tangent space :)

Balarka Sen

11:43 PM

Cool!

So if $X$ is this vector field I want to consider $f_t : M \to \Bbb R$, $f_t := f \circ \Phi_X^{-t}$

Ted Shifrin

Oh, I'm not done yet?

Balarka Sen

That's the last step of the construction. I'm getting a little weirded out by how $f_t$ behaves on $\partial M$ but I suppose if you agreed with my picture so far there's no issue in what I want to do

Ted Shifrin

So locally just linearize everything @Balarka. I don't see a problem from my picture.

Balarka Sen

What's a little strange is $f_t(x) = (f \circ \Phi^{-t}_X)(x) = f(\Phi^{-t}_X(x)) = f(x)$ because flow of $X$ preserves the level sets of $f$

I guess $f_t = f$ then

Ted Shifrin

Yes, since you constructed it to move along level sets, this isn't shocking.

Balarka Sen

11:50 PM

Yeah fair enough

Hm

BigSocks

ok so we got 2 lemmas

Lemma $1$: Let $f \in k[x,y]$ be any non-constant polynomial. Then the ideal of $k[x,y]$ generated by $f$ is not maximal.

Lemma $2$: Let $A$ be any factorial domain with field of fractions $K$. Let $M$ be a maximal ideal of $A[y]$ that is not principal. Then $M \cap A \neq (0)$.

Now we got a proposition

Prop: Let $M$ be a maximal ideal of $k[x,y]$. Then $\exists (a,b) \in \Bbb A^2(\overline{k}$ s.t. $M$ is generated by $(x-a)$ and $(y-b)$

This guy starts out with the old "Let $P := M \cap \overline{k}[x]$. Lemmas $1$ and $2$ show that $P$ is a non-zero prime ideal of $\overline{k}[x]$ "

-so far so good-

"Then $P = (x - a)$ for some $a \in \overline{k}$".

but why though. how are we so sure it doesn't have some constant lurking about?

bear in mind this is in hopes of showing (or equivalent to showing) weak Nullstellensatz, so I don't think we can use that

Prop should have $\Bbb A^2(\overline{k})$,

missed the parenthesis

polite proofs

Set $f(x) = \pi, g$ to floor(x). Then a = 0.5, where f will approach $\pi = b$. g will approach $3$ as x approaches $\pi$, and g(f(x)) will approach $0$ as x approaches 0.5

Good bye..

Thorgott

how is $M$ gonna be generated by elements that aren't even necessarily contained in it

BigSocks

I mean $P$

I guess you are implying it, bc what is in $P$ is in $M$?

Fargle

@politeproofs $g(f(x)) = g(\pi) = 3$, so $g(f(x)) \to 3$ as $x \to 0.5$. Another hint: your $f$ is fine, but you need $g$ to somehow be discontinuous at $\pi$.

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$Mathematics$ Mathematics