Mathematics - 2023-10-27

Obliv

03:48

I have a question about an existence proof, There is a set $X$ for which $\mathbb{N} \in X$ and $\mathbb{N} \subseteq X$

I'm wondering whether I can define $X = \{\mathbb{N}\}\bigcup \mathbb{N}$

Idk that looks weird

I guess would you say $\mathbb{N} \in \mathbb{R}$?

I mean I'd say $\mathbb{N}$ intersects R for sure, but I wouldn't say it's an element of.

PSA: read problems carefully before spending 30 minutes trying to prove in general $a \equiv b (\mod p)$ for some $a\in \mathbb{Z}$ and prime $p$ implies $a^3 \equiv a (\mod p)$ because it's supposed to be $a^p \equiv a (\mod p)$

I'm not even sure if I proved it right anyway, like $a \equiv b (\mod p) \implies a-b = pq$ for some $q \in \mathbb{Z}$ then $a = pq + b$ , raise to power p to get $a^p = (pq+b)^p$ rewrite in binomial theorem expansion $a^p - (pq)^{(p-1)}b = \sum_i^p \binom{p}{i}(np)^{p-i}b^i - (np)^{p-1}b$

and claim $(pq)^{(p-1)}b \equiv b (\mod p)$ means $a^p - (pq)^{(p-q)}b \implies a^p \equiv b (\mod p)$

That was a last ditch effort to finish this proof on time so it's likely gibberish

Thomas Finley

05:24

When a question says, "Show that the limit of the function does not exist at a point (say,) c." Does it mean that the question asks us to show that the limit cannot be real number "only"? Or do they also want us to show that, the limit can neither be $\pm\infty$ ?

@leslietownes , @TedShifrin Can you please help with this?

Alessandro Codenotti

06:51

@Obliv sure

PNDas

07:54

Hi! We have $\rho<r$. The book said there exists $\eta\in C_c^{\infty}(B_1)$ such that $\eta=1$ in $B_{\rho}$, $\eta=0$ in $B_r^c$ and "$|\nabla \eta|<\frac{2}{r-\rho}$".

Partition of unity gives a function which satisfies the first two requirements but I don't know how do we get the last condition.

Thomas Finley

10:10

0

Q: Prove that $\lim_{x\to\frac {\pi}{2}+}\tan x=-\infty.$

Prove that $\lim_{x\to\frac {\pi}{2}-}\tan x=\infty$ and $\lim_{x\to\frac{\pi}{2}+}\tan x=-\infty.$ The domain of $\tan x$ is $A=\Bbb R\setminus\{2n\pi \pm \frac{\pi}{2}: n\in\Bbb Z\}.$ In order to prove $\lim_{x\to\frac {\pi}{2}-}\tan x=\infty$ I need to show that for any $M\in\Bbb R$ we have a,...

real-analysis

Need help with this one :/

@robjohn I need your calculus skills to solve this :)

@leslietownes Can you help me with this, please?

Jakobian

11:12

Both approaches that Martin mentions in the comment and answer work

the same for $\pi/2^-$ as for $\pi/2^+$

Thomas Finley

11:40

But I am interested in proving this, in similar lines of the proof that I did on OP.

Mad

13:18

https://math.stackexchange.com/questions/2493197/identity-operator-isnt-bounded
so i understand the proof, what is confusing me is, that norms are continious, so if i am taking the identity, wouldnt i get a continious functional, thus bounded.
ie : $ \lVert x \rVert_1 =\lim_n \lVert x_n \rVert_1 = \lim_n \lVert Id x_n \rVert_\infty $ $

i am trying to prove the statement that all norms are equivalent, by constructing a continious operator, that is bounded between two normed vector spaces R^n

PNDas

@Mad That statement actually proves that in $(C[0,1],||\cdot||_{\infty})$ the map $x\mapsto ||x||_1$ is continuous but in $(C[0,1],||\cdot||_{1})$ the map $x\mapsto ||x||_{\infty}$ is not continuous.

Sine of the Time

@Mad the identity is not always continuous

Jakobian

13:40

@SineoftheTime huh

Sine of the Time

as an operator

Jakobian

I suppose you mean, with different topologies but with the same underlying set

?

Sine of the Time

yes, with different norms

Jakobian

Yeah, sorry for the confusion, I have Mad ignored because he's been a prick to me in the past

Soumik Mukherjee

Should it be called identity if the underlying structure is different?

Mad

13:43

@PNDas @SineoftheTime no i understand what you mean, i understoo the proof, but i dont know "why" what i wrote is not right

Jakobian

I guess its still the identity function

just not the identity morphism

Sine of the Time

@Jakobian if you ignore a user, you don't see his messages?

@Mad what are you referring to?

Jakobian

@SineoftheTime yeah

@SoumikMukherjee so I'd say its fine to call it the identity... but it was something I was wondering myself

Mad

@SineoftheTime well, look what i wrote, what is the mistake in it?

Sine of the Time

the part $\|x\|_1=\dots$?

Mad

13:53

yes

Sine of the Time

Mad

that does not really answer my question, but yes you are right.

Xander Henderson

@Jakobian I very much disagree. The identity function on a space / set / whatever $X$ is a function $\operatorname{id} : X \to X : x \mapsto x$, where the domain and codomain are identically equal (e.g. isomorphic).

If the underlying structure is different, it isn't the identity function, since the domain and codomain are not the same space.

I suppose that you could define things differently, but I, personally, don't like it.

Jakobian

Its not really a thing to disagree on

The ambiguity comes from the fact that the word "identity function" doesn't provide enough data

it boils down to how we understand things

and on what we agree on

Xander Henderson

@Jakobian That was the point I just made. You can define things however you like. I dislike the notion that the identity "function" is somehow different from the identity "morphism", because in the world I work in, we don't talk about morphisms.

Jakobian

14:01

I see how what you're saying is valid, but also, I'd just prefer something more precise. Identity function as just the function between underlying sets, is passable to me

Xander Henderson

The language in mathematics is not universal, and we define terms in whatever way is useful in a particular context.

But I have never heard an operator theory person make a distinction between the identity as a function vs morphism. Maybe some Lie algebra folk do this?

Jakobian

Yeah thats not okay

Xander Henderson

Who are you to pass judgement?

I'm talking about the way people actually use language.

Jakobian

I don't need to be anyone in particular to pass judgement

Xander Henderson

Competent users of mathematical English, in certain contexts, don't distinguish between these things...

Jakobian

14:05

If you have "identity operator", this now needs to be from the same space to itself, since thats what operators are

Xander Henderson

Comments like "Yeah thats not okay" make it sound like you have appointed yourself the arbiter of Good Math™. It's kind of arrogant, don't you think? The point I made is that I, personally, don't like the idea of an identity function between spaces which are not identical. This was an expression of taste and opinion, in contrast to your statement of opinion, i.e. "I guess its still the identity function".

As long as we are expressing opinions, I am happy to disagree (we should all be happy to disagree).

Jakobian

I'd see it as really awkward to have it otherwise

Xander Henderson

But when you go on to say "Thats not okay", I feel like you are crossing a line, and telling other people what they should or should not do.

Jakobian

I am telling them that

You shouldn't say "identity operator" for identity function between vector spaces with different norms

Xander Henderson

I am telling you that my perception of what you have said is arrogant and judgmental. If that is not the way you wanted to come across, you might consider rephrasing.

@Jakobian I wouldn't say "identity function" for such an operator.

Jakobian

14:11

well anyway

I think it'd be interesting to have some kind of characterization of subsets of $\mathbb{R}^n$ homeomorphic to a regular closed set

I think it might just be as simple as $\overline{\text{int}(A)} = \overline{A}$

Lemon

Why are formal linear combinations written as $f = \sum f(x_i)x_i$ instead of $f = \sum f(x_i) \delta_{x_i}$? The $x_i$s are elements of an arbitrary set $X$, where the space of formal linear combination is a set of maps.

Jakobian

@Lemon Think of it as the free module generated by $x_i$

I'm out of context but I think this is what's happening here, given you said formal linear combinations

PNDas

14:30

Any help for chat.stackexchange.com/transcript/message/64636514#64636514 will be appreciated.

Lemon

@Jakobian I think your construction is a formal linear combination of a module (which has addition already). I am asking about formal linear combinations of any set, where "+_S" may not have any meaning. Most of the sources I am reading define it like this, but I just found one link that basically says if $ \sum f(x_i)x_i$ is not already value I want, then we force it to be $\sum f(x_i) \delta_{x_i}$

Jakobian

@Lemon No I don't mean that, I mean a free module

Say you have polynomials, $\mathbb{R}[x]$

you don't wonder what $x$ is, right?

it comes out of nowhere... and same with other variables you could add

here its the same

the set of polynomials has a construction, but we know it exists, but its not predefined

Lemon

You are right, I never actually questioned what is "addition" doing in the polynoimal rings. It just bothered me to see $f = \sum f(x_i)x_i$ and if i symbolically write $f(x_i) = (\sum f(x_i)x_i)(x_i)$ just did not make sense to me unless we really mean to write $\sum f(x_i) \delta_{x_i}$ since the free vector space is a space of functions.

Jakobian

You could construct it as a space of functions, but you don't want to think about it this way

Lemon

it here being the space of formal sums?

Jakobian

14:43

You might think of $f(x_i)$ as the coefficient next to $x_i$

@Lemon yes, which is the free module

When you construct polynomials, you also don't want to think of them as, say, infinite sequences which are eventually zero

you want to think of them as polynomials

You don't want to think of $x$ as $(0, 1, 0, 0, ...)$

and of $x^2+x+1$ as $(1, 1, 1, 0, 0, ...)$

Thomas Finley

2

Q: Let $V$ be a vector space over $C$ with dimension $n.$ Prove that if $V$ is now regarded as a vector space over $R,$ then $\dim V =2n.$

Let $V$ be a finite-dimensional vector space over $C$ with dimension $n.$ Prove that if $V$ is now regarded as a vector space over $R,$ then $\dim V =2n.$ I tried solving the problem as follows: Given, $\dim V =n$ when $V$ is vector space over $C$ and so, we assume $b_1,b_2,...,b_n$ to be a basis...

linear-algebra solution-verification vector-spaces

Need help with this :/

Jakobian

You got two answers though

Lemon

Should I not even bother thinking about expressions like $f(x_i) = (\sum f(x_i)x_i)(x_i)$? Because thinking about the free vector space as a space of function makes more symbollically sense than the polynoimal ring example.

Jakobian

15:01

You decide what to think about

I don't think its important as long as you understand it

Jakobian

15:28

@AlessandroCodenotti do you see any easy way to show that "open disk with a beard" $\{z\in \mathbb{C} : |z| < 1 \text{ or }|z| = 1\text{ and }\text{Im}(z)\leq 0\}$ isn't homeomorphic to a closed subset of $\mathbb{R}^2$?

such set would need to have $[0, 1]$ as its boundary

Thorgott

open disk with a beard is a funny name

can't you argue as follows? the open disk with a beard has a compact subset s.t. any neighborhood thereof contains sequences without convergent subsequences. this property is preserved by homeomorphism, but can't hold for a closed subset of euclidean space.

Jakobian

Yeah I think that would do it

thanks

Jakobian

16:19

@AlessandroCodenotti Suppose that $A\subseteq\mathbb{R}^n$, $\overline{\text{int}(A)} = \overline{A}$ and $A$ is homeomorphic to a closed subset of $\mathbb{R}^n$. Is $A$ homeomorphic to a regular closed subset of $\mathbb{R}^n$?

the answer should be yes

Jakobian

16:42

Apparently, the subsets of $A\subseteq \mathbb{R}^n$ homeomorphic to a regular closed subset of $\mathbb{R}^n$ are precisely those sets homeomorphic to a closed subset of $\mathbb{R}^n$ for which $\overline{\text{int}(A)} = \overline{A}$.

I consider this problem pretty much fully explored now, since "homeomorphic to closed subset" cannot really be simplified, as far as I can see

Sine of the Time

18:22

can someone help me with this exercise on Cauchy's integral formula? The integral is : $\oint_{|z-1|=3}\frac{dz}{z(z^2-4)e^z}$. The problem is that the singularity $z=-2$ is on the boundary of the curve. How to proceed?

Ted Shifrin

@Sine That's generally considered a mistake in the problem. The integral does not exist, although one can talk about a Cauchy principal value.

Sine of the Time

so it doesn't make sense computing the integral?

Ted Shifrin

What did I just say?

Sine of the Time

I don't know what is Cauchy principal value

Ted Shifrin

Ask your professor if this is a mistake in the problem.

Sine of the Time

18:30

ok thank you

I have also a doubt about Laurent series: I have to write the series for the function $f(z)=\frac{1}{(z-1)(z-2)}$ for $|z|<1$. I used partial fractions but I don't know how to treat $1/(z-2)$

Ted Shifrin

Make it look like $1/(1-u)$.

Sine of the Time

yes, in this case I divide by $2$ and the formula works if $|z/2|<1$, namely $|z|<2$.

Ted Shifrin

Which means it holds in particular when $|z|<1$

Sine of the Time

yes, in this case is true

Ted Shifrin

18:45

That will always be the case if there isn't another singularity inside $|z|=1$.

Sine of the Time

ah ok makes sense

but for example at infinity it seems like I'm adding hypothesis on $|z|$

Ted Shifrin

Whoa. That was a big leap. What are we talking about now?

Sine of the Time

let me make a concrete example

$f(z)=\frac{1}{(z+1)(z+3)}$ find the expansion for $|z|>3$

so since $|z|>3$ I can use the formula for $|z|>1$

Ted Shifrin

Ah, OK. So it's going to be the same idea. You could have used your same function as before, and asked about $1<|z|<2$ and $|z|>2$.

Sine of the Time

makes sense, it's a typo in the book

thank you, now it's clearer

Ted Shifrin

18:50

It's just basic logic every time, along with the geometric series.

It's just that when $|z|>a$, you have to factor out the $z$ instead of the constant. Right?

Sine of the Time

I factor so I have something < 1 and then I apply the formula for the geometric series

Ted Shifrin

Right.

Jakobian

19:29

When considering tagged partitions of $[a, b]$, why can we assume that no tag, except possibly $a,b$, is an endpoint?

Ted Shifrin

What is a tagged partition and why?

Jakobian

Like for Riemann integrals, partitions with choice of points

Ted Shifrin

So your question makes no sense if you wish to use all the left-hand endpoints to do your Riemann sum.

So "why can we assume" needs some context.

Jakobian

I think by "we can assume" they mean the Riemann sums are the same

Ted Shifrin

Clearly not.

Jakobian

19:38

I'm sure they do mean that because they're talking about how if the tag is in the interior of the subinterval you can divide the subinterval so that this tag is an endpoint of both subintervals, without changing the Riemann sum

Maybe its an error in the book?

Ted Shifrin

It sounds like they want to change the partition to assume that the tag is an endpoint.

You're asking a question out of context, so it really makes no sense to me.

Jakobian

No, that's not it. They're talking about three convenient assumptions about tag partitions, one of them is that every tag is an endpoint, but the other one is that no tag is an endpoint except for maybe $a$ and $b$

There's also third one, that no tag is an endpoint of two distinct subintervals

The other two make sense to me, but I'm not sure how to show this one

Ted Shifrin

This continues to make no sense to me. What precisely are they asserting?

Jakobian

That with dealing with tagged partitions we can assume that ...

Its the phrasing they used

Ted Shifrin

We can assume that .... in order to assure what?

Jakobian

19:47

This is after showing that we can make a tag in interior an endpoint, and conversely, if a tag is endpoint of two subintervals then we can make it an interior point, without changing the Riemann sum

Ted Shifrin

So we are allowed to make a totally different partition?

Jakobian

It seems so, yes

As long as the Riemann sums stay the same

But I suppose this has to work for any function $f$

Ted Shifrin

I go back to my left-endpoint Riemann sum (which every calculus student in the US ends up having to compute at least once).

I don't see what they expect to do with that.

I can make the tag point an interior point by changing the partition, but the Riemann sum will change.

Jakobian

Yeah, I don't know

amazon.com/Modern-Integration-Graduate-Studies-Mathematics/dp/…

its this one

Ted Shifrin

That doesn't help. Bartle is a well-respected author, so I suspect what he actually says is correct.

Jakobian

20:10

Ted Shifrin

21:15

Clearly he has established i and iii, and ii is wrong.

sunny

Consider $$f(x,y)=\begin{cases} \sin(x-y)& \text{if } 0<x-y<2\pi \\ 0 & \text{else}.\end{cases} $$ It is claimed this function is continuous on $\mathbb R^2$. What are some ways in which I can verify this? I'm not sure where to begin.

user123234

can maybe someone help me here: math.stackexchange.com/questions/4795387/…

Ted Shifrin

@sunny What is the boundary of the region where it’s nonzero?

Jakobian

@sunny use gluing lemma

$f$ is continuous on $0\leq x-y\leq 2\pi$ as well as outside of $0 < x-y < 2\pi$, both are closed sets whose union is $\mathbb{R}^2$

sunny

21:31

interesting

Jakobian

this is because on first set this is $\sin(x-y)$ and on the second its $0$

Ted Shifrin

Sunny needs to answer my question.

shrug

sunny

I'm thinking, Herr Shifrin.

^ that's German.

Lukas Heger

Herr Shifrin kann deutsch...

Ted Shifrin

I know what Herr is, thanks. What are you thinking about?

Das stimmt. Er kann.

sunny

21:38

:)

leslie townes

bonjour, M. Shifrin (this is spanish)

2

Ted Shifrin

Hasta la vista, señor.

Lukas Heger

I think he prefers to be called Monsieur Approche Géométrique

Ted Shifrin

LOL

leslie townes

his name on the street is "Dessins D'Enfants"

Ted Shifrin

21:45

Sunny should have had an answer in 10 seconds. Now it’s 15 minutes.

sunny

I'm a slow reader.

Ted Shifrin

Reader of what?

Jakobian

Text

Xander Henderson

22:04

Ugh... I have a student who specifically made an appointment to meet me for office hours this afternoon. I am here for him, on a Friday, when I would rather be going home, and HE ISN'T HERE. :(

FINALLY showed up.

Just as I was about to go.

Ted Shifrin

Aww. Be nice.

Jakobian

I think my diarrhea finally subsided. I've been through things today

actually never mind

Xander Henderson

22:52

@Jakobian T

@Jakobian M

@Jakobian I

Jakobian

sorry

leslie townes

just got back from the bathroom, i feel about six pounds lighter. what a relief.

i was putting 3L of hand soap refills from costco under the sink. should have mentioned that first

Xander Henderson

X(

leslie townes

what's new with you?

Xander Henderson

Nothing much. I'm going home.

It is time for the week to be over.

Ted Shifrin

23:09

Nighty night.

Transcript for

$Mathematics$ Mathematics