Mathematics - 2025-01-04

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leslie townes

01:07

ilikemathematics if you choose any basis for a vector space (such as V^*) and any function from that basis to another vector space (such as V^double star) then there is a unique linear map that extends that function to all of the vector space

so if you just want some random map it's not interesting

ILikeMathematics

Ah, you're right

Thanks

leslie townes

there's probably questions on main or MO about what it means to define a 'natural' map from one thing into another, there is always something from V to V star star but not into V star

although formalizing and proving that kind of stuff is not something that a healthy person should care anything about

1 hour later…

handan_toddler

02:41

How are the munchkins getting along? @leslietownes

leslie townes

02:56

every day a little bit better!

handan_toddler

03:13

cool

nickbros123

04:00

@leslietownes I love the double dual

Xander Henderson

04:24

@nickbros123 It doesn't love you, and it asked me to tell you to stop sending love notes and flowers.

It's creepy, dude.

handan_toddler

>_>

Bob

04:49

good evening

I would like to know why I cannot reopen this question:

3

Q: Solving a system of two differential equations with no initial conditions given

Problem: Use the operator method described in this section to find the general solution of the following system of differential equations. \begin{align*} x' + y' - 2x - 4y &= e^t \\ x' + y' - y &= e^{4t} \end{align*} Answer: \begin{align*} Dx + Dy - 2x - 4y &= e^t \\ (D-2)x + (D-4)y &= e^t \\...

ordinary-differential-equations

nickbros123

05:26

@XanderHenderson I thought we had something special, you mean to say when it gave me the canonical basis to V from V*, it was just being nice? come on Xander

handan_toddler

<_<

2 hours later…

one potato two potato

07:18

peep

handan_toddler

08:03

a boo

Dannyu NDos

08:46

I've just realized that the real line admits multiple different complete uniform structures.

In particular, the fine uniformity on $\mathbb{R}$ effectively identifies the hyperbolic cosine function as uniformly continuous.

psie

09:40

> Let $\mu$ be a finite measure on $(\mathbb R,\mathcal{B}(\mathbb R))$. For every $x\in\mathbb R$, let $$F_\mu(x)=\mu((-\infty,x]).$$ The function $F_\mu$ is increasing, bounded, right-continuous and such that $F_\mu(-\infty)=0$.

I struggle with a small detail of the right continuous part. Suppose now that $x\in\mathbb R$ and that $(x_n)$ is a decreasing sequence with limit $x$ (e.g. $x_n=x+1/n$). Then $(-\infty,x]=\bigcap_{n=1}^\infty (-\infty, x_n]$, and so by continuity from above we have $F_\mu(x)=\lim_{n\to\infty} F_\mu(x_n)$. Thus $F$ is right continuous, since for any $x<y<x_n$, we have $|F(y)-F(x)|\leq|F(x_n)-F(x)|$.

Question The inequality $|F(y)-F(x)|\leq|F(x_n)-F(x)|$ shows that $|F(y)-F(x)|<\epsilon$ for $n>N$, right? But what definition of continuity is this satisfying? I'm confused about the $n>N$ part, since there's nothing in $|F(y)-F(x)|$ that depends on $n$.

The argument "Thus $F$ is right continuous, since for any $x<y<x_n$, we have $|F(y)-F(x)|\leq|F(x_n)-F(x)|$." appears in Cohn's book, and I'm paraphrasing from there.

Dannyu NDos

Actually, the act of choosing $y$ is dependent to $n$, for $x_n$ acts as an upper bound.

psie

true :)

hmm, the whole argument, by writing $|F(y)-F(x)|\leq|F(x_n)-F(x)|$ for $x<y<x_n$ seems unnecessary

Dannyu NDos

This utilizes the sandwich theorem.

psie

If one just left it at $|F(x_n)-F(x)|<\epsilon$ for $n>N$, then at least this would be the sequential definition of continuity, but I'm puzzled at what introducing $y$ and writing $|F(y)-F(x)|\leq|F(x_n)-F(x)|$ brings to the table.

Dannyu NDos

The $y$ must be involved because the sequence $x_n$ doesn't cover a right-open interval $[x, x + \delta)$.

Though, yeah, the sequential continuity coincides with the (usual) continuity when the space involved is metrizable.

Yet still, this fact cannot be used for one-sided continuity.

psie

09:56

@DannyuNDos what do you mean by $x_n$ doesn't cover a right open interval $[x,x+\delta)$? How does introducing $y$ help us cover $[x,x+\delta)$?

Dannyu NDos

By "doesn't cover", I mean there does not exist a $\delta > 0$ such that $[x, x+\delta) \subset \{x_n\}_{n=1}^\infty$.

So we need $y$ to, intuitively speaking, "fill the gaps" of the sequence $x_n$.

By introducing $y$, we're basically turning the sequence $x_n$ of points into the sequence $[x, x_n)$ of intervals.

psie

ok 👍

2 hours later…

Thomas Finley

11:47

user image

user image

I don't get how do they claim that, "Each $P_i$ is isomorphic to $P^1$ ..." (highlighted part in the last image). Can someone please help me with it?

Vladimir Lysikov

12:07

The conjugation map $x \mapsto \sigma^{-i} x \sigma^i$ is an isomorphism from $P_1$ to $P_i$

Sine of the Time

12:31

libgen is not working?

Vladimir Lysikov

12:42

Looks like it

Sine of the Time

thanks

nickbros123

13:08

has anyone read the book "Introduction to probability models" by Sheldon Ross?

would you recommend?

Xander Henderson

13:28

@nickbros123 yes.

@nickbros123 for what?

nickbros123

I have a course on probability and stochastic processes

course outline seems rather application based rather than measure theoretic based

Thomas Finley

14:14

@VladimirLysikov Fine... I get it. But can you explain the line that follows after the highlighted text? I think there might be confusion with the phrasing tho.

leslie townes

to say that a permutation f of integers "influences" an integer k is presumably to say that f(k) is not k

there is an implicit claim that if f and g are permutations and they "influence" disjoint sets of integers, then they commute

the compositions fg and gf will both be "do f on the set that f influences, and do g on the set that g influences." you can find versions of this on main and on the web

proofwiki.org/wiki/Definition:Disjoint_Permutations

Xander Henderson

@nickbros123 Is that the book assigned?

psie

15:02

Consider an increasing, right continuous and bounded function $F:\mathbb R\to\mathbb R_+$ such that $F(-\infty)=0$. Then this function gives rise to a unique finite measure $\mu$ on $(\mathbb R,\mathcal B(\mathbb R))$ such that $F(x)=\mu((-\infty,x])$. Now, my book then goes on to define the Lebesgue-Stieltjes integral for every bounded Borel measurable function $f:\mathbb R\to\mathbb R$ $$\int f(x)\,\mathrm{d}F(x):=\int f(x)\,\mu(\mathrm{d}x).$$

I wonder, why only for bounded Borel measurable functions?

Or put differently, why do we require boundedness?

leslie townes

perhaps to avoid any issues around "convergence"

psie

yeah

leslie townes

i wouldn't assume that any hypotheses in a treatment like this are necessary in any strict sense, it's just, there's a whole world of math around potentially divergent integrals and if that's not the subject of today's lecture then you maybe add hypotheses to avoid it

psie

right 👍

I just thought the word "bounded Borel measurable function" popped up quite suddenly in the sequences of definitions and theorems in the book, but as you say, it's probably just to reserve the integral for finite things (for the moment being), which we've discussed before :)

leslie townes

this is something that very roughly separates introductory algebra from introductory analysis, if there is a hypothesis in an intro algebra book it is more likely to be there for some "real" reason other than convenience

whereas almost everything in analysis is convenience oriented and the vast majority of people who somehow care about what hypotheses are both necessary and sufficient for the FTC to hold aren't analysts

as one randomly chosen example

PDE people are more than happy to prove some inequality with like 10,000 more hypotheses than necessary to deal with one equation and you won't learn anything useful by asking why they didn't assume some weaker thing

psie

15:14

:D

Thomas Finley

15:43

@leslietownes Thanks for the clarification!

nickbros123

@XanderHenderson it's in the list of recommendations in the course catalogue, yes

ILikeMathematics

How do you justify that $\begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix}1 \\ 0 \end{pmatrix} + \begin{pmatrix}0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}$ can't be given by $x \otimes y$ in $\mathbb R^2 \otimes \mathbb R^2$? We know that a basis of it is equal to

$$B = \left\{\begin{pmatrix}1 \\ 0\end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}1 \\ 0\end{pmatrix} \otimes \begin{pmatrix}0 \\ 1\end{pmatrix}, \begin{pmatrix}0 \\ 1\end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1\end{pmatrix} \otimes \begin{pmatrix}0 \\ 1\end{pmatrix}\right\}$$

I mean, we probably can't say "you just see it from the basis"

I guess you can say it's because of the distributivity, we have $e_1$ as first vector once and $e_2$ afterwards. You can't obtain that with $x \otimes y$. Is there a different argument too?

Nevermind, I should've picked something like $e_1 \otimes e_2 + e_2 \otimes e_1$; like this, it's obviously expressible by $x \otimes y$.

psie

16:38

Reading on Wikipedia about support of a function. I always forget what is what here, but if $f:X\to\mathbb R$ (where $X$ is some topological space) is continuous, then everything simplifies it seems. The set-theoretic support of a function is then simply the open set $f^{-1}(\{0\}^c)$ and the support is the closure (in $X$) of this open set. (Correct me if I'm wrong.)

I'm trying to figure out, if $f:X\to\mathbb R$ is continuous and $X$ is a locally compact metric space, and $f$ is compactly supported, if then it vanishes outside a compact set? But I find this difficult, since the support of the function doesn't seem to be strictly speaking the set on which it is nonzero.

Jakobian

16:59

@psie I don't know what "set-theoretic support" is, but yes, support of a function is defined as the closure of the set of points $x$ for which $f(x)\neq 0$

@psie that $f$ is compactly supported means that $\text{supp}(f)$ is compact

By definition, $f$ vanishes outside of $\text{supp}(f)$, and so if its compactly supported, then it vanishes outside of a compact set

and conversely, if $f$ vanishes outside of a compact set $K$, then the set of the points for which $f$ is non-zero is contained in $K$, so support of $f$ is contained in $K$, so support is compact

psie

@Jakobian ok, makes sense, but it is possible for $f(x)=0$ for $x\in \text{supp}(f)$, or?

Jakobian

yes

it's possible

If the set of points for which $f$ is non-zero is not closed, and it most often isn't, then there will exist some $x\in \text{supp}(f)$ with $f(x) = 0$

psie

ok 👍

2 hours later…

Anthony

19:00

Hello, I was trying to solve the following exercise :

Let M be a compact manifold of *positive dimension*, and let p \in M . Show
that M is homeomorphic to the one-point compactification of M\{p}

I think I have solved the exercise but at not point I had to use the hypothesis that M has *positive dimension*. So I tried to study the 0-dimensional case in particular and unless I am mistaken in that case, since M is compact and discrete, it is a finite set with the discrete topology, and in that case, I think the one point compactification of M\{p} is also just a finite space with the discret…

leslie townes

19:15

anthony all of that seems right, i would expect that the hypothesis is simply to exclude a category of spaces that don't resemble what normal people think of as manifolds

i'm not 100% sure of whether people would use 'compactification' to describe what is happening in the case where you have a finite set of points (i.e. a space that is already compact) and add another one

maybe they are also avoiding the case for that reason

"manifold of positive dimension" is just a bizarre way of saying "manifold" to me, i wouldn't think of the 0 dimensional case as having anything to do with what i think of manifolds as being

but there's no shortage of bizarre people in mathematics

Sine of the Time

@leslietownes the concept of "manifold of positive dimension" was introduced by Ted

Anthony

19:41

@leslietownes Thank you for your help, you are probably right when it comes to the reason why the additional hypothesis was included in the exercise. Your confirmation is enough for me to consider the exercise solved and to move on to another one.

Just for the sake of completeness, the definition of the one-point compactification of a space X (assumed to be locally compact and Hausdorff) that the author uses is to add a new point called \infty to X, and then define the open subsets of this new space X* = X \cup \infty to be the open subsets of X, and the subsets U such that X*\U are compact subsets of X.

leslie townes

yeah, that seems fine to me, i just don't know why anybody would consider a compactification of a space that is already compact, and maybe that is where the exercise writer is coming from. this may be a situation where definitions and intuition just depart from one another (another way of rephrasing my remark above, is that the assumption that M has positive dimension, is exactly what ensures that M \ {p} is not compact and hence could be thought of as in want of a compactification)

from a pedagogical point of view, it was maybe a mistake for this writer to introduce 0-dimensional manifolds in the first place. offhand, i can't think of how you improve life by doing this, but there must be some reason for it

Anthony

@leslietownes To be fair, it seems like a lot of definitions in the field of manifolds/topology can vary slightly depending on which author is writing them, so it might explain why people tend to be extra careful and present exercises/propositions in such a way that they are true in one definition and wrong in another. But I agree that sometimes some choices of definition are quite high on the bizarre scale

@leslietownes I get your point and I agree

ModularMindset

is there a way to instantly get emails when you get a notification on the main site?

Anthony

@leslietownes Thank you again for your help

leslie townes

@ModularMindset the speed of light is an obstacle to 'instantly' :)

ModularMindset

19:50

I guess an alternative option is to get desktop notifications

leslie townes

i have generally found SE notifications (i only use it on desktop) to be pretty sporadic and, if i were planning on relying on them, unreliable (i use the network only for recreation so it basically doesn't matter). e.g. sometimes i'll get a notification of something that happened like a week ago, and anybody who assumed i was getting notifications in real time would have their expectation frustrated

Xander Henderson

20:20

@nickbros123 At every institution I have ever interacted with, books come in two categories: required texts, and recommended texts. Instructors generally teach along the outline of the required text(s), but may recommend exercises or exposition from the recommended texts from time to time.

If it is a required text, I don't see as it matters if it is any good or not. You will need it for the class. If it is a recommended text, I would generally wait to see how the instructor is using it.

In any event, I think that Ross' book is perfectly fine. It has been 20ish years since I've opened it, however, so my memory could be wrong.

leslie townes

sometimes the choice of book matters less than the fact of making a choice of book

Xander Henderson

@leslietownes Indeed.

handan_toddler

TLDR; questions from the required textbook show up on tests :P

Xander Henderson

@handan_toddler If you want to be cynical, sure.

Though I've had instructors who pulled exam problems primarily from the recommended texts, rather than the required one.

handan_toddler

Ouch 😳

Without any mention of it in the class notes?

ILikeMathematics

20:42

Let $\varphi: V \to W$. Then there exists a unique $$\hat \varphi: U \otimes V \to U \otimes W \quad \text{with} \quad \hat \varphi(u \otimes v) = u \otimes \varphi(v).$$ Show that if $\varphi$ is injective, so is $\hat \varphi$.

Does my solution look fine? We will instead show $\ker(\hat \varphi) = \{0_{U \otimes V}\}$. So let $f \in \mathcal L(U, W; K)$ be arbitrary. Then $$(u \otimes \varphi(v))(f) = f(u, \varphi(v)) \overset != 0_K.$$ Since $f$ is arbitrary, we can assume $\ker(f) = \{0_{U \times W}\}$. So we get $u = 0_U$ and $\varphi(v) = 0_W$. Since $\varphi$ is injective, we get $v = 0_V$.

Xander Henderson

@handan_toddler I mean, generally the understanding is "These are recommended texts. They cover the same material as the required text(s), but may take a slightly different approach, or give different exercises. You might want to read them."

If a book is "recommended", it is nigh certain that the instructor will be using it for something, including exercises which might show up on an exam.

Seems perfectly fair and reasonable to me...

handan_toddler

Sure if that statement is made upfront in the notes.

Vladimir Lysikov

@ILikeMathematics First, you need to consider $\hat{\varphi}$ on arbitrary tensors, not only on tensors of the form $u \otimes v$. Second, what is $\ker(f)$ for a bilinear map?

psie

If a simple function $f$ is in $L^p$ for $1<p<\infty$, is it in $L^1$?

I know the converse holds, but I'm not sure how to argue about this direction.

Xander Henderson

@handan_toddler I don't understand. A book is recommended. That, all by itself, is a statement that the material in that book might be fair game.

ILikeMathematics

20:54

@VladimirLysikov $\hat \varphi$ and $\varphi$ are both linear

Ah

You mean $f$ is bilinear

Ok, I should've written "Since $f$ is arbitrary, we can assume it only maps to $0$ for $u, v$ equal to $0_U$, $0_V$ respectively where $f(u, v) = 0_K$.

handan_toddler

@XanderHenderson ok

Vladimir Lysikov

@ILikeMathematics There are no such maps. We always have $f(u,v) = 0$ when $u = 0$, $v \neq 0$

ILikeMathematics

The first thing you addressed makes the proof unusable

@VladimirLysikov Ah

Vladimir Lysikov

Yeah
You need to consider arbitrary tensors and choose the bilinear map $f$ accurately depending on the tensors

Jakobian

21:17

@leslietownes as far as I know, the concept of "one-point compactification" as extension by one isolated point in the case of the space is compact is useful in algebraic topology etc.

while not compactification, it's still okay to name it this way, I guess

@Anthony for most people, a space is supposed to be dense in its compactification. While this will give you some new space if $X$ is compact, it's a bit sketchy to call that one-point compactification, because $X$ won't be dense in it

ILikeMathematics

@VladimirLysikov Ok, let $u \otimes \varphi(v) = z \otimes \varphi(w)$. We need to show $u \otimes v = z \otimes w$. Let $B_U = \{u_i \mid i \in I\}$ be a basis of $U$, $B_V = \{v_i \mid i \in I'\}$ a basis of $V$, then $B_W = \{\varphi(v_i) \mid i \in I'\}$ is a basis of $W$ since $\varphi$ is injective and so maps a basis to another basis.

Now, $$u \otimes \varphi(v) = \dots = \sum_{i \in I, j \in I'}s_i \ell_j (v_i \otimes \varphi(v_j)) = z \otimes \varphi(w) = \dots = \sum_{i \in I, j \in I'}s_i' \ell_j'(v_i \otimes \varphi(v_j)).$$ Now, comparing coefficients, we get $s_i\ell_j = s_i'\ell_j'$ and thus $u \otimes v = z \otimes w$.

Jakobian

> The one-point compactification is usually applied to a non-compact locally compact Hausdorff space. In the more general situation, it may not really be a compactification and hence is called the one-point extension or Alexandroff extension.

this is from nlab, whetver for reasons of density of non-Hausdorffness

ILikeMathematics

Does this look fine to you, @VladimirLysikov ?

ModularMindset

21:38

math math math math math

i love math

Jakobian

If there wasn't math then what would there be

Existence only makes sense with math

2 hours later…

Zac U.

23:18

Might there be reason for my finding learning math from online and books as resources incredibly taxing?

Thorgott

23:32

learning math is always incredibly taxing

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