Mathematics - 2024-08-25

ephe

00:01

@leslietownes Unfortunately everyone just says that such a thing exists but no one shows what it is. I suppose this is just one of those "well known fact"s... Supposedly the proof should be in chapter 4.10 of ams.org/books/memo/0558 which I cannot access but that is what the previous two papers in the reference chain has said

leslie townes

00:12

have you looked in lack's "a 2-categories companion" and/or its references? math.uchicago.edu/~may/IMA/Lack.pdf it mentions not including a definition of bicategory, which is maybe not promising

ephe

@leslietownes Yes I've checked the reference 44 they mention but that turned out to be only about monoidal bicategories.

leslie townes

surely someone around here has an academic credential that can grab 4.10

ephe

I'd be very grateful if anyone could do me that favor

zeta space

00:39

I'm about to post a question

pre-save now at this link

Pre-release of the title: "One metric to rule them all?"

Lord of the rings reference

Thorgott

I'm under the impression that this result is relatively straightforward to obtain by using the bicategorical Yoneda lemma since the bicategorical functor category from a bicategory to a strict 2-category is again strict

The Gordon-Power-Street paper proves something more general one dimension higher

zeta space

0

Q: One metric to rule them all?

Consider a surface of revolution $S$ and an embedding $e:S \hookrightarrow X^3$ for $X^3=[0,1]^3$ with points $p,q$ elements of $\partial X^3$ where $\partial X^3=X^3-(0,1)^3$ for $\mathrm {sup}~ \mathrm{dist}(p,q)=\sqrt{3}$. It follows from geometric reasons that $S$ is unique, and if one wishes...

ephe

@Thorgott I'll try to see how far I can get with this. I've been trying to construct a strict 2-category using diagrams of morphisms from the bicategory

Thorgott

yeah, that sounds even more cumbersome, but it will be cumbersome either way

Debug

03:50

Orthogonal complement for abelian groups is defined as ?

@Shaun

leslie townes

if someone is doing that at the level of abelian groups i would just consult their definition and go with it, without any expectation that others would share it or know what it meant. i do not know of some general thing that is just 'out there.'

you see a lot of authors define a 'complement' of a subgroup in a group, as in e.g. en.wikipedia.org/wiki/Complement_(group_theory) but this is not uniquely determined in general and there is no notion of 'orthogonality' (or anything close to it) that it corresponds to. as far as i know.

certainly if you add in more structure than 'abelian group' one can come up with some sensible conception of what that ought to be. but not just at the level of abelian group.

even for subspace of a vector space over a field, in general i know only 'complement' and not 'orthogonal complement'

ephe

04:08

@leslietownes This is relevant: mathoverflow.net/questions/6701/…. If one allows for self-orthogonal vectors then it is easier to find orthogonal complements of vector spaces over arbitrary fields (And orthogonal complements are an important definition in coding theory which works with vector spaces over finite fields where the inner product is just coordinatewise multiplication)

leslie townes

well, OK, but that post strikes me as some very niche thing and not some out of the box construction that works for any abelian group that anybody working with abelian groups would eventually know. there's a lot of distance between 'can somebody in some context make some kind of sense of this' and 'what's X defined as for Y.' the latter question presupposing some ambient context that might not be there.

the assumption of the characteristic not being 2 in that post is an important restriction which people working in characteristic 2 are very familiar with and not surprised by. i think if i were coming to the subject for the first time i would have no idea why the characteristic of the field should matter, let alone when it should matter

one message i do get out of that is, if you were looking for some very general notion of 'orthogonality,' maybe you don't want to approach it via quadratic forms, because quadratic forms are going to make the characteristic matter.

ephe

05:24

@leslietownes Oh definitely. I didn't mean to imply that there was such a construction for abelian groups, my intent was only to show that even this much simpler question was hard to answer and point out some weird but useful example

Debug

08:16

@AnneBauval You had it right. Sorry about my attitude on that last post. I will try to control myself next time :)

Pizza

09:40

Hi 👋

Sine of the Time

hi

Pizza

I have a question, I tried to find the absolute extremes of this function $2x^3 + 6xy + y^2$

I found the critical points (0,0) and (3,-9)

Calculating the function at these points I find that at (0,0) the function is 0 and at (3,-9) it is -27

The absolute minimum is -27, I don't understand if 0 is an absolute maximum

Wolfram tells me there is no absolute maximum

Sine of the Time

if $x,y\gg 0$, the function is positive

Sahaj

@Pizza how is the absolute minimum -27? you can keep y=0 to get $f(x,0) = 2x^3$ and put an arbitrarily small x to go even smaller

Sine of the Time

$f$ is defined on a compact or on $\Bbb R^2$?

Pizza

09:49

@Sahaj Oh no sorry, I meant local minimum

@SineoftheTime on $\Bbb R^2$

Sine of the Time

no theorem assures that min/max are global

Pizza

I found that (0,0) is a saddle point and (3,-9) is a local minimum

@SineoftheTime then I can say that there are none

Sine of the Time

yep

the function is not bounded

Binky McSquigglebottom

Hi

Pizza

@SineoftheTime Oh ok thanks 👍

Binky McSquigglebottom

10:04

$f(x,y)=\frac{1}{\sqrt{7x+4y-2}}$

Help pls I need to do the first and second partial derivatives

$f(x)=\sqrt[n]{x} \quad n\in \mathbb{N}, \quad f'(x)=\frac{1}{n}\cdot\frac{1}{\sqrt[n]{x^{n-1}}}, \begin{equation} \text{for } x>0 \text{ if } n \text{ even} \\ \text{for } x\neq0 \text{ if } n \text{ odd} \end{equation}$

Shaun

@Debug I don't know, sorry; I'll look at the question you emailed me, at a later date. I'm at a conference all week.

nickbros123

11:16

is there a name for this subgroup? for $a \in G$, $H_a:=\{x \in G: xa=ax\}$

psie

On page 45 in Folland's text on real analysis, he writes that we define Borel sets in $\overline{\mathbb R}$ by $\mathcal B_{\overline{\mathbb R}}=\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$. Then he remarks that this coincides with the usual definition of the Borel $\sigma$-algebra if we make $\overline{\mathbb R}$ into a metric space with metric $d(x, y) = |\arctan x - \arctan y|$.

I see how $d$ is a metric on $\overline{\mathbb R}$, where e.g. $\arctan(\infty)$ should evaluate to $\pi/2$, but I don't see how $\mathcal B_{\overline{\mathbb R}}=\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$ coincides with the usual definition of $\mathcal B_{\overline{\mathbb R}}$ being generated by the open sets in ${\overline{\mathbb R}}$. I don't see what the open sets in ${\overline{\mathbb R}}$ "look like".

Does anyone understand in more detail what Folland means by "coinciding with the usual definition"?

Soumik Mukherjee

@nickbros123 centralizer of a

Thorgott

@psie $\arctan$ is a homeomorphism between $\overline{\mathbb{R}}$ and $[-\pi/2,\pi/2]$

nickbros123

@psie it may be useful to know that open sets in R are countable union of open intervals

actually never mind

Jakobian

@psie Borel sigma-algebra is the one generated by open sets

psie

11:27

ok, but what are the open sets in $\overline{\mathbb R}$?

Thorgott

unions of open sets in $\mathbb{R}$ and intervals of the form $[-\infty,x)$ or $(y,\infty]$

Jakobian

The basis for open sets of $\overline{\mathbb{R}}$ are open sets of $\mathbb{R}$ as well as sets of the form $(M, \infty]$ and $[-\infty, M)$ for some $M\in \mathbb{R}$

Thorgott

(I personally prefer to define it this way, but in your case, this is a consequence of the homeomorphism and what open sets in $[-\pi/2,\pi/2]$ look like)

psie

11:42

ok, hmm. I still do not quite understand why in $\mathcal B_{\overline{\mathbb R}}=\{E\subset \overline{\mathbb R}: E\cap\mathbb R\in \mathcal B_{\mathbb R}\}$, we are taking the intersection of a subset of $\overline{\mathbb R}$ with $\mathbb R$?

I am familiar with restricting a $\sigma$-algebra to a subset $E$. The Borel $\sigma$-algebra of $[-\pi/2,\pi/2]$ looks like $\{[-\pi/2,\pi/2]\cap B:B\in\mathcal B_\mathbb R\}$, but here it seems like we are extending the $\sigma$-algebra of $\mathbb R$. I have not seen a definition yet of what it means to extend a $\sigma$-algebra. This is new to me.

one potato two potato

What do you do these days @Thorgott ?

Thorgott

12:24

@psie cause that's all that matters

think about $[-\pi/2,\pi/2]$, whether a subset of it is measurable or not depends only on the intersection with the open interval $(-\pi/2,\pi/2)$, the endpoints can be added or removed at will without changing measurability

same idea here, measurability is decided by the intersection with $\mathbb{R}$ and the endpoints $\pm\infty$ can be added or removed at will

@onepotatotwopotato mostly just learning $\infty$-categories

lynx_s

12:37

Just have a quick question whether my answer is correct. So consider the statement:

[\forall y \in Y: \exists x \in X: f(x)=y]\implies [\forall B_{1}\in P(Y): \forall B_{2} \in P(Y): \Phi(B_{1})=\Phi(B_{2}) \implies B_{1}=B_{2}]

We had to give the negation of this statement and I said the statement is of the form P \implies (Q \implies R) so the negation is P \wedge (Q \wedge \neg R) and then I got this as the negation:

[\forall y \in Y: \exists x \in X: f(x)=y]\wedge [\forall B_{1}\in P(Y): \forall B_{2} \in P(Y): \Phi(B_{1})=\Phi(B_{2}) \wedge B_{1}≠B_{2}]

ephe

13:59

@Thorgott Oh, I am also preparing to give a presentation about infinity categories (well, $(\infty,1)$-categories) and am currently pushing through Lurie's HTT which is undeniably painful as an undergrad. Do you have any source suggestions with the same constructions or otherwise?

Curious One

can anyone help me to solve this question math.stackexchange.com/questions/4656304/…

Thorgott

@ephe I just finished reading the body of Markus Land's Introduction to Infinity-Categories yesterday (still have some exercises left to ponder). It's not perfect, but a very concise and readable text one can learn a lot from. Undeniably more humane to approach than HTT, which tethers on unreadable for mere mortals. I use HTT only as a reference to look up details in case they become necessary; same with Kerodon (which I still find more down-to-earth than HTT for the most part).

Rezk and Haugseng also both have neat sets of Notes

zeta space

14:26

I only have 28 rep :(

I can't even comment

Pizza

Hi

zeta space

Hello pizza

Sine of the Time

@zetaspace 28 rep with 400+ questions and 30+ answers is crazy

2

zeta space

@SineoftheTime yes lol :P

Sine of the Time

75 Offered bounties for 9,250 reputation

Pizza

14:34

guys im in this situation $\arcsin(x^2-y-1)$ , and asks me to find the absolute extremes in the circle with center (0, −1) and unit radius. But I found that there are no internal critical points. So now I had to analyze the edge, but since the arcsine is only defined between -1 and 1, can't I say that the absolute maximum is π/2 and the absolute minimum is -π/2?

If I parameterize then: $\arcsin(\cos^2(\theta) - \sin(\theta))$

Sine of the Time

you don't know a priori if there is $(x,y)$ that "reaches" $\pi/2$

it's like saying $-1\le \sin x\le 1$, but if you want to find max and min of $\sin x$ on the interval $[0,1]$, you don't reach those values

note that $t\mapsto \arcsin t$ is increasing

Pizza

Then I can calculate the derivative and set = 0

$\cos(\theta) (-2 \sin(\theta) -1) = 0$

Sine of the Time

did you use shifted polar coordinates?

Pizza

Yes

$x = \cos(\theta) , y = -1 + \sin(\theta), \theta \in [0,2\pi]$

Sine of the Time

ok

Pizza

14:49

@SineoftheTime So I can only study the derivative of $\cos^2(\theta) - \sin(\theta)$

Sine of the Time

yes

Pizza

$\cos(\theta) = 0$ when $\theta$ = π/2 and 3π/2

$\sin(\theta) = -\frac{1}{2}$ when $\theta$ = 7π/6 and 11π/6

When $\theta$ = π/2 arcsin(-1) = -π/2

When $\theta$ = 3π/2 arcsin(1) = π/2

For the others I found 5/4 so > 1

So absolute minimum -π/2 and absolute maximum π/2

Sine of the Time

15:05

do you also have to find the points where the function has the max/min?

robjohn

@Pizza $-\frac12$

Sine of the Time

long time no see rob

Pizza

@robjohn yes, sorry I forgot the -

robjohn

I hop in once in a while, but things have been slow. I’ve been occupied IRL for quite a while. Now we’re in Mammoth Lakes on vacation.

Pizza

@SineoftheTime Calculate the function definition set: $f(x, y) = \arcsin(x^2-y-1)$ and calculate the absolute extrema in the circle with center (0, −1) and and unit radius.

robjohn

15:16

@Pizza I fixed it

Pizza

This was requested

@robjohn Oh ok thank you very much!

Sine of the Time

@robjohn it seems a relaxing place

@Pizza ok

Pizza

@SineoftheTime So is this the correct procedure?

Sine of the Time

yes, I did not check the computation

Pizza

I hope I can remember all the values of the sine and cosine in these angles and others

@SineoftheTime 👍

Sine of the Time

15:21

aren't you allowed to have a calculator?

Pizza

No, we can't use anything

only the calculator but not the scientific one

Sine of the Time

strange

ephe

@Thorgott I have the Land book available but never properly checked it out. Will do now, thank you. I think I should have learnt my lesson when I read Douglas Ravenel saying "Definition 1.1.5.5] baffled me for several years" in his summary paper

Pizza

@SineoftheTime In analysis 1 we could bring for example the Taylor developments, the elementary derivatives, the elementary integrals, the notable limits

Sahaj

hi @robjohn

Sine of the Time

15:27

that sucks, expecially for the DE

Pizza

obviously those of the prof

@SineoftheTime what Is DE?

Sine of the Time

differential equations

Pizza

in analysis 2 the professor changed and we can't use anything except the normal calculator

Sine of the Time

is the written exam hard?

Pizza

@SineoftheTime and in fact one must remember everything

Thorgott

15:32

@ephe lol, haven't heard that

Pizza

@SineoftheTime Normally there is an exercise like this one that I sent now that is to find the domain and calculate the absolute (sometimes relative) extrema, a Cauchy problem that is almost always second order

@SineoftheTime Then a double integral to evaluate both directly and with Gauss Green

Thorgott

the coherent nerve is quite complicated at first, but I think it's very reasonable once you write down enough examples

Sine of the Time

that's a standard exam

Pizza

@SineoftheTime Then a surface integral and a differential form

Sometimes the power series also come out, it depends

@SineoftheTime yes

Oh then there are 2 theory questions

Sine of the Time

proofs or what?

Pizza

15:35

I'll give you some examples

Sine of the Time

3h time, right?

@Pizza yes, I'm curious

Pizza

6. Sequences of functions.
7. Divergence theorem and Stokes formula

@SineoftheTime 2

I think you can write whatever you want... These are a bit general questions.

Sine of the Time

these are not even questions

Pizza

and in fact I was thinking about what would be best to write

Because they are not specific questions, they are topics and you have to write what you know

Sine of the Time

for 7, I'd write the statement of the theorems

Pizza

15:39

This is another example: 6. Differentiability and Schwarz Theorem
7. Existence and uniqueness theorems for Cauchy problem

Sine of the Time

these are more specific

Pizza

But the professor said that the exercises are more important

In the score

Sine of the Time

obviously

there's also the oral exam?

Pizza

Yes but it is not mandatory, however if you want to do it you can increase or decrease the score by only 20%

The maximum for the written part I think is 27/28

ephe

@Thorgott Oh no I don't do examples. I'd rather keep believing I understand it

Sine of the Time

15:43

@Pizza this does not make a lot of sense. If you get a low mark, you can improve it only by a couple of marks whereas if you got a higher mark you have more margin of improvement

Pizza

@SineoftheTime No, I think it's a maximum of 3 points for score

Sine of the Time

$\pm 3$?

Pizza

Yes

Sine of the Time

that makes sense

Pizza

Instead, in analysis 1 you could also get the minimum in the written exam and then get a lot of points in the oral exam

Sine of the Time

15:47

that happens everywhere

Thorgott

@ephe understanding is just a platonic ideal anyway, but examples build intuition

Pizza

@SineoftheTime Yes, only in this case there is this "restriction"

But with the other subjects the oral exam was mandatory, here it is not

ephe

@Thorgott I'm stealing that. But all jokes aside I wouldn't survive without any examples here. Intuition is all that I have about all this.

Sine of the Time

I see

Pizza

Yes, I was just telling you this thing that I have to know practically everything even about the angles etc.,

Thorgott

15:52

yeah, that's very understandable, but also a well-built intuition can get you pretty far in this subject

Sine of the Time

@Pizza you can draw a picture or use the formulas for sum of angles

Pizza

@SineoftheTime What do you mean by drawing a picture? For now I'm using an image from the internet where the various values at the respective angles are indicated

Sine of the Time

you actually have to know only the values in the first quadrant

Pizza

@SineoftheTime and how do I do for example with 11π/6, 5π/3 etc

I saw that they are reversed

Sine of the Time

cosine is even and sine is odd

Pizza

16:07

and the sin has a negative sign

Sine of the Time

$\sin(11\pi/6)=\sin(-\pi/6)=-\sin(\pi/6)$

Pizza

Yes

ephe

@Thorgott Honestly I'm pretty okay with how much I did manage to learn. I'm going to hope that the audience will be content with the intuition

Thorgott

16:24

yeah, hoping it goes well for you

ephe

16:39

@Thorgott thanks!

robjohn

@Sahaj hello, belatedly.

Jakobian

17:00

@robjohn new word for my dictionary

robjohn

@Jakobian Belated birthday wishes, etc are a common sentiment in greeting cards here.

Jakobian

@Thorgott I suppose we all have this idea of what a platonic ideal should be, a model of sorts, but not all of us believe that it really exists. This is what separates platonist from other people

I think this model can be deceiving though

Thorgott

the platonic ideal of a platonic ideal

robjohn

17:25

@SineoftheTime it probably has to do with the bounties offered.

@SineoftheTime Ah, I see you noted this.

zeta space

19:05

You can get the bounties back

If you buy out the network

I don't care about the builder base (math s.e.) I just focus on the main village (mathoverflow)

psie

19:25

@Thorgott is it correct that, since $\arctan(x)$, or really, the continuous extension of it (which I'll denote $f$), is a homeomorphism between $\overline{\mathbb{R}}$ and $[-\pi/2,\pi/2]$, the open sets in $\overline{\mathbb{R}}$ are precisely the sets of the form $f^{-1}(U)$, where $U$ is open in $[-\pi/2,\pi/2]$ and thus of the form $U=[-\pi/2,\pi/2]\cap V$, where $V$ is open in $\mathbb R$?

Ben Steffan

@psie That's correct, yes.

Homeomorphisms induce a bijection between the topology on the domain and that on the codomain.

psie

Ok, cool. I was afraid that it was a slightly convoluted way of putting it, but my aim was to characterize the open sets of $\overline{\mathbb{R}}$ through something I know of well. I know the open sets in $[-\pi/2,\pi/2]$, but only through the open sets in $\mathbb R$.

Now I know what the open sets in $\overline{\mathbb R}$ "look like". Great!

Jakobian

19:55

@psie The gist is that the open sets of $\overline{\mathbb{R}}$ are those which come from the open intervals i.e. sets of the form $\{x : x > a\}$ or $\{x : x < b\}$ or intersections of both

Here something like $(0, \infty] = (0, \infty)\cup \{\infty\}$ would be an open interval in $\overline{\mathbb{R}}$

We would say that the topology of $\overline{\mathbb{R}}$ is that of order topology, because it's induced by the ordering

psie

ok 👍

Jakobian

To prove that that $\mathcal{B}_{\overline{\mathbb{R}}} = \{B : B\cap \mathbb{R}\in\mathcal{B}_\mathbb{R}\}$, the inclusion $\subseteq$ follows because $\mathbb{R}\subseteq \overline{\mathbb{R}}$ has the subspace topology. And the other inclusion $\supseteq$ follows because $B\cap \mathbb{R}$ is then a Borel subset of $\overline{\mathbb{R}}$ from transitivity, as $\mathbb{R}$ is Borel. And any singleton is Borel here too

Right, if $B\subseteq X$ is Borel in $X$ and $Y\subseteq X$, then $B\cap Y$ is Borel in $Y$

And if $X\subseteq Y\subseteq Z$ and $X$ is Borel in $Y$, $Y$ is Borel in $Z$, then $X$ is Borel in $Z$

In fact, $\mathcal{B}_Y = \{B\cap Y : B\in\mathcal{B}_Z\}$

this is basically a corollary of this post applied to the inclusion map $Y\hookrightarrow Z$

Now hopefully it's clear why transitivity is true for example, since $X = B\cap Y$ for some $B\in\mathcal{B}_Z$ and $Y\in\mathcal{B}_Z$, we see that $X\in\mathcal{B}_Z$ as their intersection

Singletons of $\overline{\mathbb{R}}$ are Borel because they are closed sets

psie

20:23

ok, I really struggle with this. How does $\sigma(\{\text{open sets in }\overline{\mathbb R}\})\subseteq\{B\subseteq \overline{\mathbb R}: B\cap\mathbb R\in \mathcal B_{\mathbb R}\}$ follow from the subspace topology? Somehow, the topology of $\overline{\mathbb R}$ must be a subset of the latter set, which I don't see.

My attempt at an argument goes like this: since $f^{-1}(U)$ are the open sets in $\overline{\mathbb R}$, where $U=[-\pi/2,\pi/2]\cap V$, $V$ open in $\mathbb R$, we have that an open set in $\overline{\mathbb R}$ is of the form $\overline{\mathbb R}\cap f^{-1}(V)\subseteq\overline{\mathbb R}$, but I don't know if $\overline{\mathbb R}\cap f^{-1}(V)\in\mathcal B_\mathbb R$, since as far as I know, a homeomorphism only maps open sets to open sets in the subspace topology (or vice versa).

Here $V$ is open in $\mathbb R$, which complicates my argument I feel like.

psie

20:39

@psie ah ok, I think I understand: if $B$ is open in $\overline{\mathbb R}$ then by definition of open subset, $B\cap\mathbb R$ is an open set in $\mathbb R$, thus belongs to $\mathcal B_\mathbb R$, so hence the inclusion $\subseteq$

the other inclusion, $\supseteq$, is slightly trickier I think, have to think some more on this

Jakobian

20:57

@psie by definition of subspace topology I would say

psie

yeah

Jakobian

sorry it is what that means

psie

right, if $\mathcal E\subset \sigma(\mathcal A)$, then also $\sigma(\mathcal E)\subset \sigma(\mathcal A)$

Jakobian

@psie the function $f$ is borderline irrelevant, you just want it to be able to say something about relation between $\overline{\mathbb{R}}$ and $\mathbb{R}$, possibly, but not much else

i.e. that $\mathbb{R}$ is an open subspace of $\overline{\mathbb{R}}$

psie

ok

Jakobian

21:01

@psie yeah I was just confusing $\sigma(\text{basis of a top. space})$ vs $\sigma(\text{topology of a space})$

where the two aren't necessarily the same because of the countable sums restriction

@psie Well, like I said the other direction follows from $B = (B\cap \mathbb{R})\cup (B\cap \{\infty, -\infty\})$ being a union of Borel sets

finite sets are Borel - they're even closed

the trickier part is $B\cap\mathbb{R}$

psie

@Jakobian right, $(B\cap \mathbb R)$ is a borel set of, note, $\mathbb R$

Jakobian

$B\cap\mathbb{R}$ being a Borel set of $\mathbb{R}$, and $\mathbb{R}$ being a subspace of $\overline{\mathbb{R}}$ implies that $B\cap\mathbb{R} = B_0\cap \mathbb{R}$ where $B_0$ is a Borel set of $\overline{\mathbb{R}}$

and since $\mathbb{R}$ is a Borel set of $\overline{\mathbb{R}}$, this implies that $B\cap\mathbb{R}$ is Borel in $\overline{\mathbb{R}}$

This is because $\mathcal{B}_Y = Y\cap \mathcal{B}_Z$ (written slightly informally) when $Y\subseteq Z$

psie

@Jakobian yeah, that is simply the restriction of a sigma algebra to a subset in the sigma algebra

Jakobian

In other words, since $B\cap\mathbb{R}$ is a Borel set of $\mathbb{R}$ and $\mathbb{R}$ is a Borel set of $\overline{\mathbb{R}}$, it follows that $B\cap\mathbb{R}$ is a Borel set of $\overline{\mathbb{R}}$

@psie I mean the intersection goes into the generators

$Y\cap \sigma(\text{open sets of }Z) = \sigma(Y\cap \text{open sets of }Z) = \sigma(\text{open sets of }Y)$

Of course the intersection with $Y$ is supposed to indicate that we are taking intersection with every element of $\sigma(\text{open sets of }Z)$ and its not literally an intersection

psie

21:16

ok, so writing $\mathcal B_Y=\{A\cap Y:A\in \mathcal B_Z\}$ is something different than what you meant by $\mathcal{B}_Y = Y\cap \mathcal{B}_Z$?

Jakobian

No its the same

I was just saying that $Y\cap\mathcal{B}_Z$ is not supposed to be treated as intersection of $Y$ and $\mathcal{B}_Z$, but as $\{A\cap Y : A\in\mathcal{B}_Z\}$

psie

makes sense

ok, finally grokking this, thanks very much!

psie

22:04

@Jakobian if we equip $\overline{\mathbb R}$ with $d(x,y)=|f(x)-f(y)|$ (this is what Folland does), where $f$ is the continuous extension of $\arctan x$, doesn't then $f$ determine kind of the open sets in $\overline{\mathbb R}$? I'm trying to understand in what sense $f$ could influence your arguments above in proving both inclusions.

Ben Steffan

22:50

To repeat what I said a few hours back, yes, $f$ determines the topology on $\overline{\mathbb{R}}$ fully.

In fact, you should think of $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, \infty\}$ as coming as set. To define a topology on it, you could put a metric like Folland does, and this makes $f$ into a homeomorphism.

Thus, topologically speaking, $\overline{\mathbb{R}}$ is exactly as good as $[-\pi/2, \pi/2]$, or any other closed interval for that matter.

In particular, since the Borel sigma algebra (although I'm not quite sure why we're discussing it in the first place) is a topological invariant, you can just as well ask what $\mathcal{B}_{[0, 1]}$ is.

In this case, instead of $\mathbb{R} \subset \overline{\mathbb{R}}$ you're looking at $(0, 1) \subset [0, 1]$, but that's pretty much the only thing that should change.

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$Mathematics$ Mathematics