Mathematics - 2024-02-19 (page 1 of 4)

Jakobian

00:01

So it will be $f_*([p]) = [f\circ p]$ where $p:[0, 1]\to X$ is a path with $p(0) = p(1) = x_0$ is the base point

technically you want to talk about $f:(X, x_0)\to (Y, y_0)$, a morphism of based spaces

>it just seems a little strange that the group morphism is essentially just the based topological space morphism

it isn't one, a morphism of based spaces induces a homomorphism of fundamental groups

Thorgott

indeed, fundamental group is a functor on the category of pointed spaces, not on the category of spaces, that's an important distinction

Jakobian

yeah. If you want $f:X\to Y$ to just be a continuous function, then it induces a morphism $f_*:\Pi_1(X)\to \Pi_1(Y)$ between the fundamental groupoids

i.e. $f_*$ is a functor

(here, more or less my knowledge about the subject ends :P)

Dannyu NDos

00:29

Is there a standard terminology for the induced morphism from the coimage to the image?

Note that such morphism is always a bimorphism, but not necessarily an isomorphism.

Jakobian

@DannyuNDos I think its called pushforward sometimes

Thorgott

never seen that tbh

id just call it the canonical morphism

Jakobian

Direct image functor

In mathematics, the direct image functor is a construction in sheaf theory that generalizes the global sections functor to the relative case. It is of fundamental importance in topology and algebraic geometry. Given a sheaf F defined on a topological space X and a continuous map f: X → Y, we can define a new sheaf f∗F on Y, called the direct image sheaf or the pushforward sheaf of F along f, such that the global sections of f∗F is given by the global sections of F. This assignment gives rise to a functor f∗ from the category of sheaves on X to the category of sheaves on Y, which is known as the...

or pullback not sure. But I think I've seen that terminology used

Thorgott

00:45

i know pushforward/pullback of sheaves

i dont see what they have to do with the question

Jakobian

just similar

Ted Shifrin

WtF is coimage?

Thorgott

$A/\ker(f)$

cokernel of the kernel

Jakobian

I misunderstood the question

Ted Shifrin

@Thorgott Huh?

Jakobian

00:51

The map $f:A\to B$ induces an isomorphism from $\text{coim}(f) = A/\ker(f)$ to $\text{im}(f)$

its one of the isomorphism theorems

Ted Shifrin

Well, yes, obviously I’ve known that for 52+ years.

Jakobian

actually we're not talking about modules or anything so its more general definition

Ted Shifrin

still, I’ve never heard of coimage. Cokernel, yes, 52 years ago.

Dannyu NDos

@Thorgott That works only for algebraic structures. In general settings, the coimage is the coarsest quotient object of the domain that induces a morphism.

Jakobian

ncatlab.org/nlab/show/coimage

Ted Shifrin

00:54

I hate categorical s**t.

Thorgott

sure, i was just thinking of a context that's not useless

Jakobian

its useful when you're not doing category theory and actually applying this "s**t"

that's what category theory is about after all, introducing ideas of concepts in analogy to other already existing concepts

Ted Shifrin

I am opposed to formalism in mathematics for formalism’s sake.

Jakobian

e.g. we have products, lets generalize this notion... etc.

Thorgott

well, the salient point is that there's context where you can talk about coimage and image, but the natural map between them is not an isomorphism

Ted Shifrin

00:57

my taste is not going to change at this point in my life.

Thorgott

i.e. theres contexts in which the isomorphism theorem fails

Dannyu NDos

@TedShifrin Aw come on, I hope you know the beauty of coproducts.

Jakobian

me too, Ted. But it might be useful if you know what you're working with

Thorgott

e.g. for topological abelian groups, that's a very concrete example

Dannyu NDos

@Thorgott Just topological spaces themselves already provide a counterexample. Namely, the identity function but the codomain has strictly coarser topology than the domain.

Ted Shifrin

00:59

Thor, I’m happy with the natural isomorphism from row space (image of the transpose) to the column space (image of the map).

@Jakobian I’ve worked with plenty of notions that plenty of people consider abstract formalism …. Like sheaf cohomolgy.

Thorgott

@DannyuNDos that's not a pre-abelian category, these notions are hardly interesting outside of pre-abelian categories

@TedShifrin derived functors are excellent formalism :P

Ted Shifrin

I don’t think in Hartshorne style ….

But I think that’s natural if you try to see where SES go ….

Thorgott

yes, i recently left a comment on a post about that

Jakobian

@AlessandroCodenotti Any idea about this? If $f:X\to Y$ is a Borel surjective map from Polish space to regular space $Y$, then assuming PFA, $Y$ is separable. What about in ZFC?

Silly Goose

01:14

@Jakobian I think I mean that because one is defining the induced group morphism in terms of the based topological space morphism in a sort of trival (not in the sense of obvious or easy, but just almost like you are doing nothing) way, it seems like a strange example of a functor, and I thought I was maybe missing something in my understanding of the definition

Jakobian

its not strange at all

you take a continuous map and you lobotomize all the information from it (unneeded from algebraic topology perspective)

thus obtaining group homomorphism

Silly Goose

Lol is lobotomize official math terminology

Jakobian

the point of this functor is to be able to tell something about topological spaces

Silly Goose

maybe i was thrown because i thought the associated exercise was to sort of get familiar with functors and the fundamental group to me seems like a poor choice of an example to get familiar with functors. since there are a lot of identifications in this case which are sort of special to this case.

@Jakobian okay this is making sense now then

Jakobian

well, this is basically the idea behind categories

Silly Goose

01:21

well i do have the impression that functors are the natural sort of concept you'd want to turn a problem in math structure $1$ to a related problem in math structure $2$

Jakobian

we translate from setting of topology to the more algebraic setting of group theory

Thorgott

the word is "tautological" rather than "trivial"

Silly Goose

ah yes that is a better choice of word

Thorgott

many good definitions are tautological

are you familiar with Hom-functors?

cause if you squint your eyes, that's all the fundamental group functor is

Silly Goose

i am not but i will look into it

Jakobian

01:23

Hom bifunctor

Thorgott

01:39

man, i derive an unhealthy amount of joy from proving utterly useless results

just in: if $(X,A)$ is a pair of spaces, $R$ a coherent ring, $M$ a flat and $N$ a finitely presented $R$-module, there is a convergent 2nd quadrant cohomological spectral sequence $Tor_{-p}(H^q(X,A;M),N)\Rightarrow H^{p+q}(X,A;M\otimes_RN)$

Jakobian

02:03

@AlessandroCodenotti Forget what I said above with $Y$ being separable under PFA

Stan Shunpike

@TedShifrin hey bud! 😂 Back doing math. Any chance you could look at this question? math.stackexchange.com/questions/4865139/…

Sorry the link is here math.stackexchange.com/q/4865139/197705

Ted Shifrin

02:23

Hi @Stan. The answers are bunk. You need an orthonormal basis with $v$ as one of the vectors.

Jakobian

@AlessandroCodenotti Can an $L$-space be Borel image of a Polish space?

Ted Shifrin

The usual thing is to find one vector $u$ orthogonal to $v$ by linear algebra (solve dot product =0), then use cross product to get $w=v\times u$. Then make unit vectors.

robjohn

I give an extension for higher dimensions in $(8)$-$(10)$ of this answer.

Ted Shifrin

02:42

LOL, I was trying to be as simple-minded as possible. :)

Jakobian

03:14

I don't see why if $f, g$ are Borel maps then $(f\times g)(x, y) = (f(x), g(y))$ wouldn't

am I missing something

oh I think I get it. Its because of sigma-algebra on the image

X4J

03:42

Suppose $P_{f}(x) = (x+1)(x-1)(x^2 + x + 1)$ is a characteristic polynomial for a linear operator $f$ where the space is $V$ and the field is $\mathbb{R}$. If for any polynomial $Q$ that divides $P_{f}$ we can find $f$-invariant subspace $W$ such that $P_{f|_W} = Q$, does it imply that $W$ is unique? (Namely there is not $W' \neq W$ with the same property for $Q$)

*Any monic polynomial Q

Silly Goose

can i intuitively think of a deformation retraction as keeping a subset of the topological space the same, but "deforming" everything else (in a nice way)?

Ted Shifrin

Deforming it to the subset by homotopy, yes.

@X4J What is special about your $P_f$ here?

Sahaj

04:00

Any opinions on this question? math.stackexchange.com/questions/4865494/…

I haven’t studied Mobius inversion so I can’t say, but could this actually resolve the twin prime conjecture?!

冥王 Hades

This habit of staying up late will result in my death via sleep deprivation someday

Jakobian

@Sahaj no

Sahaj

Oh, ok.

sad

Jakobian

Its nothing new that Daniel is posting about Collatz

oh its twin prime this time

well regardless, don't give online posts much hope

X4J

@TedShifrin Homework question and that's what I am trying to figure

Ted Shifrin

04:16

You can’t notice anything special about this polynomial, X4J?

@冥王Hades I thought it already had.

冥王 Hades

@TedShifrin It will again, and then I'll once again reincarnate 243 years later.

Ted Shifrin

Well, I should be gone by then .

what ever happened to your reading Pedoe’s beautiful book?

Jakobian

I always learn the most when I'm interested in solving a problem

冥王 Hades

Procrastination happened

Ted Shifrin

Jakobian, my adviser, a most eminent 20th century geometer, said that’s how one should learn.

@冥王Hades Not just death?

X4J

04:29

@TedShifrin Obviously $P_f$ is a multiplication of two linear factors and an irreducible polynomial of degree 2. The uniqueness of the invariant spaces that correspond the linear factors is indeed clear but, for example, why would $Q(x) = (x-1)(x^2 + x + 1)$ be unique?

Ted Shifrin

You still haven’t answered my explicit question.

And the direct sum of unique subspaces is unique.

X4J

Yes it clicks to me now

Ted Shifrin

So if you won’t answer my question, give me an example of non-uniqueness.

X4J

04:47

I understand intuitively why it is true but I struggle to reason it by the knowledge I've gained during the course so far

Sahaj

@Jakobian so apparently he likes to spend his time on unsolved problems and posting them on MSE

Im guessing that must attract a lot of downvotes too

X4J

Like what would be the space that corresponds $Q = x^2 + x + 1$? I'm trying to think how I could generally find a $w \in W$ so $(w, f(w))$ forms a basis to $W$ but I fail

Ted Shifrin

05:03

Yeah, that won’t be the case, because you have complex eigenvalues.

Think about $f(x)=x^2+1$.

Jakobian

05:21

@AlessandroCodenotti What is the family of images of Borel maps $f:\mathbb{R}\to\mathbb{R}$?

I've seen it mentioned those are analytic sets, but I feel like this might be wrong - here $f$ is not continuous but Borel

Alessandro Codenotti

06:18

@Jakobian is still analytic sets. There is a finer Polish topology on $\Bbb R$ that makes $f$ continuous

There is a section in Kechris called something like "turning Borel sets into clopen sets" where this is explained in chapter 12 or 13 by memory

Daniel Donnelly

@mick

See my most recent post

That is what I have a question about

And vise versa an intro to you of my "attack style"

Alessandro Codenotti

@Jakobian I checked, this is 13.11 in Kechris's book

psie

11:51

I've been looking for a beginner's friendly proof of the open mapping theorem in functional analysis. I found the following proof on Youtube and would be very grateful if someone could just take a peak at it and say if it looks legit or not.

> Theorem: Let $X$ be a normed space over $K$, and $f$ be a nonzero linear functional on $X$. If $E$ is an open subset of $X$, then $f(E)$ is an open subset of $K$.

> Proof: Since $f$ is nonzero, there is some $a \in X$ such that $f(a)=1$.

> [Existence of $a$: $\exists y \neq 0$ in $X$ such that $f(y)=p \neq 0 \implies f\left(\frac{y}{p}\right)=1$. Let $a=\frac{y}{p} \in X$ ].

> Note that $a \neq 0$. Let $x \in E$. Since $E$ is open, there is some $r>0$ such that $B(x, r) \subset E$. If $|k|<\frac{r}{\|a\|}$, then $x-k a \in E$, since $\|x-(x-k a)\|=\|k a\|=|k|\| a \|<r$. Thus $f(x)-k=f(x)-k f(a)=f(x-k a) \in f(E)$. And hence $\left\{k^{\prime} \in K:\left|f(x)-k^{\prime}\right|<\frac{r}{\|a\|}\right\} \subset f(E)$.

> In other words, $B\left(f(x),\frac{r}{\|a\|}\right)\subset f(E)$, which implies $E$ is open.

Thorgott

that proof looks good, but this isn't really the open mapping theorem now, is it?

psie

yeah, it's something softer :)

thanks for taking a look at it though. I rarely look for proofs on Youtube, so always good to have it double checked.

mick

12:09

@DanielDonnelly ah to attack the sieve idea of twins ? or ?

Jakobian

12:48

@AlessandroCodenotti thanks

Jakobian

14:00

@AlessandroCodenotti Do you know Four Poles theorem?

prac.im.pwr.edu.pl/~ralowski/science/Cantor/unions41.pdf

user 85795

Alessandro Codenotti

@Jakobian never heard of it

Jakobian

I've seen someone claiming you can use it to prove that Borel image of Polish space has countable tightness

but I want to consider separable metric spaces instead, so I'm trying to figure out how they did it, and what could modifications be

Suppose $I\subseteq \mathcal{P}(X)$ is a proper $\sigma$-ideal of Polish space $X$ with Borel basis. Let $\mathcal{A}\subseteq I$ be a point-finite cover of $X$. Then there is $\mathcal{B}\subseteq\mathcal{A}$ such that $\bigcup \mathcal{B}\not\in\sigma(I\cup \mathcal{B}(X))$, the $\sigma$-field generated by $I\cup \mathcal{B}(X)$.

If this were true, then I could prove that under proper forcing axiom, if $f:X\to Y$ is a map such that $f^n:X^n\to Y^n$ are Borel and $X$ is a separable metric space, then $Y$ is hereditarily separable

The assumption on $X$ can be dropped a little from Polish space to an analytic set, but that's still not a lot

Claudio Menchinelli

14:45

I have a domain $D$ in which $f(z)$ is regular. $\gamma$ is a simple closed curve lying in $D$ and surrounding a point $\zeta \in D$. To show that $$f'(\zeta) =1/{2\pi i}\oint_{\gamma} \frac{f(z)dz}{(z-\zeta)^2}, $$ The author arrives at the requested form, plus an additional term $I$. I need to prove that $$\vert I \vert \to 0 \text{as } |h| \to 0$$ (the author assures me it does :P); where $$I = \frac{1}{2\pi i}\oint_{\gamma}\frac{hf(z)dz}{(z-\zeta)^2(z-\zeta -h)}$$

Now I can an upper bound for the numerator using the fact that $\vert \oint_\gamma f(z)dz\vert \le Ml_{\gamma}$

now I'd like to find a lower bound for the denominator, and I do not know how to do that, I have very little experience with proofs like this in general

Dang, I feel more and more ashamed for asking this :P

Jakobian

15:12

@AlessandroCodenotti Another way, do you see if given a Borel map $f:X\to Y$ from separable metric space $X$ to $T_3$ space $Y$ (or $T_{3.5}$ if you want) could be extended to a map $\tilde f:\tilde X\to \tilde Y$ where $\tilde X$ is Polish and $\tilde Y$ is $T_3$?

and $X\subseteq \tilde X, Y\subseteq \tilde Y$

Fargle

17:33

Just noticed something neat while reading some stuff --- I saw $re^{i\theta} \mapsto \frac{r}{1+r}e^{i\theta}$ given as an example of a homeomorphism $\Bbb C \to $ the open unit disk, and I realized that, restricted to the positive real line, this is just the relationship between interest rate and rate of discount in compound interest!

Sahaj

hi

Ted Shifrin

17:45

@Fargle This is the standard homeo from $\Bbb R_+$ to $[0,1)$.

Fargle

I figure --- I just didn't realize that it showed up in the context of interest/discount.

Ted Shifrin

It’s the natural formula to turn $\infty$ into $1$.

Fargle

I don't disagree. Just had never seen it from this angle before, and it strikes me that it could be used as a "concrete" way to introduce that map as a motivating example of a homeomorphism, for those inclined to think in terms of money.

Balarka Sen

@Fargle Hah, that's pretty funny!

Ted Shifrin

Are you suggesting that math majors would ever be inclined to think in terms of money?

Fargle

17:56

@TedShifrin You've got me there.

But I dunno, sometimes I find myself infodumping upon unsuspecting bystanders in my life. I am sure they love that.

Ted Shifrin

I'm sure they're no longer so unsuspecting.

Stan Shunpike

@TedShifrin hi ted!

Ted Shifrin

hi @Stan

Stan Shunpike

how r u? :)

Jakobian

I don't think that whoever came up with this thought of it as compound interest

Ted Shifrin

17:59

We had three power outages in the area last night, so I'm trying to do some bookkeeping on my computer ... :)

Stan Shunpike

@TedShifrin you're advice was great! i was able to figure out the equation myself with just that piece of advice!

Ted Shifrin

Of course not, @Jakobian.

Stan Shunpike

@TedShifrin so thanks!

Ted Shifrin

Ah, good to hear, @Stan. Yeah, the answers you got were either wrong or worthless.

Jakobian

If you have a metric $d$ then you also make it bounded by letting $\frac{d}{1+d}$. Its just one of those natural ways of making something bounded

Ted Shifrin

18:00

You should write up an answer to your question.

Balarka Sen

@Jakobian The most natural way is the boring $\min\{1, d\}$

Stan Shunpike

@TedShifrin I will as soon as I finish. I first need to find an equation of the circle in a custom plane. Stuck on that part.

Fargle

@Jakobian I guess what I mean is that the interest-discount case is just a special case of demanding such a bound --- an arbitrary nonnegative interest rate makes sense, but discount rates are only sensible below 100%

Ted Shifrin

@Stan What do you mean by "an equation"?

Jakobian

@BalarkaSen In an analysis class, I had a moment bounded metric had to be used as $\frac{d}{1+d}$ for an argument to work

Fargle

18:02

And they furthermore are exactly related by that formula rather than the min function or arctan, or whichever other you like

Stan Shunpike

@TedShifrin let me get a picture and show u

Ted Shifrin

@Fargle Well, that's sorta silly. Of course, a ratio is going to be a ratio and not a min or an arctan.

Jakobian

the transformation $r\mapsto \frac{r}{1+r}$ is more natural from analysis perspective I suppose

Ted Shifrin

Seeing it's a metric is a bit easier with the min, though, @Jakobian.

Stan Shunpike

Ted Shifrin

18:05

Cool picture, @Stan, but it doesn't answer my question :)

Stan Shunpike

@TedShifrin ofc, give me a second to type what i mean

Balarka Sen

Reminds me of the double bubble theorem

Jakobian

I usually just write $\min(d, 1)$ myself too, but $\frac{d}{1+d}$ does seem a lot more useful. For one, as noticed, $r\mapsto \frac{r}{1+r}$ is actually a homeomorphism so I'd expect it to change the balls in a nicer way

Ted Shifrin

@BalarkaSen Indeed.

Claudio Menchinelli

Just a quick doubt to see if there's a mistake in my book: I have two points $z_1,z_2$ in the complex plane, and a circle $C$ of center $z_1$ and I require that the radius of $C$ be $\rho \ge 2|z_1-z_2|$. if $z$ lies on the circle, then it's impossible to have $|z-z_2| = 1/2\rho$, isn't it?

Ted Shifrin

18:09

You know that $|z-z_2|\ge \rho-|z_1-z_2|$.

Stan Shunpike

@TedShifrin So I currently have paramterized this circle such that I can in vector form specify any point I want on it with some angle $\theta$. This is great because when I put it into a 3D plotting tool, I can spin $\theta$ and it will move the point around in the circle specified.

The problem is, I also want to plot the circle itself. So in a standard 2D plotting tool, I would just plot a circle like $(x-a)^2 + (y-b)^2 = r^2$, but that won't work here because this circle is in 3D dimensions.

Ted Shifrin

Parametric plots are by far easier than plots of equations. I'm confused.

Fargle

@TedShifrin I don't mean that it's incredible, just that, whereas the example I always held in my head of such a homeo was arctan for some reason, $\frac{d}{1+d}$ is a very much simpler one, and it falls out of a common real-world interaction.

Ted Shifrin

Yeah, @Fargle, I've almost never used arctan :P

Claudio Menchinelli

@TedShifrin that's what my intuition told me but then $|z-z_2| = |z-z_1+z_1-z_2| \le |z-z_1|+|z_1-z_2| \le \rho +1/2\rho$

Stan Shunpike

18:11

@TedShifrin I'm probably just not understanding. To me $\mathbf{x}$ is just a single point. It's not a set. So I'm having trouble envisioning how it will end up looking like a circle like the one in my image. to me, $\mathbf{x}$ would just be a point on that circle right?

Balarka Sen

@TedShifrin I was surprised to hear that Michael Hutchings of contact homology fame was one of the people who co-wrote the proof of double bubble.

Ted Shifrin

@Claudio That's not helpful. You need to use what's called the reverse triangle inequality.

leslie townes

balarka: he's double trouble

Ted Shifrin

@Balarka I didn't know that. I wonder if he was in one of Frank Morgan's summer research groups as a youngster.

Balarka Sen

Yeah I think so

Claudio Menchinelli

18:12

@TedShifrin I see, thanks for the help

Balarka Sen

I remember reading about that

Ted Shifrin

@Fargle What is $\mathbf x$?

Fargle

@TedShifrin Great question. @Stan? ;)

Ted Shifrin

Oops. Sorry, Fargle.

@Stan, what is $\mathbf x$?

Stan Shunpike

@TedShifrin @Fargle LOL Oh i thought i defined it above. It's a point that lies on the circle in my particular unique plane. This was created with the orthonormal basis ted said I would need

Ted Shifrin

18:15

But you're giving equations $\mathbf x=\mathbf x(\theta)$, where $\theta$ is the parameter.

Any decent graphics CAS will do a 3D plot of that.

You don't want to have it do an implicit plot of a sphere and a plane. Computers can't plot intersections just by themselves, as far as I know. My years of using Mathematica I've always had to figure out the parametric equations if I want that.

Jakobian

In particular, with $\min(1, d)$ we lose, say $B(x, 2)$, but with $\frac{d}{1+d}$ the set of balls is still the same

so this should be useful if you don't consider only the more local structure

Ted Shifrin

@Jakobian You're beating a dead horse.

Just let it go.

Stan Shunpike

@TedShifrin if im understanding correctly, then exactly. I need to figure out that intersection. as i understand it, $\mathbf x(\theta)$ is not the parametric equation of the intersection right?

Ted Shifrin

What else would it be, @Stan? You have the data for that. You know the normal vector to the plane and (by Pythagoras, I presume), you know the radius of the circle.

Stan Shunpike

@TedShifrin thinking...trying to reason lol give me a minute

Jakobian

18:22

An interesting question is, if given a metric $d$, can we obtain a uniformly equivalent metric $d'$ such that $d'$-bounded sets coincide with sets bounded with respect to the uniformity generated by $d$

Stan Shunpike

@TedShifrin LOL ofc ur right. I guess I was going in circles you might say....ba dum tsk

I guess I just didn't understand the phrase parametric equation very well

Jakobian

I think this is related to Heine-Borel

Ted Shifrin

@Stan Tsk tsk. Surely you had that in your multivariable calculus course years ago. If not, you can watch a few minutes of one of my videos for a refresher :P

Anyhow, you got it now?

Stan Shunpike

yes, i believe so.

@TedShifrin i was just thinking the other day i better do that

don't want to get even more rusty than i am now

Jakobian

@TedShifrin I'm just drawing inspirations

Stan Shunpike

18:27

@TedShifrin I was walking down the street the other day and realized i'd been being silly and making a 3D version of my thesis model was trivial so i've been trying to work out how to do the equations the last couple days

Ted Shifrin

That's a good exercise, doubtless.

EE18

Is this an error in the naive set-theoretic definition of a product of sets?

Stan Shunpike

@TedShifrin I've been tutoring my friend in math cuz he just resumed school again. never finished college and wants to try again. he said "do you ever need help?" and I said to him, "all of us need help. Every time I get stuck, I have to go ask my friend Ted." :')

EE18

This from Wikipedia I follow:

Ted Shifrin

LOL @Stan.

EE18

18:29

But my first picture doesn't seem to "mix" $X$ and $Y$ so I'm not sure how it works out.

Jakobian

@EE18 error?

Ted Shifrin

$\mathcal P(\mathcal P(X)\cup\mathcal P(Y))$ surely is a mouthful. It's too much for me.

EE18

That is, I can't follow how $\{x,\{x,y\}\}$ us a member of the set so constructed

Ted Shifrin

Oh, do we really need to think of ordered pairs that way? I sure don't want to.

EE18

Indeed I can't see how I get anything "mixing" $x,y$ in said set, if that makes sense?

Jakobian

18:32

So you're confused about $\{x, \{x, y\}\}\in \mathcal{P}(\mathcal{P}(X)\cup\mathcal{P}(Y))$?

EE18

Definitely not, but when passing through something in a book which I fail to follow I do usually try to stop ;)

Exactly :) @Jakobian

Stan Shunpike

@TedShifrin ok so separate question. So i understand that if i have the normal vector $\mathbf n$ to a plane and I decide to choose an arbitrary vector $\mathbf u$ in my plane, then its trivial to find an orthogonal vector $\mathbf v$ that can span 3-space as well as with $\mathbf u$ and $\mathbf v$ together, they can span my plane.

any suggestions for how to choose $\mathbf u$?

Jakobian

@EE18 first thing means that $\{x, \{x, y\}\}\subseteq \mathcal{P}(X)\cup \mathcal{P}(Y)$

EE18

We have $x \in X$, $y \in Y$ to begin. Then $\{x\} \in P(X)$ and $\{y\} \in P(Y)$, and thus both $\{x\}, \{y\} \in P(X) \cup P(Y)$.

Ted Shifrin

I gave you my suggestion. You have to read off a solution from the linear equation $\mathbf n\cdot\mathbf x=0$. The basic linear algebra algorithm given a reduced echelon form of a matrix gives you an algorithm for that

Jakobian

18:34

@EE18 oh okay this is an error

it should be $\mathcal{P}(\mathcal{P}(X\cup Y))$

the order of operations is wrong

Fargle

@EE18 Yeah --- with this definition, where you just define $p = \{\{x\},\{y\}\}$, the sets $X \times Y$ and $Y \times X$ are literally identical, but you would like them not to be the same set.

EE18

Agreed I think :) and that's what wikipedia says too! Zorich is killing me... thanks very much for confirming Jakobian!

thank you too Fargle!

I wasn't sure if it worked out as another "model" for ordered pairs but seems it's just flat wrong

Jakobian

Then we get $\{\{x\}, \{x, y\}\}\subseteq \mathcal{P}(X\cup Y)$, that is $\{x\}, \{x, y\} \in \mathcal{P}(X\cup Y)$

Ted Shifrin

It seems like Zorich doesn't really have the patience to do this boring foundational stuff correctly.

Fargle

It does work as a model for unordered pairs at least

Stan Shunpike

18:38

@TedShifrin excellent! thank you!

Jakobian

in other words, $\{x\}, \{x, y\}\subseteq X\cup Y$, that is $x, y\in X\cup Y$

EE18

@Fargle touche :)

Ted Shifrin

@Stan Anyhow, unless one of $n_1$ and $n_2$ is zero, just use $(-n_2,n_1,0)$.

Fargle

(Sort of a goofy model though. The set $\{x, y\}$ would work just as well >_>)

EE18

Ya I'm gonna stay with him because all of the reviews are rave. Trying to use it to give my book Amann Escher a little more context too, but am always grateful I can ask you all some questions :) thanks again everyone!

Stan Shunpike

18:40

@TedShifrin $n_1$ and $n_2$ won't be fixed. and at the moment, i'm not sure what values they are restricted to. but maybe i'll just try that and if i get errors in my plots, then i'll figure out what to do

Ted Shifrin

Of course they're not fixed.

Jakobian

Also the axioms used here, is the axiom of union for $X\cup Y$ being a set, axiom of power set for $\mathcal{P}(\mathcal{P}(X\cup Y))$ being a set, and axiom schema of comprehension for $X\times Y$ being a set

Ted Shifrin

Just think about the equation $\mathbf n\cdot\mathbf x = 0$.

EE18

is comprehension also referred to as specification sometimes (i think that's what Halmos calls it)?

Jakobian

also axiom of extensionality is used for set-builder notation

EE18

18:41

basically you can choose a subset of a given set with some property

Jakobian

alternatively to axiom of power set, you can use axiom schema of replacement here

Stan Shunpike

@TedShifrin well, $\mathbf x$ lies in my plane and my plane is perpendicular to $\mathbf n$. is that relevant?

Jakobian

oh and of course axiom of pairing for definition of order pair to make sense

Ted Shifrin

@StanShunpike You have forgotten everything? What is dot product? Of course it’s relevant.

Thorgott

Jakobian this closes to just listing out all the axioms

Soumik Mukherjee

18:55

@EE18 yes

Stan Shunpike

@TedShifrin the dot product is a measure of basically how close the vectors are. when the dot product equals 0, they are orthogonal.

if the parametric definition of the circle is done correctly, ensuring that the circle's center and its radius are defined with respect to my plane, then any point $\mathbf{x}$ generated by this parametric equation will inherently satisfy the plane's equation. This means that the constraint $\mathbf{n} \cdot (\mathbf{x} - \mathbf{p}) = 0$ would be naturally met for all points on the circle. This constraint would be then a confirmation that the circle indeed lies within the specified plane.

that's what i got out of that

Ted Shifrin

No, I reject your measure of closeness entirely.

EE18

@SoumikMukherjee If so then isn't extensionality entirely different and just used to define set equality? @Jakobian

Jakobian

@Thorgott not all of them... just the ones you need to do set theory

Ted Shifrin

Maybe you should go back to my second lecture. Dot product has an algebraic definition and then geometric properties. From the algebraic definition, you easily find a basis for the plane orthogonal to a nonzero vector $\mathbf n$.

Jakobian

18:59

@EE18 extensionality is not a definition of equality

Transcript for

$Mathematics$ Mathematics