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12:51 AM
wahts up gamers
 
1:10 AM
A={a,b,A} is a set.
what is wrong with this statement?
Does it violate axiom of extension?
I think no.
 
In mathematics, the axiom of regularity (also known as the axiom of foundation) is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A. In first-order logic, the axiom reads: ∀ x ( x ≠ ∅ → ∃ y ( y ∈ x ∧ y ∩ x = ∅ ) ) . {\displaystyle \forall x\,(x\neq \varnothing \rightarrow \exists y(y\in x\ \land y\cap...
violates this one no?
 
There are sets A such that $A\in A$.
 
if you're in zfc, it violates regularity i think
 
yeah, you're right. ZFC
We always use ZFC.
2
thanks a lot shintuku. : )
 
np!
 
1:21 AM
My profile on mse reads 'visited 961 days, 499 consecutive'.
 
impressive
 
 
1 hour later…
2:45 AM
@robjohn Hi robjohn! I think $A$ can take both real and complex eigenvalues depending on the entries of A (the first column of A).
If $x$ is an eigenvalue of $A$, then $1+x+x^2$ is an eigenvalue of $I+A+A^2$
?
If $x \in \Bbb{R}$, then $1+x+x^2 > 0$
hmm tricky when $x \in \Bbb{C}$
 
@BAYMAX yes
What is the relation of the eigenvalues to the determinant?
 
Indeed, determinant is product of eigenvalues and hence if $x$ is a real eigenvalue of $A$ then det($I+A+A^2$) > 0
I am thinking now what would happen if x is a complex eigenvalue?
If it is compex we cant comparte them with inequalities
I think we have a problem when eigenvalues of $A$ are complex? @robjohn
 
3:31 AM
@robjohn I think even if eigenvalues of A will be complex .. but they will occur in complex conjugate pairs and hence
same for matrix $I+A+A^2$, so when we multiply the complex conjugate eignevalues we get psoitive number
hence $det(I+A+A^2) > 0$
The picture might be different in the case when entries of $A$ are no longer real numbers
 
4:08 AM
yes, even the 1x1 case is different then
 
4:45 AM
How was sushi? Any Munchkin sashimi?
 
5:28 AM
@robjohn Well, $\det A=\det A^top=\det(-A)=-\det A$ does that
 
ted: it was great. munchkin opted for PB&J.
 
5:44 AM
*$A^\top$
Missed a slash
 
6:38 AM
Fun reading
 
i don't like the font he's using for section titles
really bad match with computer modern for main text
also lol
 
Haggstrom should try Stack Overflow if she wants to see what top class arrogance is about.
Never realised there was a Zermelo Football Club
 
6:55 AM
the weird ego shit is by no means confined to the internet, it's basically a small but potent mass of people in any math department ever
nice to see a jeremiad focused like a laser beam on MSE tho
 
he must have been suspended at some stage for arguing...
 
i'm really wondering about that font choice for the sections. that's not a default option. that was done on purpose
 
looks like default latex to me?
 
for "1 Introduction" ?
they've been fiddling with package defaults since i used them, if that's the section font
 
This isn't the first time I've heard someone say they dislike Stack Exchange. Not to discount their experience, but I've personally never had a bad experience with it
Then again, all the questions I ever asked here have been low-stakes idle curiosities and not, say, help with a class I'm taking
My frustration would be a lot higher if the situation were, "I need to understand this concept before the test but the internet is being unhelpful"
 
7:02 AM
for me the high point was being able to answer a question that helped my daughter. not much return, one answer in 3 years.
the low point was my suspension
 
I have used Stack Exchange to help me with my homework, but through finding answers that have already been written rather than creating new posts
All of my homework questions have been asked in the past
 
of course, the suspension is now something i am proud of
 
How did you get suspended
 
people tend to find what they seek on internet stuff. if you want to take offense or get into fights you can definitely have that experience. i don't think it's that probative of any particular forum or subject matter
MSE is better than the cesspit of facebook or twitter
 
@AkivaWeinberger i was arguing with someone about the closures of some questions.
they accused me of being a rep hunter when, in reality, all i want is the coveted MSE jumpsuit
as granddad knows
 
7:05 AM
The whatnow
 
ongoing silly joke
 
i have a friend in real life who is sometimes a lot, and she once complained about twitter. i asked what the issue was. she has thousands of followers, and quote-tweeted some random person to dunk on him on some legal issue, which she was unfortunately wrong about, and people were replying and telling her that
it was an awkward moment for me, because she is a real friend
 
when you hit a certain rep you get a t-shirt & a mug (or at least you used to, now i think its crotchless underwear or something)
 
i said "there's no nice way of saying this, but you're wrong, and it's not 'mansplaining' for people to think that it's messed up that you dragged some random person up from the depths of twitter for anticipated mockery only to be wrong about what they were right about"
our friendship survived it but she stopped tweeting about law stuff for about two weeks
 
@leslietownes Oh I avoid that by not having a spine
 
7:08 AM
modern society likes perfect people
 
I have opinions but I've generally decided that Twitter does not need to know them
 
i am not modern
 
twitter enables the worst aspects of 'call-out culture'
i'm not sure i've ever tweeted text
i sometimes tweet images
 
my brothers call twitter megaphones for morons
when i see twitter i think of noam chomsky's warnings
 
akiva you should tweet your two arcs = three arcs
i'll QT "this dude doesn't know the difference between 2 and 3 lol im dead at how dumb this dumb dumb is"
 
7:10 AM
i'm down with that, whatever been down with something means
 
@leslietownes I did, no one cared lol
 
despite following your brother and a number of math accounts i don't think i've ever seen your tweets on my feed
you need to raise your twitter profile
 
I actually got a decent amount of attention (39 likes) on my tweet explaining Niven's proof of the irrationality of pi
 
@copper_hat
 
which is odd because in retrospect I think I explained it really poorly
 
7:12 AM
surprise usename
zero content, except for Bart whines
and replies to bobert
 
haha i love these bart whines
 
passive aggressiveness at its best
i used to write letters and have some impact
speaking of passive aggressiveness, i only ate one slice of little star in a totally ineffectual protect.
wow, johnny nash is playing on the radio, i can see clearly now
awesome uplifting song, almost like mufasa
 
i honestly wouldn't mind higher prices. i do object to 'surcharge' on principle. it's too close to a bait and switch
most of my family work in fairly menial jobs in the food service industry, it's important that they be paid well
 
ever since pushed back on the $600/hr para in 1999, its all bait & switch
 
but don't have a section on your webpage where you make it out like Government Regulations are the reason for you having to pay people to work at your business
haha we bill professional staff at our firm between that and $800/hr, but this is 2022
and they're very professional
 
7:17 AM
luckily my friends are epicures, so the salad (misnomer) was totally awesome
as was the steelhead mousse
that's the problem with wealthy friend lawyers
 
clients do push back on some stuff but not the hourly rates of senior paralegals
more like, my rates
haha and our firm tried to start charging clients x/page for scanning documents even when there was no printing or other outlay of anything. that went over like a lead balloon
 
yeah, but $600 is ok for real legal advice, but for photocopies...
besides, most of what we needed was boilerplate
AOI, byelaws, etc
 
yeah that is cookie cutter stuff
 
wow, i've hit the whine button already
 
anything outside of a legal dispute where you might lose a zillion if you lose the fight is overpriced
but i'm not overpriced
 
7:22 AM
i'll tell you what i want, what i really really want
i seem to have typing turets
let me say, steelhead mousse is pretty close to what i imagine crack is like
i almost licked my friend's bowl
i am trying to imagine what a one party America would be like
 
with crack, you would have licked your friend's bowl
 
well, his wife was looking... seemed weird
 
we already have a one party america. neo-fascists, and boomer conformists who triangulate and pivot to whatever the 'center' is as determined by neo-fascists
it's going to be ugly for a while, at least until all of the baby boomers die of old age
 
we have the neo-puritans
apparently squeeze.com is where it is at, according to 101.3.
i leave the radio on as a low budget burglar deterrent
haven't got around to extinguishing it yet
 
is there a lot of that where you live?
in my old neighborhood in oakland, parked cars were free for the taking but nobody broke into houses
 
7:29 AM
not on our street, fingers crossed
 
one time when my now-wife was visiting me, they broke into her car in front of my building, and they ended up leaving a knife in the car
so after insurance we profited from the exchange
 
to be fair, we have little in the house worth stealing
unlike cyril ramaphosa who had $4m stashed in his furniture
 
we were broken into a few times at our old home and they didn't take anything
haha
imagining the disappointment as they see our old TV, damaged furniture, no.... gems? gold? what were they expecting?
 
i just wear army surplus pullovers as a deterrent
 
it was very much an "if we had money we wouldn't be living here" situation
 
7:34 AM
probably some of the irish guy thing going on
some friends will not believe me when i say i have no guns in the house
 
our new neighborhood is almost completely free of crime
there are a few uninhabited houses that people broke into about a year ago. it was obvious nobody was living in them and so naturally some people decided to see what was inside
i'd probably have something in the house if we still lived where we used to. we had a few close calls where the cops would not have come in time if the people hadn't run off
 
a gun might be good as a visual deterrent, but can backfire in many ways
 
with a kid in the house i wouldn't dream of it
 
a expandable baton is probably the most effective non lethal deterrent.
but one needs to know how to use it
and the bigger issue for most is knowing when to use it
 
i pepper sprayed someone off of our front yard once at the old place, but had they been armed it would have been a different story
i'm glad i had enough money to move to a better neighborhood
 
7:43 AM
i have had surprisingly few contacts in the usa.
i think they probably took a look and thought 'unstable, better leave now'
my first was two guys in mcdonalds on spa & university ave in berkeley tried to steal my badck pack when i was trying to help them with directions
to help them more i reached for my backpack and found the other guy pulling my pack away
omg, i went live
they left pretty quickly when the realised the mouth was not going to stop
 
it's always something on san pablo
 
i think the mcdonald's manager (my next target for not calling 911) was my next target
this was in 1983, i was still a naive young fellow
 
the last time i saw a street hooker was on san pablo
 
generally i try to get them to a side street first
jk
i used to meet some ladies when i rollerbladed down spa
one was willing to trade favours for my blades
but how would i get home
they were funny, tough & sad all at the same time
 
the one i saw just looked cold
it was november
 
7:52 AM
my blading was generally in the dry months. had a few folks spit at me, some stones
but mostly good
i'm off, good night!
 
goodnight
 
 
1 hour later…
9:08 AM
@leslietownes I never saw a street hooker. All I've seen were hanging out near forests
Sometimes when I travel to my sister, there is this forest and prostitutes like to stand there. Actually that's the only time I ever see hookers
 
what does it mean to say two vector spacecs are " Abstract isomorph"?
 
@MadSpaces who knows
 
Can i block this dude please? everytime i ask a question he just replies with nonesense
nothing more than an annoying troll
If you have nothing to say why bother write dude, i am too busy for such child stuff, bye. jesus christ get a hobby or something
 
You got pretty emotional there. It's funny that you say those kinds of things to people older than you
I would never disrespect you, and I do want to help if you even care to elaborate on what you're saying. I understand you might have some problems, but I'm not a person to take them out on
 
9:27 AM
@MadSpaces if you click on any user's avatar there is a "ignore" option...
Also, "hide posts."
Actions
So yes, you can block any dude you please.
 
@Jakobian by the merit of being older, you under no circmustance gain my respect, you gain it by a being a decent person, irrelevant from your age. and since so far your replies have been from my prespective just trolling. i am going to have to ignore you. do not take it personally
@user4539917 thanks
 
You do what you want
 
np
☮️
 
10:02 AM
(•‿•)
 
10:24 AM
Is there a way to create links more easily in chatrooms?
Usually I use the "[]()" format whenever I want to create a link. But when I want to create 2-3 links, it becomes quite hard.
 
10:34 AM
I would just post them on separate lines.
But yeah, if you try and squeeze them all in one post it gets hard.
 
Are the topologies of uniform convergence on two different saturated families of sets, different topologies?
Hmm... actually, yeah. This makes sense. Because a saturated family of sets, makes it so that all the translations of polars generate the topology, I think.
Actually maybe not. I'm just wondering why in the Mackey-Arens theorem they are actually compact, even if one of the direction of the proof doesn't use that.
I'm guessing they don't have to be, but they always can be chosen so that they are.
 
 
1 hour later…
12:27 PM
does anyone know why this is true?
here $\mu_{w}, \overline{\mu}$ are probability measures supported on a compact subset of the plane (supported in the same compact subset)
and $w(z)$ is some nonnegative, upper semicontinuous function defined on this compact subset, where it happens to also be bounded from below (and above by compactness)
i think its something obvious, but the only thing I can think of is that the author implies if $\mu$ is some finite signed measure with total mass zero, then integrating anything bounded against it gives you zero
but im not even sure if what I just stated is true
actually im sure what I just said is false, it makes no sense for a lot of reasons, one being that if $\mu$ were such a signed measure, having total mass zero does not imply its total variation $|\mu| = \int \chi_{P} - \chi_{N} d \mu$ is zero
here $(P,N)$ is a hahn decomposition of $\mu$
there is a lot more context to this problem.. I was hoping that I was just being oblivious to some simple fact
the context is the following: $\mu_{w}, \overline{\mu}$ are both probability measures supported in $\Sigma_{\epsilon} = w^{-1}[\epsilon,+\infty)$ (which is compact) that minimize the quantity $\int \int \log[|z-t| w(z) w(t) ]^{-1} d \mu(z) d\mu (t)$ , over all probability measures supported in $\text{dom}(w)$, $w \geq 0$ is USC and 'admissable', in the sense that $w^{-1}(0,+\infty)$ has positive capacity, and $w(z)|z| \rightarrow 0$ as $|z| \rightarrow \infty$ if $\text{dom}(w)$ is unbounded
everything is taking place in $\mathbb{C}$ as well
positive capacity here can just be taken to mean that $w^{-1}(0,+\infty)$ contains some compact set which is the support of a probability distribution $\mu$ whose 'energy' $I(\mu) = \int \int \log[|z-t|]^{-1} d\mu(z) d\mu(t) \in (-\infty,+\infty]$ is finite
 
12:46 PM
@porridgemathematics Can't you go like, $$\int \log[w(z)]^{-1} d(\frac{1}{2}(\mu_w-\overline{\mu}))(t) = 0$$ and similar for the other one, use Fubini?
 
omg lol, indeed you can
thanks a lot! @Jakobian
I think thats exactly what youre supposed to figure out
so the result I was too stupid to see was that integrating a constant against a signed measure of total mass zero is always zero
 
 
4 hours later…
4:47 PM
@BalarkaSen Let $A_1$ and $A_2$ be two loops in 3-space such that their projection onto the $xy$-plane is regular. Prove that the total number of crossings is even; prove that if $A_1$ crosses over $A_2$ an odd number of times (and therefore under an odd number of times) then they are linked.
(This is easy)
@BalarkaSen Conclude that the complete graph on 6 vertices, $K_6$, is intrinsically linked - that is, every spatial embedding of $K_6$ contains a nontrivial link.
Notation: let $\omega(A_1,A_2)$ refer to the number of times $A_1$ crosses over $A_2$, mod 2. (This is defined even if $A_1$ and $A_2$ are arbitrary spatial graphs whose projection is regular, not just loops. If they are loops, $\omega(A_1,A_2)={\rm lk}(A_1,A_2)$ mod 2.)
 
5:46 PM
In the zonoid algebra, stemming from the monoid structure of zonoids in a 3-dim. real vector space, there is a notion of being able to "multiply" zonoids. Does anyone know if there is closure? That is, does the multiplication of two zonoids yield another zonoid?
 
What's a "Zonoid"?
 
it's a shape built from the Minkowski sum of vectors
 
I get it conceptually now, I guess. Seems like something I'd have to get into the details to really understand. Won't help you here
 
6:07 PM
I was gonna say I thought it was called a "zonohedron", but apparently that's only in 3D
 
I will start by reading about how they prove that zonoids form a monoid
 
Thinking about the phrase "growth mindset"
I wonder if maybe it's something a teacher (or a learning environment) provides for a student rather than a student just having
so that if the student doesn't have a growth mindset, it's the fault of the educators
I dunno, I'm just rambling
 
Hi guys, can someone help me in this one?

I need to prove that $$P(X = k) = \frac{\theta}{(1 + \theta)^{k+1}}$$, where $$X ~ Poisson(\lambda = y) and Y ~ Exponential(\theta) $$

I know that $$P(X = k) = \frac{e^{-y} y^k}{k!} $$ and $$ P(Y = y) = \theta \cdot e^{-\theta y} $$
I don´t understood how to relate these two formulas.
 
6:38 PM
0
Q: Check if $\Bbb{Z}[1/2]$ is local or not.

Wave Let me consider $R=\Bbb{Z}[1/2]$. I need to check if it is local or not. I know that $R=\{p(1/2):p\in \Bbb{Z}[X]\}$ but I don't see where to start when I want to show if there is a unique maximal ideal or not. I know that the the maximal ideals of $\Bbb{Z}[X]$ are $(p)$ where $p\in \Bbb{Z}[X]$ ...

can someone help me here?
 
Let $K$ be either $\mathbb{R}^2$ , $\mathbb{S}^2$ or $H$(Hyperbolic space) . If $c$ lies in a geodesic segment $[ab]$ in any of these spaces, then does it follow for any other $p$, we must have $\angle apc + \angle bpc = \angle apb$?
For $K=\mathbb{R}^2$ this is clear
 
@leslietownes perfect thanks!
 
7:20 PM
@AkivaWeinberger I think like always, the fault lies at both sides
 
 
1 hour later…
8:29 PM
They are inevitable, but random, inexplicable, and unexplained downvotes are disappointing.
I note that inexplicable sort of explains the unexplained part.
 
i agree
 
9:09 PM
Hi guys, can someone help me in this one?

I need to prove that $$P(X = k) = \frac{\theta}{(1 + \theta)^{k+1}}$$, where $$X ~ Poisson(\lambda = y) and Y ~ Exponential(\theta) $$

I know that $$P(X = k) = \frac{e^{-y} y^k}{k!} $$ and $$ P(Y = y) = \theta \cdot e^{-\theta y} $$
I don´t understood how to relate these two formulas.
 
9:27 PM
Hi
If you are given $x<3, x>5$ to raw on a number line is it correct to mark both the ranges for $x<3$ as well as $x>5$?
 
9:45 PM
there is no element that is smaller than three and bigger than 5 at the same time...
 
it's all about what ", " means.
i think interpreting it as "or" makes sense in context. otherwise, as mad notes, there's nothing to draw.
 
I'm reading the proof of Mackey-Arens theorem, and I have trouble to see why $|f|\leq 1$ on $G^\circ$ implies that $f\in G^{\circ\circ}$
This is when they're trying to prove that $f\in E^*$ implies $f\in F$. But isn't $G^{\circ\circ}$ by definition a set of $g\in F$ such that $|g|\leq 1$ on $G^\circ$?
 
Hmm yeah, @MadSpaces and @leslietownes
That means, we can't take "," to be "or" always, but it should be defined in the question, isn't it?
Usually when we solve a quadratic equation, say, (x-1)(x-2)=0, we write x=1, x=2 and there it means x=1 or x=2
like that usually doesn't "," mean "or"?
 
there is no universal convention about what ", " means.
it should have been provided in the question.
 
10:01 PM
Hmmm, okay
Thanks a lot @leslietownes and
@MadSpaces
 
So I think, we can treat $G^{\circ\circ}$ as the bipolar for the dual pair $\langle E, E^*\rangle$ instead, and then it makes sense to say that it equals to $G$. I was just stuck thinking about $\langle E, F\rangle$ instead.
 
jakob, which book are you working through?
 
The Banach space theory: The Basis for Linear and Nonlinear Analysis
the one with a bunch of Czech authors
 
oh, hm
have you looked at pietsch, history of banach spaces and linear operators
it is a good book
 
So far I looked at Flaming, Jaminson - Isometries on Banach spaces, Lindenstrauss Tzafriri - Classical Banach spaces and the one above. So far I like the one I'm reading so I'll be sticking to it
Lindenstrauss, Tzafriri seems a little too hard for me right now, too much things left untouched
Same with the book about isometries
 
10:14 PM
i don't know a huge amount of the general theory. i learned from handwritten course notes and papers mostly. the pietsch book is funny. he is somewhat sarcastic in a characteristically german way where you can't tell if he's actually trying to make fun or only giving that impression.
 
I enjoy reading books from cover to cover, and since I'm still a student, I can find time for it
 

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