Mathematics - 2018-05-05 (page 2 of 3)

Leaky Nun

15:07

RIP

Alessandro Codenotti

(removed)

0celo7

@AlessandroCodenotti I have a stupid question for you

It's probably retarded

skull

(removed)

Alessandro Codenotti

I hope I have a non stupid answer for it

0celo7

@AlessandroCodenotti Clearly the class of all compact metric spaces is not a set, right?

Or maybe it is a set!

Balarka Sen

15:14

can't you create a Russell's paradox like situation by imposing a compact metric on that set

i dunno set theory

Alessandro Codenotti

Hmmm, that's not a stupid question at all, I don't see a simple way to make compact metrizable spaces of arbitrarily big cardinality

Mike Miller

a compact metric space has i think cardinality bounded by that of the reals

Alessandro Codenotti

Ah, wait, all compact metrizable spaces are the continuous image of the cantor set

Mike Miller

but it's not clear to me if the class of sets in bijection to a quotient of [0,1] are a set

irritating category stuff

Balarka Sen

@MikeMiller ah

0celo7

15:16

There's a theorem that says the space of compact metric spaces with Gromov-Hausdorff distance is separable. But that supposes it's a set to begin with.

Mike Miller

no it doesn't, you've quotiented by isometries

Balarka Sen

It's the set of all isometry classes of compact metric spaces

Well flip

I should just retire

Mike Miller

no, we should just merge into one being, with limbs flailing everywhere, begging for death

2

0celo7

So it's clear that the isometry classes form a set?

Mike Miller

and if you're willing to do that this is clearly true: by Alessandro's result you're asking about the data of $C$, an equivalence relation on $C$, and a metric $C/\sim \times C/\sim \to \Bbb R$. There is at most, say, $\Bbb R^{\Bbb R}$ of all of this

skull

15:20

i don't want to beg for death :(

0celo7

Yeah, given Alessandro's result it's clear. Although there's probably something easier.

Alessandro Codenotti

All compact metric spaces are separable and all Polish (separable+metrizable) spaces embed into $[0,1]^{\Bbb N}$ so you have at most as many metric compact spaces up to isomorphism as subspaces of $[0,1]^{\Bbb N}$, which is a set

0celo7

In terms of sets, you can only have finite sets, N, or R

And then the metrics are functions on these guys

Mike Miller

yeah, I'm phrasing it with the intent of avoiding continuum hypothesis

if you allow choice but not the continuum hypothesis you could say "functions on subsets of R"

0celo7

Ah yeah you can't say what the cardinality is from my argument but it does give you a set

Alessandro Codenotti

15:23

ismorphism was supposed to be homeomorphism

0celo7

So the next question is, is the proper pointed category small? Probably not. You can have arbitrarily large bois in there, presumably.

Alessandro Codenotti

I don't even know what those words mean, sorry

0celo7

Proper means closed balls are compact

Pointed means you consider $(X,x_0)$

Is the collection of these guys a set?

skull

bois means boys

0celo7

bois means metric spaces

Balarka Sen

15:26

@MikeMiller horrible

Mike Miller

i wanted to tag my message with that

Alessandro Codenotti

So, the proper bit is no problem since it actually decreases the number of spaces we're interest in, right?

Mike Miller

but it wouldn't let me

Balarka Sen

The Fly - Deleted Baboon Scene

►

0celo7

@AlessandroCodenotti Yeah

Alessandro Codenotti

15:27

Ah, wait, are you considering metric spaces or compact metric spaces now?

0celo7

Regular metric spaces. Oviously compact ones are proper.

Mike Miller

that's attractive

0celo7

Regular not in the topological sense

Alessandro Codenotti

lol I figured

Mike Miller

@0celo7 No, I think this is still ok. Now a proper metric space is defined by the diagram $B_1 \to B_2 \to B_3 \to \cdots $

So you need a $\Bbb N$'s worth of compact metric spaces, and maps between them

Alessandro Codenotti

15:28

(there aren't many metric spaces which aren't regular anyway :P)

Silent

@LeakyNun , why is this true: Let $g(x)$ be polynomial with real coefficients and $\alpha\in\Bbb R$, then reminder of division of $g(x)$ by $x-\alpha$ is $g(\alpha)$?

0celo7

@AlessandroCodenotti Yeah I realized

@MikeMiller Hmm, does this work for any first countable space?

I guess you need here that each ball is compact so we can use the previous argument.

Secret

Balarka Sen

looking back, The Fly was probably Croonie's most 80s movie. The aesthetics is super over the top like most 80s films

(still a great movie)

Secret

Mercio: If anything, the calculation is not very illuminating

Mike Miller

15:33

Not gonna think that hard about this :) The claim is that the topology on the union is the direct limit topology, I guess, and there is probably a name for this property

Compactly generated, maybe

Balarka Sen

@Secret What are you trying to do

Mike Miller

Wikipedia says that first-countable is compactly generated, so yes

Secret

@BalarkaSen Mercio's ellipse problem

turns out it is a lot messier than I thought

Balarka Sen

what is the problem

Secret

He is trying to find the condition of the coefficients $a,b,c$ of the ellipses such that given slopes of tangents $\pm i$ all intersections are real

mercio

15:36

no I'm asking where do the tangents of slope $\pm i$ to a real ellipse intersect

skull

@BalarkaSen have you seen the original fly movie?

mercio

and especially the real intersection points

Balarka Sen

@skull Yup, it's a good movie

skull

Yup

Mike Miller

There was another?

Secret

15:38

My 4th attempt computed:

Mike Miller

Oh, a 1958 one. Interesting

skull

the original is in black and white

mercio

maybe i shouldn't have asked for the real intersection points because you're so much fixated on doing complicated things because of it

Mike Miller

@skull Sure, flies only evolved color in the early 70s

skull

apparently

the raging bull was done in black and white

Secret

15:40

$$\text{Intersections} = \left(\frac{m(z_i-z_j) + w_i + w_j}{2}, \frac{z_i+z_j}{2}+ \frac{w_i+w_j}{m}\right)$$ for $(i,j) \in \{(1,2),(3,2),(3,4),(1,4)\}$. And $z_i,w_i$ is given by:

Alessandro Codenotti

@MikeMiller Before the whole world was in black and white so they didn't need colours to camouflage

mercio

maybe you can pick $(i,j) = (1,2)$ and replace $w_1,z_1,w_2,z_2$ with their expressions and see if any simplification happens ?

Secret

$$(z_i^2,w_i^2) = \left(\frac{a^4c^2}{\left(\frac{a}{b}\right)^2+\left(\frac{b}{m}\right)^2},\frac{b^4c^2}{m^2 \left(\frac{a}{b}\right)^2+b^2} \right)$$

Ok will do that next:

Leaky Nun

@Silent division algorithm

mercio

(and please don't use nested fractions they are ugly)

Mike Miller

15:45

oh god agree

Secret

Actually I have no idea how to take the squareroot of this thing since squareroots in complex numbers are not necessary of the form $\pm$ stuff

and given how ugly it is, converting that to polar form without computer will be hell

Mike Miller

I think this convinces me There Is A Better Way

Balarka Sen

So we're looking at the picture of four lines of slope $\pm m$ that are tangential to a given ellipse?

If I apply an affine transformation to it to make the ellipse into a circle, you get that the cyclic product of the proportions of the tangent segments (given by the tangential points) is 1

If that counts for anything

Mike Miller

I think he was passing to $\Bbb C^2$ somehow first

Balarka Sen

Oh.

Mike Miller

15:54

The tangent (complex) lines to a (complexified) real ellipse for which the slope is $\pm i$

And I guess affine transformations break that particular form, unfortunately

Balarka Sen

Yeah I don't know about that

Mike Miller

Though I would still expect some sort of symmetry...

Secret

Sanity check: A real ellipse is an ellipse in $\Bbb{R}^2$ or an ellipse in $\Bbb{C}^2$ with real coefficients?

Mike Miller

I think an ellipse with real coefficients, for which you think of the solutions in $\Bbb C^2$

I was confused about that too though

Balarka Sen

I am not going to complexify my conics

I was born in $\Bbb R^2$ and I'll die in $\Bbb R^2$

5

Mike Miller

15:57

Don't let Ted see you

Secret

I need to lookup how to do principle root stuff, I am never familiar with them. I can see some symmetry arguments such as how there are only two possible magnitudes for the y coordinate and x coordinate no matter what m is, but I don't know how to formalise that since my current knowledge base don't know of more advanced tools

Balarka Sen

he's going to give it to me in the neck for hating on complex geometry

mercio

who is hating on complex geometry D:

but yeah i said real ellipse to highlight that its coefficients were real

and not complex

(since i'm playing with complex slopes)

Balarka Sen

what a sloppy complexity you've landed yourself in

Secret

If $a,b,c$ are real, then in theory that would help simplify the workings as the only thing that will be affected by the conjugate is m. But before the squareroot of that big scary expression is taken, I cannot be sure yet

EWWWWWWWW

mercio

16:04

just fix one square root of the thing and don't try to open it up

and yeah, a b c are real

Secret

Well, the denominator is still going to be pretty ugly but here goes...

e.g. $$(z_1,w_1) = \left(\frac{a^2bcm}{\sqrt{a^2m^2+b^4}},\frac{b^3c}{\sqrt{a^2m^2+b^4}} \right)$$

ok that means something...

mercio

no this fails dimensional analysis

$a,b,z_i,w_i$ are length, $m,c$ have no dimension

so you are adding areas with hypervolumes inside the square root

Secret

hmm....

ah great, I pulled out the $1/b^2$ on the right end of the denominator but forgot to do the same on the left end when calculating $w_i^2$. Let me quickly fix this (the tangent stuff looks fine meanwhile)

$$(z_i^2,w_i) = \left(\frac{a^4c^2m^2}{m^2a^2+b^2}, \frac{b^4c^2}{m^2a^2+b^2}\right)$$

Thus:

$$(z_1,w_1) = \left(\frac{a^2cm}{\sqrt{m^2a^2+b^2}}, \frac{b^2c}{\sqrt{m^2a^2+b^2}}\right)$$

mercio

16:29

this looks correct

Secret

that's one root, now to find the other root since the i,j in the expression for the intersections means we are adding or subtracting two different roots for each coordinate

(which I am not sure it is the minus one since m is a complex number)

that's weird... why did I only got two of the 4 possible roots

each coordinate should have 4 different values, and they form 2 $\pm$ pairs

Secret

16:55

actually, does a rectangle circumscribing an ellipse has to be upright?

Jannik Pitt

If I proved something like $pi(x^n)>\frac{2^n}{48n}pi(x)$ and I'm not sure whether all of my steps are valid, what's the quickest way to test such an inequality numerically?

Secret

ok I have no idea why I am missing two of the roots (even for the case of real slopes). I will need to think about it later (or perhaps you should find someone else in the meantime)

mercio

I'm not sure what you're trying to say with your 4 roots

roots of what ?

Secret

Looking at the diagram, for the real case, there are 4 possible values for the coordinates of the tangent points:

mercio

there are two tangent points for each slope

Secret

17:05

(p,r),(q,s),(-p,-r),(-q,s)

So I got p,r and -p,-r but not q,s for some reason...

mercio

there are two tangent points with slope m and two tangent points with slope -m

Secret

17:32

Actually, does the condition for lines $m_1 \cdot -\frac{1}{m_2} = -1$ holds for complex slope?

cause if it were, we would have a weird case where $i \cdot -\frac{1}{i} = i \cdot i = -1$ thus the line of slope $i$ is self normal

Anonymous

@Secret How do you define "complex slope"?

Secret

uh, you just do the same thing as equation of lines, except with complex numbers

Anonymous

Well, are you just considering $\Bbb{C}$, or $\Bbb{C}^2$ ?

Secret

$\Bbb{C}^2$

Anonymous

Hmm. Then you can't represent it like a 2D plane. It's 4 dimensional over the field of real numbers.

Anonymous

17:36

You can't draw it on paper

Secret

yeah, but mathematically a complex slope is well defined. I am not terribly familar with its rigorous definition though

mercio's question involves complex slopes and I felt I kept bumping into special cases

Anonymous

@Secret It's defined for the Argand plane. Which is $\Bbb{C}(\Bbb{R})$

Anonymous

I don't know how you would define a slope in $\Bbb{C}^2$

mercio

well yeah a line of slope i is orthogonal to itself

Silent

@0celo7, how do we know that $7x+7\equiv0\pmod n$ for no $n$ except $n=1,7$?

mercio

17:39

but I'm not sure why you're bringing orthogonality in the picture

Anonymous

In the argand plane complex slope is $w=\dfrac{z_1-z_2}{\bar{z_1}-\bar{z_2}}$

mercio

Blue we are talking about the complexified real plane

like we do all the real geometry normally

and we pretend things make sense if we plug complex numbers into it

Anonymous

And the equation of line is $a\bar{z}+\bar{a}z+b=0$

mercio

for example a line of slope i would be y = ix

Secret

Anonymous

17:40

@mercio That means you are working in the Argand plane! Not $\Bbb{C}^2$ as Secret said

Secret

somehow, I am missing out 2 solutions of z here because the $w'=\pm m$ got squared

mercio

no the Argand plane is when you draw $\Bbb C$ as a plane

Secret

I think I need to refresh my brain a bit before continue to hammer at this problem

Anonymous

Could you first please define " complexified real plane" properly? That phrase makes no sense in mathematical terms at all

mercio

how many tangent points do you get with a single slope m ?

Secret

17:42

it should be a pair (p,q) and (-p,-q), thus two for each coordinate

Annabelle

Hi um, I'm trying to explain something to someone and can't figure out how to do so.

mercio

well you know how the real plane is $\Bbb R^2$ and it things like lines whose equation is $y=mx+b$ with $m,b \in \Bbb R$ and things like circles whose equation is $(x-a)^2+(y-b)^2 = r^2$ with $a,b,r \in \Bbb R$

well replace $\Bbb R$ with $\Bbb C$ everywhere

Annabelle

Why can we use the fact that both 80 and 125 are multiples of 5?

mercio

yup secret

Secret

but somehow, a slope of -m also give me those two points (p,q) and (-p,-q) while I should be expecting a pair (r,s) and (-r,-s) where r=/=p, s=/=q

mercio

17:43

so you have 2 tangent points for slope m right ?

well no, (p,q) can't be both a tangent point with slope m and with slope -m

Secret

but I only got two solutions when solving for w. The only thing I found puzzling is I squared the z in terms of m in order to plug into the ellipse equation to solve for w

mercio

I think you should get (p,q) and (-p,-q) with slope m and (-p,q) and (p,-q) with slope -m

Anonymous

@mercio Still doesn't make sense unless you define what field you're working in, for the latter. You can't draw $\Bbb{C}^2(\Bbb{R})$ on paper, and slope in that case doesn't make sense.

mercio

or something like that

Anonymous

What's the original problem statement anyway?

mercio

17:46

@Blue well it's like doing an extension of scalars from $\Bbb R$ to $\Bbb C$, so we are working over the field $\Bbb C$

the original problem is taking a real ellipse (an ellipse with real coefficient that is not a circle), and look at its tangent lines of slope $i$ and $-i$ and then look where they intersect

Anonymous

@mercio You can't represent all elements of $\Bbb{C}$ on a single line or a single axis as you can do for $\Bbb{R}$

mercio

I never said you could

however, you can still see some things for example any complex line whose slope is not real has a real point

:)

Anonymous

And hence converting your geometrical problems in the real plane to $\Bbb{C}^2(\Bbb{C})$ doesn't make much sense. Slope is not properly defined for such cases. Anyhow, I can't say without seeing the actual question.

mercio

can't I define a complex point by saying it is a pair of complex numbers ?

Anonymous

Yes, but you can't draw that on paper

mercio

17:51

yeah but hypothetical aliens living in a 5 dimensional universe can

so why should they be able to talk about it and not us

Mike Miller

Slope of a plane in $\Bbb C^2$ means the unique complex number $z_0$ so that your plane is parallel to $w = z_0 z$, unless the plane is "vertical", $z = c$, in which case we could call the slope "infty"

Anonymous

Well, why would you even do that I'm not sure. Almost all $\Bbb{R}^2$ problems can be solved in the real plane itself without using any complex numbers.

mercio

because lines of slope i are really nice

Mike Miller

This fits with one natural definition: slope of a line through the origin in $\Bbb K^2$, $\Bbb K$ a field, should by definition be the corresponding point in $\Bbb{PK}^1$; we identify $\Bbb K$ with the non-infinite points by $x \mapsto [1:x]$

I dunno, if one wants to dick around with complex numbers, that's their natural-born right

Anonymous

@MikeMiller mercio is talking about slope of a line

Anonymous

17:54

I think

Mike Miller

I should have said "complex line" instead of plane above.

mercio

yeah but a line over the complex number looks like a plane over the reals

Mike Miller

I mean a real dimension 2 subspace which is closed under complex multiplication.

Which is a complex 1-dimensional subspace.

The fancy projective space language is just what I said there but over an arbitrary field

Anonymous

Fine. But I don't see any need of converting a nice 2D problem into such ugly 4 dimensional things

Mike Miller

I thought the problem he started with was something he made up with complex numbers in the statement

mercio

17:57

yeah because the answer happens to be really nice >:(

Mike Miller

which is not a 2-dimensional problem, just one you get by doing formal mathematical trickery to what we might call a 2-dimensional image

mercio

even though they are ugly real points

Mike Miller

formal mathematical trickery is fun sometimes

Silent

@Semiclassical,

18 mins ago, by Silent

@0celo7, how do we know that $7x+7\equiv0\pmod n$ for no $n$ except $n=1,7$?

Semiclassical

if something is zero mod n, what does that mean in terms of divisibility?

Silent

17:59

@Semiclassical that thing has reminder 0

Semiclassical

That's not a statement about divisibility.

If x is 0 mod n, what do we know about x?

Silent

hmm, that thing is divisible?

@Semiclassical x is divisible by n

Perturbative

Hey everyone

Semiclassical

Right. So in this case we've got 7(x+1) divisible by n

Silent

n divides x

how do we know, eg, 6 does not divide (x+1)?

Semiclassical

18:00

Now: I'm guessing what you're really after is that 7x+7=0 is only true for all x when n=1,7

Because otherwise the statement is obviously false. If x=-1, for instance, then 7x+7=0 and every n will make the equation be satisfied

Silent

@Semiclassical oh, for all! that was great. thank you!

Semiclassical

np

Silent

@Perturbative hi

Perturbative

Quick question, can someone give me an example of a polynomial ring $R[x]$ over $R$ and a polynomial $h(x) \in R[x]$ such that $h(x) = f(x)g(x)$ for some $f(x), g(x) \in R[x]$ but where $h(a) \neq f(a)g(a)$ for some $a \in R$

Tobias Kildetoft

@Perturbative How could that happen? Evaluation at $a$ is a homomorphism

Perturbative

18:07

@TobiasKildetoft It was in my abstract algeba test today

Tobias Kildetoft

@Perturbative Are you sure it was not the other way around?

Perturbative

I'm sure, that's the best I could remember it

Tobias Kildetoft

Well, there are no examples.

Secret

Ah, I now know what's going on:

desmos.com/calculator/boiialvagy

Slopes of $\pm m$ are special. They intersect at the axes

and hence, there are only one pair of values for the coordinates of the tangent points

which means...

$$(z_i,w_i) = \left(\pm\frac{a^2cm}{\sqrt{m^2a^2+b^2}}, \pm\frac{b^2c}{\sqrt{m^2a^2+b^2}}\right)$$

for $i=1,2,3,4$

and hence, the intersections are:

$$\left(0, \pm\frac{b^2c}{\sqrt{m^2a^2+b^2}}\right),\left(\pm\frac{a^2cm}{\sqrt{m^2a^2+b^2}‌}, 0\right)$$

Thus for $m = \pm i$ we have:

$$\left(0, \pm\frac{b^2c}{\sqrt{b^2-a^2}}\right),\left(\pm\frac{a^2c i}{\sqrt{b^2-a^2}‌}, 0\right)$$

and so for real intersections we have:

$$\left(0, \frac{b^2c}{\sqrt{b^2-a^2}}\right),\left(0, -\frac{b^2c}{\sqrt{b^2-a^2}}\right)$$

mercio

18:27

but what if $|a| > |b|$ ?

also that's not the intersections

unless you draw horizontal lines and vertical lines instead of lines of slope +/-m

Secret

At least for real slopes, the lines must intersect the x,y axes

since varying the $m$ gives a scissor like motion meaning, the intersections move along the axes

Algebraically, I only get the values (p,q),(p,-q),(-p,q) and (-p,-q)

for some p=/=q

If we have slopes $m,n$ instead, then we will have (p,q),(r,s),(-p,-q),(-r,-s) instead, which is what is actually drawn in my diagram (and I misread that as a $\pm$m pair)

Also you are right about the $|a| > |b|$ so we do have to split cases

If the above are intersections of the horizontal and vertical lines, then no coordinates will be zero

(I think the fact that in the algebraic workings using $\pm m$ and found the tangent points as (p,q),(p,-q),(-p,q),(-p,-q) should prove the scissors motion)

So...

If $|a| > |b|$ the real points are:

$$\left(\frac{a^2ci}{\sqrt{b^2-a^2}}, 0\right),\left(-\frac{a^2ci}{\sqrt{b^2-a^2}}, 0\right)$$

Balarka Sen

19:02

@Blue complexification is not an ugly operation in any sense.

it takes work to be able to functionally work with conics in $\Bbb C^2$.

but that it takes work to be able to see conics in $\Bbb C^2$ doesn't mean it's ugly :p

The motif for passing to $\Bbb C^2$ is that the classification of conics gets simpler: hyperbolas and ellipses are not distinct.

Anonymous

@BalarkaSen "hyperbolas and ellipses are not distinct" Could you explain that?

Balarka Sen

$-x^2 = (ix)^2$

Anonymous

Ah. Okay, the equations

Balarka Sen

Also, intersection number of curves sitting in $\Bbb C^2$ is more generic, because of fundamental theorem of algebra.

Anonymous

@BalarkaSen I see. But for what category of problems does it actually provide an improvement? Say I want to find the intersection points of an ellipse and a hyperbola

Anonymous

19:09

Would rewriting them in $\Bbb{C}^2$ make it simpler?

Balarka Sen

No, because (arguably) that's not an important problem.

Anonymous

Okay, then maybe you could give me an example

Anonymous

(of an important problem)

Balarka Sen

Counting number of points when intersecting curves is an important problem. If you have a degree $n$ curve and a degree $m$ curve in $\Bbb C^2$ they intersect, generically they have $mn$ points of intersection. Generic in the sense that if they don't have $mn$ points of intersection, perturbing a one curve a little bit by some small isometry resolves the issue.

This is garbage in $\Bbb R^2$

Because real curves have imaginary and complex points of intersections, which we don't see in the real plane (also the essence of the reason why fundamental theorem of algebra fails)

Anonymous

@BalarkaSen And that comes from the fundamental equation of algebra, right?

Anonymous

19:14

*theorem

Balarka Sen

It's a great deal harder to prove than fundamental theorem of algebra, but that is correct intuition, yes. Fundamental theorem of algebra -- more-or-less -- confirms that statement when $n = 1$.

Look up "Bezout's theorem"

Anonymous

This sounds similar to the homogenization of conics we do in the real plane

Anonymous

I need to read it up in more detail

Leyla Alkan

20:15

Hi all! e is larger than both c and d. Why it says e is larger than either c or d?

Fargle

An unfortunate use of language, it looks like. They're using "larger than either c or d" to mean "larger than c, and larger than d".

Leyla Alkan

Okay, I just wanted to make sure. Thanks!

BeginningMath

i have a question about the definition of independable events, would appreciate your help

if i toss a coin 50 times and define the following events:
a- getting 4 heads in the first 4 tosses
b - getting head 4 times in the last 4 tosses
c - getting a total of 20 tosses out of 50.
are a, b,c independant? i mean a and b are disjoint, but it can occur that get 4 in the beginning, 4 in the end and 12 elsewhere

so if i am asked if a,b,c are independant, what should i answer? a and b are not dependant of each other, but a and c are dependant, and b and c are dependant

Mike Miller

@Balarka We still get mn mod 2 I guess... The complex solutions appear in pairs. But that's not very helpful

Balarka Sen

yeah fair point

Fargle

20:29

@BeginningMath Note that the probability of C given A is (46 choose 16)/2^46, but the probability of C by itself is (50 choose 20)/2^50, which is not equal to it. Since P(C) =/= P(C | A), C and A are not independent.

Similarly, C and B are not independent.

However, A and B are independent; the first four tosses and the last four tosses have no causal relationship between each other. P(B) = P(B | A), and vice versa.

BeginningMath

thank you very much

Fargle

No problem.

BeginningMath

.

Leyla Alkan

@Fargle Btw, I am not yet convinced that there is no least upper bound for $\{a,b\}$ . Yes we can't tell from the diagram whether $c\gt d $ , $c\lt d $, or $c=d$ . how would you conclude from here that there is no such least upper bound? Is it because of the ambiguity here that proves there is no least upper bound?

BeginningMath

just to clarify, because my english is not that good, when you say C given a, you mean p(c | a)?

Fargle

20:41

@LeylaAlkan Yes. The least upper bound must be less than or equal to all upper bounds, so in particular it must be comparable with all upper bounds.

BeginningMath

$Fargle - when you said "he probability of C given A" you meant P(c|a)?

english is not my native language unfortunately, so i am just verifying

Fargle

21:16

@BeginningMath Yes, sorry I didn't reply quickly.

BeginningMath

thank you

Salehen Rahman

21:43

Let's say we have a set, and each element of the set has a probability attached that determines how likely is it that it will be picked, if we are to randomly sample from the set.

Let's say, we are to permanently remove one or more elements from the set, what happens to the probability distribution of the set?

My intuition would claim that we should just normalize the probabilities.

Thoughts?

skull

21:55

does permanent removal of one or more elements from the set change the probabilities of the remaining elements in the set?

Salehen Rahman

Each element is independent from one-another.

For the lack of a better word, the "bias" will remain the same.

The probability distribution should (I think) be determined by 1) the number of elements, and 2) the bias assigned to each element.

The sum of the probabilities should remain the same.

skull

so the bias is not related to the number of elements?

Salehen Rahman

I just gave some thought to it.

I think this is how the probabilities should be distributed within the set:

Each element has a bias; the probability that an element is picked is predicated by its bias divided by the sum of all biases.

Is my assessment correct, or incorrect?

skull

sorry, I have to leave

BeginningMath

22:13

can someone please check math.stackexchange.com/questions/2767597/… ?

0celo7

22:38

@AlessandroCodenotti ugh my professor sent me a note where she didn't make a distinction between $\to$ and $\rightharpoonup $

Alessandro Codenotti

22:57

That's truly evil

Good luck figuring out which should be weak and which shouldn't

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