Mathematics

12:03 AM

Hey guys, I think my statistics grading system is bugged, but I'm not entirely sure. I just need some re-assurance for my answer. I entered $9$ for the variance of the Exponential Distribution and it's not accepting it.
https://i.gyazo.com/9544a70d7b7e003abc48e412ac533a44.png
Help is appreciated.

MatheiBoulomenos

I'm assuming this is an inner regular measure. If the only closed subsets of $E_j$ have zero measure, then by definition of inner regularity, $E_j$ also has zero measure, so $\mu^*(E_j-F_j) \leq \mu^*(E_j) = 0$ @Leaky

Leaky Nun

@MatheiBoulomenos hmm...

alright

Kasmir Khaan

Hello guys

I got a question

if G acts on the set S , it also acts on the set of subsets of S

we can restrict the action to subsets of S with fixed number of elements

What does this mean ?

MatheiBoulomenos

You can define an action on the power set of $S$ via $gX = \{gx | x \in X\}$ for any $g \in G$ and $X \subset S$

Kasmir Khaan

@MatheiBoulomenos @LeakyNun help :)

hmm this was the intro of sylow theorem

Leaky Nun

12:10 AM

@MatheiBoulomenos I don't think that is what it means

Kasmir Khaan

I might add that to avoid comfusion

so from what i understood

MatheiBoulomenos

I do think that this is what it means

Kasmir Khaan

G is a group acting on a set S

G can aslo act on the subsets of set

so assuming that the order of the subsets are the same

Leaky Nun

@MatheiBoulomenos well he said "restrict"

which means changing the domain

MatheiBoulomenos

Yes, but there is an induced action from the action on S

Kasmir Khaan

12:12 AM

gU wont change the cardinality of the subset U

MatheiBoulomenos

That's correct

Daminark

Restriction here means, instead of looking at the whole power set, you note that the image of a k-element set is a k-element set

Leaky Nun

@Daminark hmm

MatheiBoulomenos

he meant "corestrict"

Leaky Nun

oh nvm, I completely missed a part

the part where it says "it also acts on the set of subsets of S"

sorry :P

Kasmir Khaan

12:13 AM

np leaky :D

MatheiBoulomenos

Yes, that was what I was talking about

Daminark

@MatheiBoulomenos He comeant corestriction

Kasmir Khaan

Anyway :)

So bascilly what the action does now

MatheiBoulomenos

I actually think "corestriction" is a useful word. It's a common thing you do

Kasmir Khaan

taking the subsets and shuffling them

is that right?

MatheiBoulomenos

12:13 AM

yes

Kasmir Khaan

the subsets of S are the "elements " here

Daminark

What does corestriction entail? I thought you were just memeing

Kasmir Khaan

Okay nice I dont know why this is usefull right now but at least i got what it means :D

thanks guys :D

MatheiBoulomenos

I'm thinking if you want to "restrict the codomain", I tend to call it "corestriction", as I reserve "restriction" for restricting the domain, but I guess that's non-standard

Kasmir Khaan

@MatheiBoulomenos how does one make sense of the stabilizer of U , subset of S ?

Balarka Sen

12:15 AM

I was thinking this

Daminark

Codomain?

Oh like target space

Balarka Sen

Like, "corestricting" a function $f : A \to B$ to $f : A \to f(A)\subset B$

Daminark

Yeah sure that works, I guess

Kasmir Khaan

is gU = U for u in U ?

MatheiBoulomenos

@Kasmir yes, that's right, that's the definition of the stabilizer in this case

Kasmir Khaan

12:17 AM

hmm

so we multiply by g

and we need to get back our whole set U back

one problem here is that hmm

it can be that many elements of G , sends a subset to itself

I mean do we have a partition on G here?

some elements gU= U, stablizer, other element g'U = X , X subset of S

MatheiBoulomenos

Well, the cosets $G/ \operatorname{Stab}(U)$ partition $G$, yes. The bijection that you worked so hard on between the orbit and the cosets of the stabilizer still applies

if $G$ acts on the set of its subgroups by conjugation, what will the stabilizer of a subgroup $H$ be?

Kasmir Khaan

hmm let me try to make sense of this :D

hmm

do you mean g H g^-1 = H

very good question ._.

MatheiBoulomenos

Yes, so what is the subgroup of all such $g$ called?

Kasmir Khaan

normalizer of H ?

MatheiBoulomenos

Right!

Kasmir Khaan

12:24 AM

:DDDD

Yeeey :D

MatheiBoulomenos

That's important in applying this stuff to Sylow theory

Kasmir Khaan

Hmm but in this case

the subgroups (sets) are not disjoint

they have at least 1 in them all

on my first example, we had subsets of S only

Leaky Nun

@MatheiBoulomenos how do you derive that from $\forall A \subset \Bbb R^n : \mu^\star(E \cap A) + \mu^\star(E^C \cap A) = \mu(A)$?

MatheiBoulomenos

You should forget that they are subsets and think of them as points. The partition of $\mathcal P(S)$ that you get is made of subsets that consists of points, but in this case these points are themselves subsets of some set $S$. But this is not important, it's only confusing.

@LeakyNun what exactly?

Leaky Nun

@MatheiBoulomenos that there's a closed subset such that see above

32 mins ago, by Leaky Nun

"We can find a closed subset $F_j$ of $E_j$ with $\mu^*(E_j-F_j) \le \varepsilon / 2^j$"

Kasmir Khaan

12:30 AM

@MatheiBoulomenos okay ill continue with your second example, of subgroups =P

MatheiBoulomenos

You haven't told me what $\mu^*$ is, so I assumed it was an inner regular measure

Kasmir Khaan

I mean more like continue with sylow :D

Leaky Nun

@MatheiBoulomenos just any outer measure

let's say it's the Lebesgue outer measure

user193319

I read somewhere that $C(\{1,...,n\})$ is $*$-isomorphic to the $n \times n$ diagonal matrices. What does $C(\{1,...,n\})$ denote?

Does it denote all continuous functions from $\{1,...,n\}$ to $\Bbb{R}$?

Daminark

Here's a cute problem I found: If $H$ and $K$ are normal subgroups which intersect trivially, then each element of one commutes with each element of the other

MatheiBoulomenos

12:43 AM

That's a pretty important lemma

Daminark

@user193319 I don't think so? $\{1,\ldots,n\}$ is usually considered to by default to have the discrete topology so every map is continuous. That'd mean this set is $\mathbb{R}^n$, which doesn't sound terribly isomorphic to $M_n(\mathbb{R})$

What's this * operation?

Mike Miller

"Isomorphic as a C^* algebra" presumably

algebras of functions are commutative so that's certainly not a matrix algebra

user193319

I was thinking $f \mapsto diag(f(1),...,f(n))$ would be the isomorphism.

Mike Miller

Both dami and I missed "diagonal matrices"

You are correct

and this does refer to continuous functions

Meow Mix

hi mike

long time no see

Mike Miller

12:47 AM

hi

user193319

@MikeMiller Okay! Very good! And the $*$ operation on $C(\{1,...,n\})$ would be $f(x)^* = \overline{f(x)}$, and on the $n \times n$ diagonals it would just be the conjugate transpose?

Daminark

Oh rest in dead: us

Meow Mix

i keep writing backwards lower case deltas

$\delta$

Eric Silva

sup dudes

Daminark

Yo

Balarka Sen

12:52 AM

yo

Meow Mix

playing some zelda

proving some limit theorem thing

Eric Silva

which zelda

Meow Mix

oot

Daminark

The limit theorem zelda!

Eric Silva

one of my top 4 zeldas

Meow Mix

12:53 AM

yeah the one with the epsilons and the deltas

what if the plural of zelda is zeldae

or zeldi?

Daminark

The triforce is an artifact that allows you to make every series converge absolutely on demand

Mike Miller

1:16 AM

majora's mask best zelda

Eric Silva

I 100% agree

Daminark

Huh, never heard of that

user193319

@MikeMiller Here is a quote that confuses me: "Every automorphism on $C(\{1,...,n\})$ is a composition with a homeomorphism on $\{1,...,n\}$" My question is, a composition with a homeomorphism and what else? Also, do they mean a homeomorphism from $\{1,...,n\}$ to $\{1,...,n\}$?

Faust

Morning

nbro

1:35 AM

Anyone here familiar with backpropagation?

Maybe it is not necessary to be familiar with it to answer this question.

Suppose we have functions f, g and h.

g is defined in terms of f.

And we want to find the derivative of h with respect to f.

Now, there's must be a property in calculus which relates this derivative to an expression involving a sum of the partial derivatives of g with respect to f.

Which rule of calculus is it?

I know it's the chain rule, I'm just trying to understand where that sum comes from.

Secret

5:32 AM

[Random]

Some thoughts about the ordinal collapse function as an operator:

Silent

5:54 AM

Please someone answer this!

Secret

6:44 AM

\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta 2 < \gamma
\end{align}

The original non operator form by simpleart:

We can make this larger as follows:

\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta^2 < \gamma
\end{align}

Or even larger:

\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta 2 < \gamma \wedge 2^{\eta} < \gamma
\end{align}

Therefore simpleart's $\psi_n$ function is a weakely inaccessible cardinal

while the $\psi_n$ in the bottommost block of the above message denotes a strongly inaccessible cardinal

Therefore, it seems that:

8

A: Golf a number bigger than TREE(3)

Ruby, 349 bytes, fψ0(ψ9(9))(9) where f is the fast growing hierarchy and ψ is an ordinal collapsing function described below. Try it online! def f(b,n=0)n<1?(b.class!=Array):(n.times{n+=b==1?n:f(g(n,b),n)};n)end;def g(n,r)a,b=r;f(r)?(r>1?r-1:n):r.size>2?h(n,a,b):f(b)?(b>1?[a,b-1]:b>0?a:[a,n]):...

To get a number bigger than TREE(3), somehow a weakly inaccessible cardinal is needed

As for how to get larger than TREE(3) predicatively, my guess is that the fast growing hierarchy is perhaps the only way (to be checked)

Now, the following is suspected to be highly circular:

Secret

7:04 AM

\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta 2 < \gamma \wedge \eta^2 < \gamma \wedge 2^\eta < \gamma \wedge \omega_\eta < \gamma \wedge 2^\eta < \gamma \wedge \beth_\eta < \gamma \wedge \beth_\eta < \gamma \wedge \text{$\gamma$ is a Sup…

Anyway, that's enough rambling about the ordinal collapse function for now.

Now to reprove that there cannot be a largest ordinal (because I failed to find that proof in the transcript)

Let the foundation be some generalisation of ZF with proper classes: Let $\Bbb{On}$ be the class of ordinals

Suppose there exists a proper class $\infty$ such that it is a fixed point under all possible nondecreasing map (including the powerset operation. We insert an extra axiom in this set theory so that the only object which violates cantors theorm is $\infty$, in particular $|\mathcal{P}(\infty)|=|\infty|$). We want to show that the resulting construction is inconsistent unless the proper class of ordinals form a cyclic ordering (and thus violating the axiom of regularity)

Recall that in ordinal multiplication, $\alpha0=0\alpha=0$ because the cartesian product is empty if either input sets is an empty set

Now by the definition $\alpha \infty =\infty$ for any $\alpha \in \Bbb{On}$. what about $\infty \alpha$? If $\infty \alpha > \alpha$ by the nondecreasing property of ordinals, then this contradicts the fact that $\infty$ is the maximum. If $\infty \alpha =\infty$ we have another problem:

Note that for the second condition: $\infty \alpha = \alpha \infty$ for all $\alpha \in \Bbb{On}$. This means, $\infty$ act as a (two-sided) absorber. But we knew in semigroup theory if there is a two sided absorber, then it must be unique

Therefore $\alpha\infty = \infty = 0\infty = 0$ and thus we have $\infty = 0$

Jenna Sloan

7:22 AM

@Secret aren't you getting tired from all that typing?

Secret

this contradicts with the extra axiom that is added because $\mathcal{P}(0)=\{0\}=1$. Therefore the extra axiom is inconsistent thus we discard it and go back to ZF extended with proper classes

Jenna: Not when it is on the main computer

(cont.) Now let's check the resulting structure we end up with:

0,1,2,....$\omega$,...,$\omega_{\omega}$,...,$\beth_{\beth_{\omega}}$,..........‌....,0

This violates the axiom of regularity, hence it is inconsistent in ZF

and therefore there cannot be a largest infinity in ZF

QED

Jenna Sloan

Isn't $0\infty$ undefined?

Secret

Uh, in the above proof, $\infty$ is not the infinity in analysis, but it is assumed to be an ordinal which is a fixed point of any nondecreasing map

the proof want to show that a largest infinite ordinal cannot exist in ZF even if we don't care about the powerset axiom

Now whether the existence of $\infty$ is consistent for ZF-Regularity I have no idea, cause I am not good at working with non well founded set theories

Jenna Sloan

I have no idea what "nondecreasing map" and "powerset axiom" and "ZF" mean.

Secret

A nondecreasing map is a function $f$, such that $x < y$ means $f(x) \leq f(y)$.

Powerset axiom is an axiom in some set theory which said that given a set S, the set of all subset of S, which is denoted as $\mathcal{P}(S)$ exists.

ZF is Zermelo–Fraenkel set theory, which is tied to the foundation of mathematics and trying to express every mathematical objects as sets

Jenna Sloan

7:40 AM

I'll just assume you know what you're doing

Secret

Some update: It is known that $\alpha^0=1$ because the empty cartesian product is a singleton because the only function that maps from the empty set to itself is the empty function. This means at the present definition, even under a non well founded version of ZF the existence of $\infty$ is still inconsistent because then $\alpha^0=0\neq 1$. Even if we ignore that and redefine $\alpha^0=0$, it will not help because then ordinal exponentiation will be broken as the inductive case means

$\alpha^{\beta}=0$ for all $\beta \in \Bbb{On}$

Now, whether having "the largest infinity" (technically incorrect since we have shown above that it will lead to $\Bbb{On}$ taking a cyclic ordering and thus became a non well founded set) by defining it to be fixed under any nondecreasing definable map except exponentiation by zero is consistent needs to be checked more closely before this question is to be asked on main

Alessandro Codenotti

7:58 AM

@Secret In ZF-P you should be able to get $[0,\omega_1)$ as ordinals but no further

Secret

For starters, we can still have cantor's theorem being valid since $2^0=1$ and we can let $0$ to be at least a strongly inaccessible cardinal so that no power set of any ordinal larger than it in the ordering can reach it.

@AlessandroCodenotti Ah right (because $\omega_1$ existence need powerset), but now I am wondering ZF-Regularity

it seems it is consistent and construcible that the ordinals becomes a cyclic ordering with 0 being both a finite ordinal and strongly inaccessible

wait a sec..., I think in doing so I will violate the property of ordinals being heredity transitive sets (because the emptyset has no elements, let alone being transitive). or the definition of emptyset (because the emptyset has no elements)

unless making 0 strongly inaccessible (thus unreachable by any powersets) somehow helps evades the problem

Alessandro Codenotti

Ordinals needs to be well ordered by $\in$

And stuff violating regularity most likely isn't

Secret

37 mins ago, by Secret

0,1,2,....$\omega$,...,$\omega_{\omega}$,...,$\beth_{\beth_{\omega}}$,..........‌....,0

right, so this construction will not be ordinals by definition, but just some cyclically ordered set

Secret

9:21 AM

Is it possible to define transfinite induction without using heredity transitive sets. That is, does there exists a structure with an ordering isomorphic to the ordinals but without the property of being heredity transitive?

One thing I never understood about "Ordered under $\in$" is why we are doing that besides the reason that the will ensure every number can be written as a set and limit ordinals can be wrote in terms of unions

I am fine having all my structures not ordered under $\in$ if there is somehow a way to reproduce the well ordering of the ordinals, but throwing away the requirement of "under $\in$". Probably set theory just cannot do that job for me. I need something more expressive such as type theory

Secret

10:35 AM

in Mathworks, 32 secs ago, by Secret

I always thought a well ordered collection will mean everything is laid out in discrete chunks on the table and I can easily pick any element I want because they are all uniquely labelled somewhat by the ordering

life will be much easier if I can say $\omega +1 = \omega^+$

and not $\omega+1 = \{0,1,2,3,4,5,...\} \cup \{\omega\}$

Secret

10:55 AM

Rant: Why should I care about all the other elements when the element I am interested in is just one of them and possibly its nearest neighbours

Leaky Nun

@Secret can't you?

@JennaSloan of course he does

user21820

@Secret As @LeakyNun said, you trivially can. I think you still don't get the point that if all you want is well-orderings and don't care at all about comparing them, then you have very trivial operations. + is just concatenation.

You just have to deal with the fact that there are uncountably many well-orderings isomorphic to the naturals.

Secret

In such framework, those uncountably many well orderings are isomorphic, but cannot be shown to be order isomorphic to the naturals?

user21820

You asked this before and I already said they can be shown.

Leaky Nun

@user21820 what do you mean?

Secret

11:03 AM

ah ok, I think I got my brain tied into knots again

user21820

And I would prefer if you kept all your thoughts in one place. I don't like having two separate chats with overlapping discussions. =)

@LeakyNun Go to Mathworks and I'll show you.

Leaky Nun

@user21820 already there

Mary Star

12:05 PM

Hi!!

To create a team of 5 players out of 11 players there are $\binom{11}{5}$ ways, right?

If we have 7 starters, 15 main dish and 6 dessert, how many different
3 course menus can we create? In this case we cannot use the binomial coefficient, can we?

user84215

Hi.

Mary Star

We have 7 posibilities for the starter, 15 posibilities for the main dish and 6 possibilities fir te dessert. Since they are independet, the number of combinations is equal to the product $7\cdot 15\cdot 6$. Is this correct?

Secret

Marystar: seems so

Mary Star

Ok, thanks!

Transcript for

$Mathematics$ Mathematics