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Kasper
Kasper
12:00 AM
How would you choose on element within the set without the axiom of choice?
By the axiom of foundation?
 
Leaky Nun
Leaky Nun
well it's nonempty
so by definition there's an element inside
 
MatheinBoulomenos
MatheinBoulomenos
so there is an element, but you can't choose one, right? @LeakyNun
 
Leaky Nun
Leaky Nun
if you try to formalize what we mean by "choosing an element from a nonempty set"
you'll realize that classical logic only really deal with propositions, so we would need to phrase it as $\forall A: [A \ne \varnothing \to \exists x : [x \in A]]$
for which proving it is an exercise in logic
@MatheinBoulomenos in particular, "choosing one" isn't a well-defined notion
 
MatheinBoulomenos
MatheinBoulomenos
you need extensionality, depending on your definition of $A \neq \varnothing$, right?
@LeakyNun I was just kidding
 
Leaky Nun
Leaky Nun
yes, you need extensionality
@MatheinBoulomenos what one usually has in mind when they doubt that you can choose an element from a nonempty set, is, e.g. you can't have a program that takes a nonempty set as input and outputs an element
 
MatheinBoulomenos
MatheinBoulomenos
12:04 AM
I don't know what program is
 
Leaky Nun
Leaky Nun
but that isn't really choosing an element from a nonempty set, that's really the axiom of choice, i.e. choosing an element from each set in a collection of nonempty sets
computer program
 
MatheinBoulomenos
MatheinBoulomenos
how are you going to feed an infinite set as an input for a computer program?
 
Leaky Nun
Leaky Nun
that's another problem
 
Kasper
Kasper
So, if you can't assume the axiom of infinity, (for example, when you try to model the real world), you don't need the axiom of choice.
 
Leaky Nun
Leaky Nun
nobody uses ZFC to model the real world
 
MatheinBoulomenos
MatheinBoulomenos
12:07 AM
math isn't real
 
Kasper
Kasper
I think the axioms of ZFC minus the axiom of infinity are modelling quite closely what I can do with a set in a programming language.
Or what I can do with some collection of real world supermarket bags.
 
Leaky Nun
Leaky Nun
you might be interested in this
Hereditarily finite set
In mathematics and set theory, hereditarily finite sets are defined as finite sets whose elements are all hereditarily finite sets. == Formal definition == A recursive definition of well-founded hereditarily finite sets goes as follows: Base case: The empty set is a hereditarily finite set. Recursion rule: If a1,...,ak are hereditarily finite, then so is {a1,...,ak}. The set of all well-founded hereditarily finite sets is denoted Vω. If we denote by ℘(S) the power set of S, and by V0 the empty set, then Vω can also be constructed by setting V1 = ℘(V0), V2 = ℘(V1),..., Vk = ℘(Vk−1),... and so on...
 
MatheinBoulomenos
MatheinBoulomenos
So for any supermarket bag, there is a powersupermarket bag which contains all subsupermarket bags of that supermarket bag?
 
Leaky Nun
Leaky Nun
This should be a model of ZFC-I
in particular it satisfies choice as you noticed
given a finite collection of nonempty hereditarily finite sets
you can use induction (in the outside model) to show that there is a choice function which is itself a hereditarily finite set
notice that the choice function must be itself a set inside the model
2 mins ago, by Leaky Nun
given a finite collection of nonempty hereditarily finite sets
it must be finite because every set in the model is finite
and choice only concerns sets in the model
 
Kasper
Kasper
Ah interesting @LeakyNun, thanks. My mind likes thinking in finite things a lot.
@MatheinBoulomenos Hehe, yes, well, you may need to get some more supermarket bags, if you run out.
But I think $\{\}$ could be mapped directly to a supermarket bag. In that way, if you have for example the supermarket bag $\{\{\},\{\}\}$ which contains of two super market bags. Yes you can construct a power set out of other supermarket bags quite easily, no?
 
MatheinBoulomenos
MatheinBoulomenos
12:20 AM
you need a lot of supermarket bags
But $\{\{\},\{\}\}=\{\{\}\}$
I think of supermarket bags as being able to contain the same thing multiple times
 
Kasper
Kasper
That is a good point yes. Well, we don't allow that in our model indeed. It should have been this bag: $\{\{\}, \{\{\}\}\}$
 
MatheinBoulomenos
MatheinBoulomenos
yes, if you put that as a restriction, then it works. But if you need to put some arbitrary restrictions on what you can put in your supermarket bags, then this doesn't make Sets a good model, does it?
 
Leaky Nun
Leaky Nun
supermarket bags would be multisets
just quotient by the relation
 
MatheinBoulomenos
MatheinBoulomenos
that's as arbitrary as putting restrictions on what you can put in the bag
when I have a supermarket bag and two times the same thing, I can put that in and it's different from just putting it in once
finite multisets model supermarket bags, yeah. But I don't see how finite sets model what you can do with some collection of real world supermarket bags
 
Leaky Nun
Leaky Nun
I see, you're using ZFC to model supermarket bags, I'm using supermarket bags to model ZFC
 
MatheinBoulomenos
MatheinBoulomenos
12:30 AM
yes, @Kasper claimed that ZFC-I "[is] modelling quite closely what I can do [...] with some collection of real world supermarket bags."
 
Leaky Nun
Leaky Nun
ok
 
MatheinBoulomenos
MatheinBoulomenos
@LeakyNun headlines in the news "Undergrad proves consistency of ZFC in the real world by constructing a model with supermarket bags."
 
Leaky Nun
Leaky Nun
lol
 
Kasper
Kasper
hehe
Luckily, this disaster is saved by mathematicians on a mission to show the world how little mathematics has to do with anything in the real world.
I'm not an undergrad btw.
 
MatheinBoulomenos
MatheinBoulomenos
I was talking about Leaky, since he was the one modelling ZFC with supermarket bags (not the other way around)
 
Kasper
Kasper
12:41 AM
ZFC - I doesn't still feel completely satisfying. Why is the atom chosen as empty sets. Why not numbers? In my mind, you first create numbers. Then you may want to put them in sets for combinatorics etc. Creating set theory before number theory feels like the wrong order of doing things.
 
MatheinBoulomenos
MatheinBoulomenos
you can do ZFC with urelements, but I don't think that's used
mathematicians like minimalism
 
Leaky Nun
Leaky Nun
well if you're doing set theory
doesn't it make sense to have sets as atoms...
but what sets, you ask
what sets do you have at the beginning except the empty set
 
MatheinBoulomenos
MatheinBoulomenos
@Kasper I get what you mean, but you can also argue for the other direction. How do you come up with numbers in the first place? You realize that you can compare finite collections (i.e. sets!) regardless of what kind of actual elements they contain. So you come up with a notion of two collections having the "same number" of elements which is based, in some sense on bijections
so natural numbers are just isomorphism classes of finite sets
sets are more fundamental in some sense. You can just take some elements and consider them as a collection, but saying that this collection has "five" elements and another has "seven" requires an abstraction
 
Kasper
Kasper
1:01 AM
Hm.. I'm going to sleep about those thoughts.
Good night everyone
 
MatheinBoulomenos
MatheinBoulomenos
Good night @Kasper
@LeakyNun what do you think? Which of these are more fundamental: numbers, sets or endofunctors of the category of formal groups with coefficients in the Witt vector ring of an algebraic closure of $\Bbb F_p((x_1, x_2, \dots))$?
 
Leaky Nun
Leaky Nun
I'm going to go with endofunctors of the category of formal groups with coefficients in the Witt vector ring of an algebraic closure of $\Bbb F_p((x_1, x_2, \dots))$
 
user548331
user548331
ok so if I cant post anything can I just put the one from last night up here?
We seek to prove:

$a | b {\rm \: \Rightarrow \:}\phi(a) | \phi(b)$ (1)

$\phi(a) | \phi(b) {\rm \: \not\Rightarrow \:}a | b$ (2)

for (1) we have:

$a | b{\rm \: \Rightarrow \:} {rad(a)} |\, {rad(b)}{\rm \: \Rightarrow \:}\prod_{p|a} p \, | \prod_{p|b} p{\rm \: \Rightarrow \:}\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr)$


$a | b{\rm \: \Rightarrow \:} {rad(a)} |\, {rad(b)}$

${rad(a)} |\, {rad(b)}{\rm \: \Rightarrow \:}\prod_{p|a} p \, | \prod_{p|b} p$
I mean it is very elementary I know but I have been trying to prove really basic things instead of failing to prove more difficult things. anyway yeah if someone can correct that for me if flawed or what ever that would be much appreciated.
 
MatheinBoulomenos
MatheinBoulomenos
@user548331 I haven't looked at the full proof but why don't you just reduce it to $a=p^k$ and $b=p^m$ for some prime $p$?
 
user548331
user548331
1:16 AM
because that's only a case example it can't prove the statements true for any $a$ and $b \in \mathbb N$
 
MatheinBoulomenos
MatheinBoulomenos
but you know what $\varphi(a_1a_2)=\varphi(a_1)\varphi(a_2)$ if $\gcd(a_1,a_2)=1$
And if you write $a=p_1^{n_1} \cdot \ldots \cdot p_k^{n_k}$ and $b=p_1^{m_1} \cdot \ldots \cdot p_k^{m_k}$ (where we allow that $m_i$ or $n_i$ are zero, so that we have the same primes $p$ for $a$ and $b$), then we get that $a$ divides $b$ iff for all $i$ $p_i^{n_i}$ divides $p_i^{m_i}$
 
user548331
user548331
I would say the one flaw I can already see is the assertion about the radical implying divisibility of the Euler products. ok again a case example for coprime pairs sure I don't dispute that is valid if we include that we are talking about coprime pairs
that is equivalent to the statement about the radicals of a and b in the first part
 
MatheinBoulomenos
MatheinBoulomenos
So now $\varphi(a)=\varphi(p_1^{n_1}) \cdot \dots \cdot \varphi(p_k^{n_k})$
 
user548331
user548331
but the question I asked myself also required to demonstrate the converse of (1) is false, so that is why I took the approach I did
 
MatheinBoulomenos
MatheinBoulomenos
$\varphi(4)=2=\varphi(3)$, but $4$ doesn't divide $3$
 
user548331
user548331
1:23 AM
yeah but come on dude you can't call that a proof
 
MatheinBoulomenos
MatheinBoulomenos
I don't understand what you did for that direction. If you want to prove that something is wrong, you just ahve to give one counterexample
it's a proof
 
user548331
user548331
it's a proof there exists a counter example, it doesn't prove that any number of counter examples can exist
look you just proved Carmichael's conjecture they have the same totient but they are different
 
MatheinBoulomenos
MatheinBoulomenos
The statement $\varphi(a)\mid \varphi(b) \not\Rightarrow a \mid b$ is just that exists at least one counterexample
 
user548331
user548331
ok so throw in $\forall a,b \in \mathbb N$ for my statement
 
MatheinBoulomenos
MatheinBoulomenos
but then it's clearly wrong
take $a=b$
 
user548331
user548331
1:29 AM
ok so what would be the notation to state that any number of counter examples can be found on $\mathbb N$
 
MatheinBoulomenos
MatheinBoulomenos
not sure about notation you can just write that out
If you want infinitely many counterexamples: $\varphi(4 \cdot 5^n)=\varphi(4)\varphi(5^n) = 2 \varphi(5^n) = \varphi(3 \cdot 5^n)$, but $4 \cdot 5^n$ doesn't divide $3 \cdot 5^n$
or for every odd $k$, $\varphi(2k)=\varphi(k)$, but $2k$ doesn't divide $k$
 
user548331
user548331
ok then insert that into my orginal post and review the content. Yeah sure ok I agree if we want we can assert that the arguments have a particular form I just prefer to remain as general as I possibly can
The proof I gave only relies on the basic principle that the factors of the numerator and denominator for the totient have known divisibility conditions when a is known to divide b, but when $\phi(a) | \phi(b)$ is only known, there is nothing restricting the case to either $a | b$ or it's negation
 
MatheinBoulomenos
MatheinBoulomenos
I don't see how your proof actually shows that there is a counterexample
@user548331 how can you remain general when you prove that a statement is wrong for some values of $a$ and $b$ when it is also true for some values of $a$ and $b$?
Showing that a statement $P(x)$ is not true for all $x$ means showing that there exists a $x$ such that $P(x)$ is wrong. This means this is an existence statement. For a proof, you need to show the existence of such elements $x$ such that $P(x)$ is wrong
 
user548331
user548331
well rather than the aim to prove a statement is wrong, it's really demonstrating that (2) is right, by proving that $\phi(a) | \phi(b)$ implies $ a | b \lor \lnot (a | b)$
 
MatheinBoulomenos
MatheinBoulomenos
but $P \Rightarrow a | b \lor \lnot (a | b)$ is true for every statement $P$
 
user548331
user548331
1:45 AM
how can you think that. what if P declares nothing about an $a$ or $b$?
 
MatheinBoulomenos
MatheinBoulomenos
But $a | b \lor \lnot (a | b)$ is always true
 
user548331
user548331
ok perhaps there is another logical symbol that I should be using there. but I was trying to speak in the context of $(a,b) \in \mathbb N^{2}$, rather than strictly a subset for which only $(a,b) s.t a | b$ are elements.
It's not always true no, how can you say something is true when you have absolutely no information about the objects the statement makes assertions about? I understand what you are saying, but in the context you are using it here it just isn't logical at all
 
MatheinBoulomenos
MatheinBoulomenos
I'm just saying that the statement "$a$ divides $b$ or $a$ does not divide $b$" is always true
and this is perfectly logical
 
user548331
user548331
sure if that is the only statement made
 
MatheinBoulomenos
MatheinBoulomenos
but this is what the notaton $a | b \lor \lnot (a | b)$ means
 
user548331
user548331
1:56 AM
but in the context of the proof I gave no, in the context I intended, a statement $a | b$ being true is an assertion that $a | b \lor \lnot (a | b)$,
 
MatheinBoulomenos
MatheinBoulomenos
I don't follow. You don't seem to use symbols like $\lor$ or $\lnot$ like they are normally used
 
user548331
user548331
is false because only $a | b$ is true, so if we think of $(a,b)$ as either belonging to a set $S_1$ such that for all elements $ a | b$, and another set $S_2$ such that for all elements $\lnot (a | b)$ is strictly true, the statement $ a | b \lor \lnot (a | b)$ implies $(a,b)$ is an element of $S_1 \cup S_2$
ok well I did just start using them off a vague understanding and kept doing so with out communicating my thoughts to people for along time so that is entirely possible, but all that means is an adjustment to my notation now that ive decided im not terrified of people
 
user548331
user548331
2:50 AM
it's actually got huge mistake in it there the radical of a divides the radical of b does not imply that the Euler product of a divides the Euler product of b, it implies that the Euler product of a divides the totient of b
so yeah that was a pretty important part and I think it still isn't a complete proof without more detail for that part
anyway
We seek to prove:

$a | b {\rm \: \Rightarrow \:}\phi(a) | \phi(b)$ (1)

$\phi(a) | \phi(b) {\rm \: \not\Rightarrow \:}a | b$ (2)

for (1) we have:

$a | b{\rm \: \Rightarrow \:} {rad(a)} |\, {rad(b)}$

${rad(a)} |\, {rad(b)}{\rm \: \Rightarrow \:}\prod_{p|a} p \, | \prod_{p|b} p$

$\prod_{p|a} p \, | \prod_{p|b} p{\rm \: \Rightarrow \:}\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr)$

$\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr){\rm \: \Rightarrow \:}{\Biggl\{\frac{b \cdot \prod_{p|b} (1-\frac{1}{p})}…
 
user548331
user548331
3:20 AM
This is the part that I still need to establish rigorously.
Show that:

$\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$

is true $\forall a,b \in \mathbb N$ such that $a |b$

Case 1:

$\lnot \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \lnot\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$

Case 2:

$\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \lnot\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$
 
GFauxPas
GFauxPas
3:39 AM
yikes
 
user548331
user548331
yeah it does get pretty dicey if you expect a rigorous proof of some things that are considered well and truly elementary it's like that combinatorics one someone mentioned in here yesterday or the day before, technically it should be a piece of cake I mean it's just a finite arithmetic progression of the 9 digits of the decimal system, but the question is actually tough as all hell to generalize
9 non zero digits oc
but if a rational function can be found for any ratio of arithmetic functions at least an approximation using interpolation polynomials, you could then split the Euler product into numerator and denominator (or approximations to them to be exact to allow for the cases that $a$ and $b$ are enormous) you could then at least show that in all three of those cases, the Euler product of $a$ will always divide the totient of $b$, but yep it's certainly got the better of for today
well that's all I've got right now anyway lol that's my standard vague answer for when I don't have a clue yet
or maybe it could be done by dealing the cases of $a \lt b$ and $a \gt b$ separately, and focusing on the greater of the two's prime factors that are left over after their division most likely being the most significant in the division of their Euler products
but no matter which way I look at it I can't see any way around making a declaration like "let $N(E_n)$ be the numerator of the Euler product of $n$, and let $D(E_n)$ be it's denominator
and then making assertions about those two expressions that are implied in each of the three cases aforementioned
 
user548331
user548331
4:15 AM
still kind of not pleased about there being a limit on how many questions a user can ask so im just going to spam your chat room with them until I get booted
$$S_N={\Biggl\{\frac{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)}{n_1}:n_1,n_2 \leq N}\Biggr\}$$

is $$\max(S_N \cap \mathbb Z)=1$$?
really is something wrong with my computers copy and paste function everytime the formatting is totally different than it was in the source document, like there were none of those spaces wtf
 
Secret
Secret
It's mathjax, which is a watered down latex
 
user548331
user548331
also what can't I copy and paste it into notepad that is the funniest shit ive seen in a long time it literally transfers everything but the latex
 
Secret
Secret
are you copying the raw latex commands or the rendered expressions?
the rendered expression is not supported by notepad
 
user548331
user548331
nah this is from a window of a one of the questions I posted here that's my latex editor lol
i press the edit button on one of those then cut and paste it into maple but because maples formatting is a little spastic i wanted to just say them all as txt but then encounter that problem
save* all of them
having said that it is a ripped copy off pirate bay
but i did actually pay full price at least 3 times in my life for the software so, yeah oh well
 
Zee
Zee
Where do I buy a blackboard from ?
 
user548331
user548331
4:26 AM
you cant buy them because they are evil
 
Zee
Zee
Lol , they are awesome
 
Secret
Secret
$$\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)^{\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)}}}}}}}$$
 
user548331
user548331
no they are for terrifying 1st grade teachers with some kind of thyroid imbalance that scare little boys for life
 
Secret
Secret
lol fail
 
Zee
Zee
Or the nice professor who changes your life after class by showing you how awesome math is
 
user548331
user548331
4:28 AM
that's not funny bro i really lose my shit over iterated expoonentiation
changes your life after class well that kind of sounds like code for an old man fetish but hey each to their own
 
Secret
Secret
$\sum_{k=1}^{n}\prod_{p|n_2} \Bigl(1-\frac{1}{p}\Bigr)$
smaller thing
 
user548331
user548331
maybe ive just been isolated for too long
and its me who is getting more and more perverse
 
Zee
Zee
Lol no I had the same joke in mind
 
user548331
user548331
yeah you think that's funny that's great i see how it is im going to be pushed out of the chat room with pointless algebra. well not the iterated exponent thing that's not pointless at all just terrifying
 
Zee
Zee
He does that all the time , you can’t let it get to you
He means well though
 
Secret
Secret
4:33 AM
I knew very little about number theory, just enough to recognise that product (though it still look different from what I expect from prime factorisation)
but all the number theory guys are asleep atm
 
Zee
Zee
Secret where are you now with your math interests
 
Secret
Secret
have not done much maths lately, very busy on my PhD
though I will say I finished reading that paper that GFaux and Semi talked about the Lebesgue integrals
and I was rereading some order theory stuff when I am stuck in the train
My supervisors and I are going to pump out a paper by July, which is why all 3 of us are incredibly busy in organising the different parts of it
Once the PhD is done (or at least I have my thesis more or less complete) I might want to investigate continua (connected complete metric spaces) and non measurable sets in more detail
 
user548331
user548331
@Zee no really i do not think anything can get to me anymore. Literally I mean the last thing that used to really do it was when they kept cutting me off welfare but i figured out a loophole i just go to a GP and say im all suicidy and they write a medical exemption that gets me out of having to continue the farce of pretending to look for employment
 
Zee
Zee
@user548331 well , if you are happy doing that , then more power to you otherwise , you’ve got to sort yourself out
@Secret what’s the paper about
 
user548331
user548331
@Zee oh sure no problem ill just quit math and have all the time i need to look for a soul destroying role in society that earns a marginally larger amount than the welfare cheque
 
Zee
Zee
4:58 AM
@user548331 it’s not marginal, find a job that makes you 50k so you can live like a human being
 
Secret
Secret
 
Zee
Zee
Eat decent food and not worry about things
And who said you need to quit math
 
Secret
Secret
It talks about while both Riemannian and Lebesgue integrals have no physical differences, the formulation of Lebesgue integral is superior since the completeness of the space of all Lebesgue integral functions allow many theorems including dominated convergence theorem to hold in general
 
user548331
user548331
because that is exactly what getting whatever job you are proposing would require. Maybe not quiting, but removing the bulk 40 hr part of my week and all of my focus away from math, so it just turning to crap and i don't get what i get out of it now improving at the rate i am, making the small amount i do completely worthlerss
 
Zee
Zee
@user548331 if you spend that much time doing math then you should have no trouble getting a PhD , so do that
You can’t do math more than like 5 hours a day anyway , so a job should not mean you need to quit doing mathematics
 
user548331
user548331
5:06 AM
i can eat decent food no drama there, sure well you get in touch with the universities in Australia and tell them you want them to reinstate my credits from 2004-2010, amounting to 3/4 of a BSc, and then maybe it's feasible. But every time i tried it was a case of them telling me they will get in touch with me when they have arranged my reenrolment, and never getting back to me. and i tried a lot of times. And there is no way i am going back and starting from freshman year that is just a joke,
especially when a Phd is essential for even standing a chance in getting a research role, which isn't guaranteed by any stretch of the imagination anyway
really? where did you pull that 5 hour figure from
I mean don't get me started here because that's only one thing that has put nail in my coffin there, i could go on with a number of things that imply a lot of things about the views of unidentified people in positions with powers of persuasion and manipulation, but it could also just be a huge string of unfortunate coincidences
 
Secret
Secret
5:39 AM
[Random]
poset
$a \leq a$
$a \leq b \land b \leq a \implies a=b$
$a \leq b \land b \leq c \implies a \leq c$
toset
$a \leq b \lor b \leq a$
min/max
Let $(S, \leq)$
$\forall a \in S \exists b \in S(b \leq a)$
$\forall a \in S \exists t \in S(a \leq t)$
minimal/maximal
$\forall a \in S \exists u (a \leq u \implies a=u)$
$\forall a \in S \exists l (l \leq a \implies a=l)$
 
Zee
Zee
@user548331 that’s just my own observation about mathematician , am not sure anybody can really do math all day long
 
Secret
Secret
Zorn's $\land S$ with certain properties $\implies \exists l (\text{defined above})$
Upper lower bounds
Let $P \subset S$
$\forall a \in S \exists u \in P (a \leq u)$
$\forall a \in S \exists l \in P (l \leq a)$
Supremum/Infimum
$\forall u \in S \exists s \in S(s \leq u)$
$\forall l \in S \exists i \in S(l \leq i)$
$\sup S = \bigvee S$
$\inf S = \bigwedge S$
Duality (not understood this well)
$\text{Dual}(\leq) = \geq$
$\text{Dual}(b,u,s) = (t,l,i)$
Joke: Dual(Bus)
Dual-mode bus
A dual-mode bus is a bus that can run independently on power from two different sources, typically electricity from overhead lines (in the same way as trolleybuses) or batteries, alternated with conventional fossil fuel (generally diesel fuel). In contrast to other hybrid buses, dual-mode buses can run forever exclusively on their electric power source (wires). Several of the examples listed below involve the use of dual-mode buses to travel through a tunnel on electric overhead power. Many modern trolleybuses are equipped with auxiliary propulsion systems, either using a small diesel engine or...
Construction
Let $(S,\leq), (T,()$
$\text{Dual}(\leq) = \geq$
$(S,\leq) \times (T,() = (S\times T, \%)$
$(x,y) \% (v,w)$
$\bigsqcup \{S,T\} = (S \sqcup T,\leq \sqcup ()$
Strict order
$x < y \implies x \leq y \land \neg (y \leq x)$
$x \leq y \implies x < y \lor x=y$
$(L,\land,\lor)$ = Lattice
Order functions:
Let $P,Q$ posets and $f : P \to Q$
$f(a \leq b) \to f(a) \leq f(b)$ order preserving
$f(a) \leq f(b) \implies a \leq b$ order reflecting
$f(a \leq b) \to f(b) \leq f(a)$ order reversing
$f(a) \leq f(b) \iff a \leq b$ order embedding
$\mathscr{P}()^{\complement}$ antitone (wtf?)
monotone + bijective + inverse is monotone = order isomerophisms
$f(P) \cong Q$
Let $g : P \to P$
$g(P)=g^2(P) \land P\leq g(P)$ closure operators
cl + Preserve unions = pretopology
$f(\sup) = \sup f, f(\inf) = \inf f$ limit preserving
Let $F : \text{Generic} \to \text{Generic}$
$F(x) \implies x$ reflective
$F(G(x)) = G(F(x))$ preserving
Pointwise order
$f(x) \leq g(y) \iff f \leq g$
 
user548331
user548331
6:24 AM
are you writing your reference sheets for an exam or something?
 
Secret
Secret
I am listing out the ingredients I need to build something
the wikipedia article is too wordy to process by staring at it, thus I extract the bits
 
user548331
user548331
what a genius invention by scholars. put a time limit on em'. all of the best ideas where manifested when they are made to fear failure with real world consequences
no those are totally reference sheets for something come on list format and absolutely no explanation for the reader
its purpose is only for as a memory aide of sorts
lol start talking to that guy from before about the strict order lemmas
second one makes sense the first one doesnt
 
Secret
Secret
Complete partial order
In mathematics, the phrase complete partial order is variously used to refer to at least three similar, but distinct, classes of partially ordered sets, characterized by particular completeness properties. Complete partial orders play a central role in theoretical computer science, in denotational semantics and domain theory. == Definitions == A complete partial order abbreviated cpo can, depending on context, refer to any of the following concepts. A partially ordered set is a directed-complete partial order (dcpo) if each of its directed subsets has a supremum. A subset of a partial order is...
 
user548331
user548331
anyway can someone please tell me how to say that i am endowing a set with an order, ie the least element is of index 1 and the maximum element is of index equal to the cardinality of the set, which is finite
say that for me
 
Secret
Secret
Let $(S,\leq)$. Then 1=min(S) and |S|=max(S)?
 
user548331
user548331
6:34 AM
yep that's the one ok so what do i call the object now?
if an arithmetic progression is one that has a constant difference between two elements ranked beside one another, what do i call just a finite integer set that is increasing, but the difference is variant
 
Secret
Secret
I don't know if there is a general name for that, other than it is a monotone finite sequence
If the difference is geometric, then you have some initial section of an arithmeto-geometric sequences
 
user548331
user548331
ok sure but i work everything out with set operations quite frequently, so i need to start dictating that precisely to allow for case that i cant work out whats happening arithmetically speaking, so i need to word everything correctly under the basis that i working a lot things out that way
what do you mean geometric
like algebraic and finite?
 
Secret
Secret
a geometric sequence is one which two consecutive terms differ by the same ratio
 
user548331
user548331
i really have no clue what people mean when they say that word it can be applied to somekind of geometry? i need to select a pet to draw as a sphere and claim it cant be brushed?
of geometric of course just say constant ratio ok i guess i will just have to go with geometric that is pretty popular
anyhow thanks i think i should at least have power nap before i have a go the next one
 
Secret
Secret
but your sequence has to satisfy a certain form in order to be called an arithmetic-geometric sequence. If your difference vary in a way that is not some nice functions, then I don't think there is a special name for them other than it is monotonic
 
Secret
Secret
6:47 AM
Domain theory
Domain theory is a branch of mathematics that studies special kinds of partially ordered sets (posets) commonly called domains. Consequently, domain theory can be considered as a branch of order theory. The field has major applications in computer science, where it is used to specify denotational semantics, especially for functional programming languages. Domain theory formalizes the intuitive ideas of approximation and convergence in a very general way and has close relations to topology. An alternative important approach to denotational semantics in computer science is that of metric spaces....
Looks like I found what I am looking for: Domain theory, topological domain theory and continuity space
 
user548331
user548331
ok. well here is my line of enquiry for it. I planned to write an algorithm that returns the number of arithmetic progressions of length k in a finite integer set of cardinality N, and want to plot this distribution for k and N for things like the average k value for that generating sequence, a i making any sense?
for example, a good question it will answer for a given sequence $a_n$, is does the maximum length for an arithmetic progression inside ${\{a_n}\}_{1..N}$ increase with N? converge to a constant length? diverge with N in what time complexity etc
 
Secret
Secret
so you want to plot k on the x axis and average($a_k$) for the y axis where $a_k$ in the progression of cardinality N?
 
user548331
user548331
like im not being specific because im tired I'm just writing out a plan for when i wake up
but those are a few of the questions i need to write code to answer for me, because a lot of the stuff that i study because of the nature i work, i will always end up working with sets of integers, like even if it something not related to integer sets, i will somehow end up doing that its probably not a good trait
but you know how it is when something fascinates you the tendency is to push it as far as you can
nature in which i work*
yeah i had my grammar for a brief 6 hours or so today but now it's gone but still better than most days
 
 
2 hours later…
user548331
user548331
9:11 AM
can we share jpegs and gif via chat?
also how where do we direct suggestions like the badge award system literally being so satirical it's insulting
 
 
1 hour later…
Leaky Nun
Leaky Nun
10:38 AM
oh someone is offended by the badge award system
 
Kasper
Kasper
11:03 AM
 
Secret
Secret
What about it?
 
jcora
jcora
fellas, given t = Arsh(x), can someone explain this transition:
should I just go around applying identities and see where I get? or could this be a mistake?
 
Secret
Secret
Why will inverse sinh has anything to do with sinh?
 
jcora
jcora
hm?
it's specifically the sh2/2 = x*sqrt(1+x^2) part that's bothering me
sh2t *
 
Anonymous
@jcora $\sinh(2\sinh^{-1}(x)) = 2x\sqrt{x^2+1}$
 
anon
anon
11:17 AM
use the double angle formula for sinh, then replace cosh using the hyperbolic pythagorean identity
 
jcora
jcora
@anon ah okay I get it now
 
mercio
mercio
just replace $x$ with $(\exp(t) - \exp(-t))/2$ and $\sinh(2t)$ with $(\exp(2t) - \exp(-2t))/2$ and check that it works out ?
 
jcora
jcora
@mercio yeah it's simpler with that pythagorean identity I just forgot about it for hyperbolic functions
 
Kasper
Kasper
11:46 AM
If you use the axiom of pairing with two times the empty set.
You then get: $\{\{\}\}$ right?
 
Kasper
Kasper
12:16 PM
Actually, what I'm trying to say, you can not just put something in a set right? This follows from pairing an a set with itself?

If I want to put x in a set, I pair x with itself, and I get $\{x,x\}= \{x\}$? Is that correct?
 
Leaky Nun
Leaky Nun
12:42 PM
yes
 
user548331
user548331
12:54 PM
Ok so who would like to earn a doubly metaphysical altruist badge and teach me how to obtain the asymptotic time complexity for an arbitrary function? I need to know how to arrive at the expression inside the big O for a given approximation, so yes of course I get the case of an easily identifiable term with the fastest growth rate being the term with the largest exponent,
but how do they find this term for all the fancies? is it simply by conversion of all terms into series expansions and then finding the largest exponent, then converting back to that function for which that term of highest exponent belonged to in its original expression?
 
Secret
Secret
Only axiom of pairing and axiom of powerset can make supersets
Also leaky some random thoughts: I wonder if it is possible to have some more fundamental axiom such that the axiom of replacement becomes a theorem
The leading reason why axiom of replacement is impredicative is because the definable function is in general an indicator function over the universe, so if that can be bypassed somehow by providing an explicit algorithm to construct these functions, then it might work
 
The Artist
The Artist
yoyo
 
Secret
Secret
But I am not sure if that is simply what the set of functions that user21820 end up with and thus we have axiom of restricted replacement instead
 
Kasper
Kasper
I try to make a javascript class that models ZFC:
https://gist.github.com/kasperpeulen/9fd19172acd20b0bf533a95681ee08cd
If you like to, you can play with it by copying the class to your chrome console.
 
Secret
Secret
hmm...
Won't mind building some nonmeasurable sets to play with
 
Kasper
Kasper
1:10 PM
Yes, the mathematical world becomes sane again if you start doing programming.
 
Leaky Nun
Leaky Nun
@Kasper nice
you might want to know that Lean has a model of ZFC here:
so Lean proves ZFC consistent
 
Kasper
Kasper
Yeah, lean is interesting. Are you involved with it?
 
Leaky Nun
Leaky Nun
somewhat
 
Secret
Secret
Sometimes I do wonder, how we can simulate a countably infinite parallel computation
We obviously cannot do it directly for memory reasons, but are there equivalents and what theorem governs the existence of equivalents
 
Kasper
Kasper
Hm.. I would guess that it is not possible to simulate such a thing.
I guess, I can just use the filter function for the axiom of subset
And the map function for the axiom of replacement.
Can I put true or false in a set?
 
Leaky Nun
Leaky Nun
1:24 PM
it's well-known that there is no computable model of ZFC
@Kasper not in ZFC
true and false are not objects in ZFC
 
Secret
Secret
I thought boolean objects can be constructed with sets?
 
Leaky Nun
Leaky Nun
sure, you can set false = 0 and true = 1
but before you do that, false and true themselves are not objects in ZFC
 
Secret
Secret
right
 
Leaky Nun
Leaky Nun
28
Q: Is there a computable model of ZFC?

skeptical scientistBackground Assuming ZFC is consistent, then by downward Löwenheim–Skolem, there is a countable model (M,$\in$) of ZFC. Since the universe M is countable, we may as well think of it as actually being the set of natural numbers, so $\in$ will be some binary relation on the natural numbers. Can...

lo.logic computability-theory set-theory
but I believe that the hereditarily finite sets, as a model of ZFC sans infinity, should be computable
hi @MatheinBoulomenos
 
MatheinBoulomenos
MatheinBoulomenos
Hi @LeakyNun
 
Leaky Nun
Leaky Nun
1:30 PM
Constructible universe
In mathematics, in set theory, the constructible universe (or Gödel's constructible universe), denoted L, is a particular class of sets that can be described entirely in terms of simpler sets. L is the union of the constructible hierarchy Lα . It was introduced by Kurt Gödel in his 1938 paper "The Consistency of the Axiom of Choice and of the Generalized Continuum-Hypothesis". In this, he proved that the constructible universe is an inner model of ZF set theory, and also that the axiom of choice and the generalized continuum hypothesis are true in the constructible universe. This shows that both...
@Kasper this might also be interesting to you
> Notice that the proof that L is a model of ZFC only requires that V be a model of ZF, i.e. we do NOT assume that the axiom of choice holds in V.
 
MatheinBoulomenos
MatheinBoulomenos
@LeakyNun If we have a model of ZFC- regularity, can we take all well-founded sets and get a model of ZFC?
 
Leaky Nun
Leaky Nun
@MatheinBoulomenos I asked someone else and he said yes
 
MatheinBoulomenos
MatheinBoulomenos
I see
 
Secret
Secret
A well founded set is one where every subset has a minimal element?
Thus a poset?
 
MatheinBoulomenos
MatheinBoulomenos
no, not a well-ordered set
 
Kasper
Kasper
1:39 PM
@LeakyNun Yes, I try to to make a model of ZFC without infinity.
 
Leaky Nun
Leaky Nun
nice
 
MatheinBoulomenos
MatheinBoulomenos
@Kasper have you thought about what I said about sets vs. numbers?
 
Kasper
Kasper
@MatheinBoulomenos Yes, I have thought about, and I still don't like, but I got admit, I tried to build my program with numbers as atoms, but it is more elegant to define equality without it. 
```
    equals(other) {
        return this.every(x => other.has(x)) && other.every(x => this.has(x));
    }
```
 
MatheinBoulomenos
MatheinBoulomenos
@Kasper but numbers are not something you find in nature
 
mercio
mercio
you can just take integers, with reunion = bitwise and.
 
Leaky Nun
Leaky Nun
1:43 PM
union
 
mercio
mercio
and x in y iff the xth bit of y in binary is 1
 
Leaky Nun
Leaky Nun
and you mean bitwise or
 
MatheinBoulomenos
MatheinBoulomenos
You don't find sets in nature, either, but it's like a set is an abstraction of things you find in nature and a number is a property of finite sets, so it's an abstraction of an abstraction
 
mercio
mercio
uh yeah, bitwise or
 
Leaky Nun
Leaky Nun
and powersets and pairing become quite a mess under your encoding
 
mercio
mercio
1:45 PM
yeah powerset is quite a mess
 
Leaky Nun
Leaky Nun
oh you can just number the set ..0001101 as the 13th set can't you
 
mercio
mercio
yes
 
Leaky Nun
Leaky Nun
so {...001101} = ...0010000000000000
this might actually be feasible
 
Kasper
Kasper
I got now all the axioms, except for powerset, choice and foundation.
https://gist.github.com/kasperpeulen/9fd19172acd20b0bf533a95681ee08cd

We talked about that axiom of choice follows from the other axioms, but I'm not sure actually how I could construct it out of what I have now.
 
Leaky Nun
Leaky Nun
@Kasper i think you should use this encoding
I'll assign each set a natural number
 
mercio
mercio
1:55 PM
and because x < 2^x, our model satisfies the foundation axiom
 
Leaky Nun
Leaky Nun
@Kasper the empty set is 0
@mercio yes
let's say our function is φ : HFS -> N
so φ({}) = 0
then φ(S) = Σ[x in S] 2^φ(x)
 
mercio
mercio
yeah
 
Leaky Nun
Leaky Nun
then you can check that φ(S U T) = φ(S) bitwise-or φ(T)
and φ({S}) = 2^φ(S)
@mercio this is dank
 
Kasper
Kasper
so I first need the power set then?
 
mercio
mercio
is it really
 
Leaky Nun
Leaky Nun
1:57 PM
you can also do that easily
 
BAYMAX
BAYMAX
Hi chat!!
 
mercio
mercio
hi
powerset is a bit clumsy
 
Leaky Nun
Leaky Nun
it's not
 
mercio
mercio
hmm
oh does it factor
lol
 
Leaky Nun
Leaky Nun
the subsets of 13 (1101) are 0,1,4,5,8,9,12,13
 
mercio
mercio
1:58 PM
$p(S) = \prod_{x \in S} (1+2^x)$ ?
 
BAYMAX
BAYMAX
can we compute the value of $\sum_{k_{1},k_{2} \in \Bbb{N}} \frac{1}{2^{k_{1}.3^{k_{2}}}$
 
Leaky Nun
Leaky Nun
@mercio maybe
 
mercio
mercio
probably not
on second thought
 
BAYMAX
BAYMAX
I fixed $k_{1} =1,2,3,..$ and varied $k_{2}$
and i got 1/2
 
mercio
mercio
maybe yes, on third thought
 
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