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12:04 AM
Okay. Any insights to here are appreciated math.stackexchange.com/questions/3406440/….
 
The introduction of numbers as coordinates is an act of violence. -- Hermann Weyl
 
Hello, quick question: Can a model interpret a formula both true and false or not interpret it in first order logic? Looking at the definition, it looks like the model is inheriting the internal logic of the meta-object/model. Just one or two sentence of guidance would be nice.
 
I've had to defind my posts lately
you can see my recent posts
People assume I'm wrong or something...
Perhaps it does indeed form a ring... go figure
I actually PROVE that it does, and still it backfires
That's what makes MSE shitty at least for me
The proof is elegant and not flawed
1
Q: Size of a boolean ring induced by a finite, commutative, boolean monoid?

Shine On You Crazy DiamondLet $X$ be a finite, commutative, boolean monoid with a $0$ element. Let $A = \{ x \subset X : 0 \in x$ and $x^2 \subset x\}$ where $x^2 = \{ab: a,b \in x\}$. Define for $x, y \in A$, $x + y = x \Delta y \cup \{0\}$. Then since $x^2 \subset x$ for $x \in A$, we actually have $x \subset x^2$ as...

These shitters man
 
12:23 AM
@KonformistLiberal a formula must be either true or false (and not both) in a model
 
@LeakyNun Where I am confused: Don't we define an interpretation of a formula recursively, so doesn't your statement needs a proof, it's not a direct property through definition?
 
sure
for a formula we recursively evaluate it
$$\bigcup_{u \in \mathcal U} U_u$$
 
@LeakyNun So can't we get a case where a formula is both true and false? I am trying to imagine that a contradiction in the model can lead to a case where the interpretation is both true and false.
 
a model can't have any contradictions
 
why not?
 
12:34 AM
because $v(\bot) = 0$
 
oh, right
so by exactly this part of the definition, we argument that an interpretation of a formula under a model cannot be both true and false, and because this interpretation is for every formula defined, we know that it must be true or false, am I right?
 
genau
 
Thank you! I want to clarify one more thing, when we say that a formula is independent of an axiom system, we mean that it has models which model this system and the formula and also this system and the negation of the formula. Am I right?
 
@KonformistLiberal well that is a consequence
 
@LeakyNun What do you mean? That it's not the definition of independence?
 
12:47 AM
independence means that the system can neither prove it nor refute it
 
@LeakyNun Aren't they the same thing? A system can prove a formula if this formula is satisfied in every model of this system, right? So, we also cannot prove the negation of this formula and this directly implies that there must be models for each case. Am I right?
 
but that's a theorem
 
That, they are the same thing?
 
yeah, it's called the completeness theorem
 
Alright, so, now, my question would be where the "law of excluded middle" lives in this context?
 
12:52 AM
in the metatheory used to prove the completeness theorem
 
So we put law of excluded middle to this kind of first order logic through using it in our metatheory if necessary, one example being completeness theorem?
And I read a phrase "we believe that first-order logic is consistent". What does it mean?
 
wo hast du es gelesen?
 
7
A: Why not both true and false?

Asaf KaragilaTruth is something specific to a particular structure, or a particular model of a theory. Sometimes, when a theory has a canonical model, we abuse the terminology and when we say "true" what we mean is "true in the canonical model". This is the case with the natural numbers and with the real numb...

 
well it means that we believe that first order logic is consistent
 
is this a mathematical statement?
 
1:02 AM
sure
it means that we can't prove $\bot$
 
So, it says empty set is satisfiable?
 
yes
 
But can't we show that by giving a finite example?
Actually, doesn't any model of any formula satisfy empty set?
 
well if you believe that sets exist then sure you can just use any model
 
If x is in a baire space and there's an injection function mapping one set of natural numbers omega to another set of natural numbers omega...is it open? I think it is because there is an unique mapping so we have interior points and open balls and neighborhood
 
1:07 AM
Woo, on arxiv
 
@Semiclassical congrats
 
Yayyyy
 
@usukidoll how many sets of natural numbers are there...???
 
Given I to be $x \in ^{\omega} \omega$ in the Baire spare there is an injection from $\omega$ to $\omega$ so there must be two sets of natural numbers
Is it open or closed? Meep. I say it's open due to the unique mapping and...interior points meep
 
either I don't understand what "Baire space" means, or I don't understand what "natural numbers" mean
 
1:11 AM
Baire space is $^{\omega} \omega$
Set of natural.numbers.is.$\omega
I could upload a screenshot
 
so $\omega^{\omega^{\omega^{.^{.^{.}}}}}$?
please do upload a screenshot
 
ok
that's the same set
 
All I can see is that we have an injection and surjection function
 
$\omega$ is the set of natural numbers
$\omega = \{0,1,2,3,\cdots\}$
 
1:14 AM
With x being in the Baire Space and yes $\omega$ is the set of natural #s
 
there are not two sets of natural numbers
I was very confused
so $x$ is a sequence of natural numbers
 
I get that part... Injection function is a unique mapping of elements isn't it?
 
$x = (x_0, x_1, \cdots)$
 
You have your set and I have mine, @Leaky.
 
Yup
 
1:15 AM
it can also be viewed as a function $\omega \to \omega$ mapping $n$ to $x_n$
 
Oh it's Ted o.o
Hiiii
 
Where?
 
Hi @TedShifrin
 
But isn't it open because there exists interior points and open balls of such things? For that mapping?
 
Oh oh, it's @topologicalmagician.
 
1:15 AM
haha
@TedShifrin how are you today?
 
I was wondering whether I am drunk
 
I'm dying in set theory đź’€
 
Cooking dinner, so not here too long.
@Leaky: Do you get drunk?
 
no
 
Cuz this is the only thing to do lel
 
1:16 AM
Let $f: [a,b] \rightarrow \mathbb{R}$ be continuous. Suppose $f(x)\geq 0$ for each $x\in [a,b]$ and $f(c)>0$ for some $c\in (a,b)$. Then $\int^{b}_{a}f>0$

My Attempt: Suppose $f$ is continuous for for $\epsilon=f(c)>0$ $\exists$ $delta_1>0$ such that whenever $x \in (c-\delta_1,c+\delta_1)$, we have $f(x)>f(c)$.

Now, how do I show that $\int^{b}_{a} f(x)$ $\geq \delta_1 f(c)
 
Mmk *eats @TedShifrin 's food$
 
LOL, @usukidoll: I'm a good cook.
 
@topologicalmagician partition
 
Yup, what @Leaky said. Show that a lower sum for some partition is at least that big.
No, you don't know $f(x)>f(c)$. That's no good.
 
oh that wasn't what I had in mind but ok :D
 
1:17 AM
What is the correct statement?
 
I am so lost as to what this is asking :/
 
I assumed contradiction, forgot to mention that
 
@Leaky, well then I'm not sure what you had in mind.
 
Besides functions.yhey
 
1:18 AM
Contradiction is not a good approach.
 
@usukidoll do you see how an element of the Baire space can be viewed as a function $\omega \to \omega$
or maybe you need some rest if you're lost in symbols
@TedShifrin proof by undergrad: assume A is false. [proof of A]. this contradicts with the initial assumption that A is false. so A must be true.
 
This boogy is due tomorrow . I did most of it though. I spend more of my time with Python
Cuz I wanna boost my resume ;)
 
@TedShifrin that's the correct statement: Let $f: [a,b] \rightarrow \mathbb{R}$ be continuous. Suppose $f(x)\geq 0$ for each $x\in [a,b]$ and $f(c)>0$ for some $c\in (a,b)$. Then $\int^{b}_{a}f>0$
 
@usukidoll the question is asking whether the set of injections $\omega \to \omega$ is open in the set of all functions $\omega \to \omega$
lmao
@topologicalmagician he means what is the correct version of $f(x) > f(c)$ in your proof
 
Which it is...... Aren't there open balls in there?
 
1:19 AM
well if you can prove it then it is!
 
hello. i have a confusion. can R treated as an R-module break up into a direct sum of other R-modules?

i felt like any submodule of R as an R-module should be free, but then it would have to have lower rank than 1, so must be trivial?

R doesn't have to be commutative
 
Damn... There is an open neighborhood too yipeee
 
so... you're asking to check?
 
Yes
 
then present your proof
@tigre yes. for example, R=Z/6Z, then R = Z/3Z + Z/2Z
 
1:20 AM
|f(x)-f(c)|<f(c) $\implies$ $f(x)>f(c)$
 
@tigre: Why must it have lower rank?
 
oh wait
lol
 
LOL is right !
 
print('I am doomed')
doom_meter = input('How doomed are you?')
 
yes
 
1:22 AM
oh it didn't have to have lower rank
oops
thanks
 
I don't even know what rank means
is it the maximum cardinality of an independent subset?
 
well the submodule was to be free, so number of generators
 
Hey hot cats.
 
well not any submodule of a free module is free!
 
I can't prove for thisssss ughhhhhh
 
1:24 AM
unless you're in a PID
 
print('Python is better')
 
@usukidoll what are you working on?
 
if user.input == "print('Python is better')":
break
 
This boogy about Baire Space, omegas, and injection and surjection functions
Getting yeeted
 
less memes, more math
 
1:25 AM
@anakhro the only thing that is driving me nuts
 
I'm glad I'm working on cooking dinner instead of this stuff.
 
@TedShifrin what are you cooking? I made a calzone tonight.
 
Chicken piccata, wild rice, veggies.
 
@anakhro here it is x.x i.stack.imgur.com/FKcco.png
print('Ted is a good cook')
 
usukidoll is printing that only because she wants me to throw her a scrap.
 
1:26 AM
:V
 
Wow. I have no idea what :V means.
 
it means :S
 
Feed me
 
LOL. Oh.
 
wait it doesn't mean :S?
 
1:27 AM
Idk
 
lol
 
okay, so $f(x)>\frac{f(c)}{2}$ for all x in a delta neighbourhood of c, so would the right way of approaching it would be by constructing a partition such $a_0,a_1,a_2....,b$ such that $a_i-a_{i-1}$ < $\delta$ for each i?
 
@usukidoll what does the notation $^\omega\omega$ mean?
 
It's the Baire Space
$\omega$ is the set of natural numbers
 
OK, @topologicalmagician. Why not take a partition like $a<c-\delta/2<c+\delta/2<b$?
 
1:28 AM
So what does it mean?
There are many Baire spaces.
 
@anakhro it's the functions $\omega \to \omega$
it's the Baire space
 
THE Baire space?
 
I had never heard of the Baire space heretofore.
 
Yeah there's a unique mapping aka injection
Dhiwnwuwmqhq d
 
if G is a finite group, a group ring k[G] for k a field is artinian, so has a decomposition into a direct sum of indecomposable projective modules. how do you actually find these modules? in practice i mean
 
1:29 AM
This is why I prefer Python. I rather code -_-
 
@TedShifrin oh wow. That cleared things up
 
@tigre that's the content of representation theory
 
@tigre: You should page @MatheinBoulomenos or one of our algebra czars.
 
Sooooo
 
it's hard to find the modules
 
1:30 AM
makes much more sense than the i was trying to construct
 
you can use Brute force
 
thank you thank you thank you thank you
 
Am I doomed?
 
so find a random element there, move it around using k[G], hope that you get a smaller module
 
People coming out of low-level calculus think everything has to be a uniform partition, @topologicalmagician.
 
1:31 AM
@usukidoll so which part are you stuck on?
 
OK, back to the kitchen for me.
 
@LeakyNun you mean take a random element, and treat it as a free module, and hope it doesnt generate the group ring?
bye ted
 
don't treat it as a free module
 
generate a free module with it i mean
 
look at the submodule it generates
 
1:31 AM
I said that I is open because there is a unique mapping and there exists interior points and open balls @anakhro
Meep
 
the module generated wouldn't be free
 
@Ted Shifrin , just out of curiosity, what do you mean by low-level calculus?
 
oops i did it again
it will be free if Ann(a) = 0?
 
This is alllllll I can think offfffff
 
it will be free if it is the whole ring!
 
1:32 AM
@usukidoll unique mapping where?
 
From omega to omega
 
Unique injection you mean?
 
I is an injection function
Ya
 
I is a set, not a function.
 
@anakhro you'll never be able to decipher the question from her words
 
1:33 AM
:S
 
I doubt @usukidoll understands the question herself
 
Well let me see how far I can get instead of discouraging people, @LeakyNun
 
@tigre you should stop concerning yourself with whether modules are free
 
I is a set?
 
@usukidoll Yes.
 
1:34 AM
So I is a set of injective functions
Set of injective functions with unique mappings
 
@tigre let's look at $\Bbb R[C_2]$ for a simple example
 
I still think this is open
 
Yes, precisely the set of all injective functions which are elements of $^\omega\omega$.
 
can you describe its elements?
 
Open for salad
 
1:35 AM
@usukidoll where are you seeing "unique"?
 
@LeakyNun it's elements are $\{a[0] + b[1]\mid a,b\in R\}$ where $[0],[1]$ are the elements of $C_2\cong(\Bbb Z/2\Bbb Z,+)$
 
I thought injection functions map one element to the next element without repeat
 
@tigre now pick your favourite element
 
$[0]+[1]$
 
what do you get if you hit it with the elements of the group?
(that's a strategy: take the G-span, and then take the K-span)
 
1:38 AM
@usukidoll if I have a function $f\colon X\to Y$, what is the formal definition for $f$ being injective?
 
Damn if f(a) =f(b) then a=b
 
Great.
 
Or in this case if f(x)=f(y) then x=y
 
i would have said the free module generated by $[0]+[1]$, but explicitly
$$(a[0]+b[1])([0]+[1])=a[0] + b[1] + a[1]+b[0]=(a+b)[0]+(a+b)[1]$$
which i think is still a free module generated by $[0]+[1]$ and i could even treat it as a R-module of rank 1 if i wanted
 
Now whaaaaaat?
 
1:40 AM
@tigre it's not a free module
 
Alright, so we have that $I = \{f\colon\omega\to\omega \mid f\text{ is injective}\}$, does it make sense what $I$ is now?
 
it's not a free $\Bbb R[C_2]$-module
it is a free $\Bbb R$-module (i.e. $\Bbb R$-vector space) since every vector space is free
 
I is the set of injective functions with the set of natural numbers
 
@tigre well you'll find it easier if you take the G-span and then the K-span
instead of both at once
 
@usukidoll what do you mean by "with the set [...]"
Do you mean "on the set [...]"?
 
1:43 AM
@tigre so the $\Bbb R$-span of $[0]+[1]$ is an $\Bbb R[C_2]$-submodule of $\Bbb R[C_2]$
can you see a complement?
 
@LeakyNun oh, its not free because (a[0]-a[1])([0]+[1]) = 0
 
precisely
 
righttttt
ok
am i right though that if ann(a)=0, then the module generated by a is free?
 
yes
 
that should deal with the other part of linear independence (which i was ignoring in the case there was one generater :S )
 
1:44 AM
but you should be able to prove that the only free submodule of K[G] is K[G] itself and the zero submodule
hint: it's a basic linear algebra argument
 
@anakhro ya
 
@usukidoll it's important to be clear about what you mean when it comes to mathematics.
 
:S
 
So what is an open set in the Baire space? @usukidoll
 
It has interior points
And open balls
 
1:48 AM
"it has interior points" describes a lot of sets, many of which are not open.
So what constitutes an "open ball" in the Baire space?
 
Infinite sequences?
Whoa wait. Many sets are not open...so there are exterior points :/ and the complement is open if it's closed
 
What is your definition of an open set?
The first one that you were given in your notes.
Not any of the equivalent ones.
 
Open interval
 
@usukidoll "open interval" is the definition of "open set"?
 
I'm stressing outtttt xc
 
1:53 AM
Why are you stressing out?
 
Too many similar definitions...
Ufhhwjjwjwjjw
I feel like guessing but I shouldn't do that since this is the only exercise left x.x
Dagh
 
To determine if something is open/closed in $^\omega\omega$ you need to know what sets constitute as open/closed sets in $^\omega\omega$.
That's really the bottom line.
 
I have it in my notes but both cases involve trees and I know the problem isn't asking for trees in here ;/
 
How do you know the problem isn't asking for trees?
 
I don't tbh. I thought the problem didn't involve trees
 
2:00 AM
You are saying that the exercise at the end of a section titled "Trees" in fact doesn't involve "trees"?
 
Right....I think I am sooo thinking of this problem wrong because I know the second one involved trees so I used one of the lemmas
 
@anakhro Poggers
 
@LeakyNun ?
 
:/
Pogges is the Pepe frog meme dude
 
every module has a generating set (take all the elements themselves, which may be very large (uncountable etc))

right?
 
2:03 AM
@usukidoll Let me give an example: suppose we are considering sets of $\mathbb R$. Is the set $(0,1)\cup (3,4)$ open?
 
@tigre correct
 
and sadly, finitely generated modules can have non-finitely generated submodules
 
that is true
 
and when a module has only finitely generated submodules we call it noetherian. is there some nice conditions of R that make R-modules always finitely generated?
 
That has to be open because if we take the union of two open sets then the union is also open because (0,4)
 
2:05 AM
@tigre every any nonzero ring R has a not-finitely-generated module
 
Great, so you notice how you found that this set is open because you knew what "open sets" look like in $\mathbb R$?
 
@LeakyNun by just taking $$\bigoplus_{i=1}^\infty R$$ ?
 
precisely
 
ok good point
was k[G] a noetherian module over itself for finite G?
 
yes
 
2:07 AM
@anakhro right
 
@usukidoll so that's a good place to start with this question. Figure out what the open sets in $^\omega\omega$ look like, and then you can better understand whether or not $I$ is open or not.
 
But what does open look like for Baire Space? There's my issue right there meep
 
That's why you go to your notes and look at the definition for open set in the Baire space.
 
Families of basic open sets
 
@LeakyNun even if k is characteristic p, and p may divide |G|?
 
2:16 AM
yes
 
how do i show this?
 
Basic open sets sounds like a good place to start, @usukidoll. What does a basic open set look like?
 
@tigre k[G] is fin. dim. over k
any k[G]-submodule is a k-subspace
so take a k-basis
this will be a k[G]-generating set
 
wow thats smart
 
Everything is open...
Unions are open and intersections are open
 
2:19 AM
Is [0,1] open in R?
 
finite intersections?
 
I just read a wiki article which claimed the following statement:
 
Noooo... [0,1] is closed and bounded...compact
 
@usukidoll why is it not open?
 
Closed interval... Closed and bounded....compact
 
2:21 AM
Oh wait there’s a typo let me try again
 
Closed sets can also be open.
 
Nuggh
 
@usukidoll the point is that you really need to learn what a basic open set looks like.
 
2:25 AM
It's a basis
 
@usukidoll What is a basis?
 
XC
A collection of subsets
 
That's all?
 
Yeah because I am getting burned out just reading past ebooks and what not x.x
 
Well I can assure you that that is not the definition of a "basis".
But since you are burnt out, you should go to sleep.
 
2:29 AM
that's what I have been saying from day 1
 
@usukidoll you write in the passive voice too much. responses are more confusing if they are written in the passive voice
 
@WilliamSun still waiting for your question
 
This is...not really a well-known fact I guess?
I mean I haven’t seen it in any analysis or point set topology book...
 
you're asking us to sample the population to find out how many people know it?
or you're asking if its true, because you haven't seen a reliable reference
 
Oh I am asking for its applications.
Is this just a useless formulation so people forget about it?
 
2:34 AM
useless???
What constitutes as "useful"?
 
It is much cleaner than the standard definition I mean.
To prove something? For example using this we can prove something interesting that does not follow directly from the usual definition.
For example use the universal property of tensor products to prove the commutativity instead of using the construction.
Then I call this an “application” of universal properties.
 
It's an "if and only if"
If you have "A iff B" and then A-->X or B-->Y, then you automatically get both A-->Y and B-->X for free.
 
So ok here is the usual definition for limit of a function:
 
@LeakyNun because when you treat k[G] as a module over itself, this corresponds to the standard k-linear representation of G, and this decomposition is somehow related to the decomposition of that representation into its irreducible subrepresentations?
 
yes
 
2:43 AM
those subrepresentations for some reason correspond to projective modules?
and the irreducibility corresponds to the indecomposability
 
And in this paragraph it doesn’t mention about the isolate points.
 
> For the group algebra of a finite group, the (isomorphism types of) projective indecomposable modules are in a one-to-one correspondence with the (isomorphism types of) simple modules: the socle of each projective indecomposable is simple (and isomorphic to the top), and this affords the bijection, as non-isomorphic projective indecomposables have non-isomorphic socles.
 
@WilliamSun I am not sure what you are asking anymore.
 
the simplest example would be $K[C_2]$ when char(K)=2
where the whole ring is projective indecomposable
and the simple module is the span of $[0]+[1]$
 
i guess i should study $k[C_2],k[C_3],k[C_2\times C_2],K[S_3]$ in $\text{char}(k)=2$, and $k[C_3],k[S_3]$ in $\text{char}(k)=3$
 
2:50 AM
sure
how about $k[C_n]$ in char(k)=0
 
Looking for n-ality transformation examples
i.e. a generalisation of dual mathematical objects
 

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