How do i prove that this family of sets is closed under compliments?

[manic kelp]

Assuming that A is part of the family, either A or its complement is finite. For the complement, either it’s finite or its complement is finite (because the complement of the complement of A is A, reference the inference rule of double negation as proof). So the family is closed under complements.

I believe that is how you can prove it, if there’s something wrong with that let me know

I can also help you formulate the proof, as this is just the reasoning behind the proof

[lost temple]

thx alot man im doing my hw rn if i need help can i dm u?

[manic kelp]

Better to write in here, I’m pre university, so there’ll probably be problems I won’t be able to help you with

[lost temple]

no problem i will ask from here

[lost temple]

I dont think i can make this more mathematical

[manic kelp]

Yep that looks pretty good

Though I’m not sure about the third line

Did you write this yourself or was it given to you?

[lost temple]

i wrote it myself

[manic kelp]

Okay so

The problem I think lies in the third line is that while any $A^c$ that’s in the family, while the proposition you write does hold true, that doesn’t mean that $A^c$ must be in the family if the proposition is true

[cobalt hollow]

FirstNameLastName

[manic kelp]

And that’s because of the order in which you arranged the parts in the third line

If you write that any A^c is in the family if the proposition is true, only then you show that A^c is in the family if A is in the family

[lost temple]

wait a sec

A is the power set for N

because it says or

means both sides included

can happen at the same time

means every possible subset of N is in A

[manic kelp]

Do you have the latex notation for that strange A?

[lost temple]

oh wait

[manic kelp]

Uneven numbers

Complement even numbers

Both infinite

[lost temple]

sry my brain glitched for a moment

xD

[manic kelp]

lol all good

[lost temple]

either or needs to be lower than inf not lower or equal

ive been doing this stuff for tha last 3 days

12 hours at a time

xD

and im only half done with the homework

but considering i started from 0 its not so bad

[manic kelp]

If you could give me this I’ll write something for you

[lost temple]

\mathcal{A}

[manic kelp]

How do you like it? I find set theory quite fun

[lost temple]

set theory is fun and all but im not used to doing proofs

i learned doing proofs in 3 days

xD

[manic kelp]

Oh wow okay haha

[manic kelp]

First year at Uni?

[lost temple]

the current version for my homework

[lost temple]

more like first year in masters where i passed my bachelors with NOT doing proofs

xD

[manic kelp]

Ohhhh lmao

That’s wild

[lost temple]

yea im statistics major and am taking statistical theory rn

this is the first time im seeing anything related to this stuff

i only learned how to do everything practical

[manic kelp]

Well I’m self taught so let’s see what masterpiece we can come up with as two people without much proof writing experience 😂

[lost temple]

oırgsjrıogsgısrg

such nice

[quiet lintel]

Question 5 (2) has an error

[lost temple]

please do tell

[quiet lintel]

The intersection of the (a-1/i, b] includes a

Unlike (a,b]

[lost temple]

O IT NEEDS TO BE UNION

wait

no

[quiet lintel]

I also don't understand why you're showing a half open interval is in the borel σ algebra

When that is given by definition

[lost temple]

but dude the question defines borel sigma algebra as the union of every half open set

so i used half open sets

actually

ur right

[quiet lintel]

But (a,b] is a half open set

It's the most straightforward set to include

Everything further on is correct

[lost temple]

so this is enough?

[quiet lintel]

Yes

In fact it can still be shortened because there's no need to write what {b}^c is

Since the only thing that matters is you're removing it

[lost temple]

nahhh i gotta use the already defined things

xD

i wrote the proof for closure under countable intersections

at the preliminaries

[quiet lintel]

Yeah but sigma algebra is closed under complements gives you {b}^C in B(R)

And then (a,b] \ {b} = (a,b] n {b}^c will do

No need to say {b}^c = ...

It's not wrong, but it's extraneous

[lost temple]

but to use closure under countable intersections i gotta show that difference can be written as an intersection no?

i mean anyone knowing what difference is would know this but still

[quiet lintel]

Still no need for the intervals

Ffs

[lost temple]

xD

[quiet lintel]

(a,b] n {b}^c will do for the intersection

[quiet lintel]

This is towards me I wrote something longer and accidentally deleted it

[lost temple]

this is the last version

[quiet lintel]

Yep

In proofs, it's good practice to not give extra information

It just clutters things up

[lost temple]

alright now lets return to the issue at hand

[quiet lintel]

Ok yeah you can't write A^c in curly A because that's precisely what you're trying to show

[lost temple]

where

[manic kelp]

For any $A \in \mathcal{A}$ it holds true that $(|A| < \infty) ∨ (|A^c| < \infty)$.
This is equivalent to $(|(A^c)^c| < \infty) ∨ (|A^c| < \infty)$, according to the law of double complementation. This shows that $A^c$ is also an element of \mathcal{A}. Therefore, $A^c \in \mathcal{A}$ for any $A \in \mathcal{A}$

[lost temple]

xD

obsidian for the win

[manic kelp]

Bullshit

[quiet lintel]

Third line here

[lost temple]

there is no curly bracket there

[cobalt hollow]

FirstNameLastName
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quiet lintel]

I mean curly A for mathcal{A}

[lost temple]

oh

[quiet lintel]

Sorry

I should have clarified

'Curly' is a common way of reading it out loud

[lost temple]

ok got it

[lost temple]

this is the same as mine tho and something is missing

i dont know what

[manic kelp]

For any $A \in \mathcal{A}$ it holds true that $(|A| < \infty) ∨ (|A^c| < \infty)$.
This is equivalent to $(|(A^c)^c| < \infty) ∨ (|A^c| < \infty)$, according to the law of double complementation. This shows that $A^c$ is also an element of $\mathcal{A}$. Therefore, $A^c \in \mathcal{A}$ for any $A \in \mathcal{A}$, meaning that $\mathcal{A}$ is closed under complementation.

[cobalt hollow]

FirstNameLastName

[manic kelp]

@lost temple @quiet lintel I think this should be good

[lost temple]

cant we just say for a where |a|<inf its in curly A and for a where |a|=inf its compliment is in curly A. Every a that satisfies |a|<inf can be found via union of these two restraints

the left side contains continuity from below and right side contains the above

[quiet lintel]

If |A| = infty it's complement isn't necessarily finite

And hence not necessarily in curly A

[lost temple]

ur right

[manic kelp]

This goes beyond the scope of my knowledge tbh. But the second part of the conjunction in your first sentence looks strange, idk if you can say that

[manic kelp]

Right, odd vs even

[lost temple]

yea as edward pointed out its wrong

this shouldnt be this hard

xD

[manic kelp]

Why not just do this @lost temple

Unless there’s something about that that makes it not be a sufficient proof

[lost temple]

i was thinking whether if they got the same restraints.. yea of course they would

ok this is good

Now comes the supposedly harder part

but maybe i can do it myself

imma try first

i need to prove that curly A is also closed under countable unions

[pallid thorn]

Which definition of closed are you using here btw?

[lost temple]

it contains every union of every combination of sets inside it

if a1, a2 in curly A then a1 union a2 in curly A

can i say something like this?

[quiet lintel]

reminder that $\mathbb N$ is the positive whole numbers, not just the positive ones

[cobalt hollow]

Edward II

[lost temple]

isnt it natural numbers

[quiet lintel]

well yes but convention on 0 varies

tbf I should have said naturals and let you decide that

[lost temple]

0 isnt in N right?

[quiet lintel]

ok let's go with 0 not in N

[lost temple]

thats why i did not write equality on epsilon < a

[quiet lintel]

(it literally is not standard at all whether it is or isn't)

[lost temple]

if i can write that equality its better

[quiet lintel]

ok, but if a = 1, what is epsilon?

[lost temple]

0 i guess

[quiet lintel]

but you said it's in N

[lost temple]

lets say a > 1 then xD

[quiet lintel]

next question: what is your goal with taking these intervals

[lost temple]

maybe like this?

[lost temple]

i will say that a-epsilon and b+delta in curly A

then say that since all unions are in N

or maybe

no

all unions are in N

[quiet lintel]

you're trying to show that all countable unions are in N

so you can't use that fact

[lost temple]

ur right

[quiet lintel]

(also if you're using latex then \epsilon or \varepsilon give actual epsilons)

(also ping if you would like a hint at any point)

[lost temple]

mcal(E) looks better imo

im thinking of yoinking from proofwiki

and substituting my circumstances

[manic kelp]

Isn’t $(A \cup B)^c = A^c \cap B^c?$

[quiet lintel]

yes it is

[cobalt hollow]

FirstNameLastName

[manic kelp]

Seems like a useful property for this proof

[lost temple]

yes

hmm

may be

[manic kelp]

Either A or A^c are finite. Same for B and B^c

If both are finite then the union is

[pallid thorn]

Is this still for your first problem, or a separate one?

[manic kelp]

Otherwise the complement of the union is, because the complement of either A or B are finite

[lost temple]

the last one which is the proof for countable unions

[quiet lintel]

are countable unions required for a field

[lost temple]

finite unions is

right?

[quiet lintel]

(note to SWR I'm pretty sure a field is what is usually called an algebra)

[quiet lintel]

yes

focus on finite

and then countable is (2)

[lost temple]

i dont know what that means

countable unions can go to inf which is uncountable

so maybe finite ones cant go to inf?

[quiet lintel]

by go to inf, do you mean 'not be finite'

[lost temple]

is there another definition?

[quiet lintel]

I mean the usual words are 'be infinite'

and also this has nothing to do with uncountability

which is why I'm asking

[lost temple]

im stuck and its 11.56pm

i want to sleep but this hw is for tomorrow 10am xD

[quiet lintel]

fun

[lost temple]

very much so

[quiet lintel]

anyways firstnamelastname's idea is correct for showing finite unions

[manic kelp]

I’m on autumn break and still doing math till midnight 🗿

[lost temple]

you are doing a great thing

i used to play pc till my death

now im doing math hws till my death :/

[manic kelp]

Yeah it’s a much better use of my time than yt for example

Because I spend time learning new concepts in order to help other people

[lost temple]

You know the reason was (i believe) the urgency and the "must do" mentality

when you have that anything else you do feel much better for some reason

xD

[manic kelp]

It’s only today that I learned what a complement is, what “closed under …” means and what a countable union is

[lost temple]

now that im in a school that does not require me to commit to anything urgently i feel like studying more

but oh boy am i studying xD

[manic kelp]

lol yeah

I had a math exam coming up last week and spent the evenings on here helping people instead of learning the stuff I needed for the exam

Just so much more fun

Anyway, should I try to write out a proof for you?

[lost temple]

please do

[manic kelp]

Is there some standard notation for a countable union?

[lost temple]

\bigcup_{n \in\mathbb{N}}

[manic kelp]

$\bigcup_{n \in\mathbb{N}}$

[cobalt hollow]

FirstNameLastName

[manic kelp]

That’s the whole countable union?

[lost temple]

change the thing in the curly brackets with n = 1 and add a ^ after it to write n

oh no

xD

[manic kelp]

I need the whole thing

[manic kelp]

So the countable union is $\bigcup_{i}A_i$?

[cobalt hollow]

FirstNameLastName

[quiet lintel]

when i is allowed to range over a countable set

[quiet lintel]

best notation is to actually give that range alongside the union symbol like you did here

but sometimes it's implied elsewhere, e.g. in the above A_i are defined for i in N, so we assume the union also ranges over N i.e. a countable set

[manic kelp]

$\bigcup_{n \in\mathbb{N}}A_n$ is the way to go then?

[cobalt hollow]

FirstNameLastName

[quiet lintel]

yep

you're jumping ahead of yourself though

[manic kelp]

Meaning?

[quiet lintel]

this argument doesn't work for countable unions

[lost temple]

this is a countable union but its not its general form

[manic kelp]

I’m not a native speaker

[quiet lintel]

oh uh what languages do you speak

[manic kelp]

German

But I’m quite proficient in English

[lost temple]

bro which city u in

[manic kelp]

Near Frankfurt

[lost temple]

ah thats far

im in dortmund

[quiet lintel]

voreilig

[manic kelp]

To some degree it does though doesn’t it

[quiet lintel]

countable union of finite sets isn't necessarily finite

the most obvious one is that $\mathbb{N}=\bigcup_{n\in\mathbb N}{n}$

[cobalt hollow]

Edward II

[manic kelp]

A finite amount of sets

All with a finite amount of elements

[quiet lintel]

that's a finite union

[manic kelp]

Shouldn’t the union of those be finite?

[quiet lintel]

not a countable union

[manic kelp]

Oh

Misunderstood the meaning then

[quiet lintel]

Suppose we have finitely many sets $A_1,\dots,A_n\in\mathcal A$. If they are all finite, then $\bigcup_{i=1}^n A_i$ is finite as a finite union of finite sets, and hence $\bigcup_{i=1}^nA_i\in\mathcal{A}$. Otherwise, without loss of generality (will explain what this means if you want), assume that $A_1$ is infinite. Then we require $A_1^c$ finite as $A_1\in \mathcal{A}$. Therefore by De Morgan's laws $\left(\bigcup_{i=1}^nA_i\right)^c = \bigcup_{i=1}^nA_i^c\subseteq A_1^c$ is finite so $\bigcup_{i=1}^nA_i\in\mathcal{A}$

[lost temple]

does closure under countable unions include closure under finite unions?

[cobalt hollow]

Edward II

[manic kelp]

Oh damn. And still doing your stuff in English?

[lost temple]

bro im not german xD

[manic kelp]

Dang

[lost temple]

i came here 2 weeks ago

[manic kelp]

Beautiful country isn’t it

👀

[lost temple]

its orderly

[quiet lintel]

every finite union is also a countable union, because countable can mean finite or countably infinite

[lost temple]

which i like

[quiet lintel]

but not every countable union is a finite union

[lost temple]

yes thx

im doing the other part of the question

xD

[quiet lintel]

this btw is a proof that we have closure under finite unions

[manic kelp]

$\bigcup_{n\in\mathbb N}{2n}$

[cobalt hollow]

FirstNameLastName

[manic kelp]

Those are the even numbers?

[quiet lintel]

yes

[lost temple]

even numbers set

yes

[quiet lintel]

which btw gives the answer to (2)

[manic kelp]

But doesn’t that mean that curly A isn’t closed under countable unions?

[quiet lintel]

to be a field, we only require finite unions

[manic kelp]

Because very ${2n | n \in \mathbb{N}}$ is an element of curly A

[quiet lintel]

to be a σ-field, we also require countable unions

[manic kelp]

I don’t even know what a field is 🥲

[quiet lintel]

not we're not being asked to prove curly A is a σ-field, only asked if it is one

[quiet lintel]

also careful, this is the set {2,4,6,8,...} of all even numbers

you mean for each n in N, {2n} is an element of curly A

[manic kelp]

Oh true yeah

[quiet lintel]

but yes

[lost temple]

umm would this work?

chatgpt gave a surprisingly good looking answer

[quiet lintel]

it missed the case where one is infinite and one is finite

[lost temple]

yea it fails on that case

[manic kelp]

What does your second to last big cup mean here? Wouldn’t it make sense to write a big cap?

[quiet lintel]

yes

that is a typo

I can't edit it anymore so will recopy

Suppose we have finitely many sets $A_1,\dots,A_n\in\mathcal A$. If they are all finite, then $\bigcup_{i=1}^n A_i$ is finite as a finite union of finite sets, and hence $\bigcup_{i=1}^nA_i\in\mathcal{A}$. Otherwise, without loss of generality (will explain what this means if you want), assume that $A_1$ is infinite. Then we require $A_1^c$ finite as $A_1\in \mathcal{A}$. Therefore by De Morgan's laws $\left(\bigcup_{i=1}^nA_i\right)^c = \bigcap_{i=1}^nA_i^c\subseteq A_1^c$ is finite so $\bigcup_{i=1}^nA_i\in\mathcal{A}$

[cobalt hollow]

Edward II

[lost temple]

can i say that: Only one of $A,B$ is finite and the other one is countibly infinite. Lets say $A$ is countibly infinite. Since $A$ is countibly infinite, by the definition for $\mathcal{A}$, $A^c$ must be finite. Since $B$ is already finite, the countibly infinite set $A \cup B$ has a finite compliment.

[cobalt hollow]

Rootsyl

[lost temple]

but no B compliment is infinite

xD

[quiet lintel]

is this the last case

[lost temple]

ye

-i) Both $A,B$ are finite. The resulting set $C = A\cup B$ is also finite and thus in $\mathcal{A}$ by the definition for $\mathcal{A}$.
-ii) Both $A, B$ are countably infinite. In this case, since finite $A^c,B^c\in\mathcal{A}$ (notice that by the definition for $\mathcal{A}$ if $A \in\mathcal{A}$ is infinite then its compliment must be finite.) exist, $C =A^c \cup B^c$ is a finite set and thus $\in\mathcal{A}$.

[cobalt hollow]

Rootsyl

[manic kelp]

Suppose we have finitely many sets $A_1,\dots,A_n\in\mathcal A$. If they are all finite, then $\bigcup_{i=1}^n A_i$ is finite as a finite union of finite sets, and hence $\bigcup_{i=1}^nA_i\in\mathcal{A}$. Otherwise, assume that $A_k$ is infinite, where $k \in \mathbb{N}$ and $k \leq n$. This means that $A_k^c$ is finite, as $A_k\in \mathcal{A}$. Because of De Morgan's laws it also holds that $\left(\bigcup_{i=1}^nA_i\right)^c = \bigcap_{i=1}^nA_i^c$. Due to $\bigcap_{i=1}^nA_i^c \subseteq A_k^c$ and the fact that $A_k^c$ is finite, $\bigcap_{i=1}^nA_i^c= \left(\bigcup_{i=1}^nA_i\right)^c $ is also finite. Therefore, either the finite union of $\mathcal{A}$ is finite or its complement is, meaning that $\bigcup_{i=1}^nA_i\in\mathcal{A}$.

[quiet lintel]

I would say you should probably explicitly say why the complement is finite

[cobalt hollow]

FirstNameLastName

[lost temple]

i was thinking of examples for one infinite and one finite sets

for exp (4,10), (15,inf)

the complement of the union of these two is finite

but idk how can i write this

mathematically

[manic kelp]

Idk why you would write it, but $|((4, 10)\cup (15, \infty))^c| < \infty$ is how I would do it

[cobalt hollow]

FirstNameLastName

[lost temple]

maybe i can write "since combination of a finite and infinite set in curly A is infinite and has a finite compliment since finite set union infinite set would still result in an infinite set that has a finite compliment"

[lost temple]

generally i mean

[manic kelp]

If one of them is infinite but part of curly A, then its complement is finite.

[manic kelp]

Remember this

If the complement is finite, then so is the intersection including the complement and therefore the complement of the union

@lost temple

[manic kelp]

This is the general solution, written using my best English

[lost temple]

i think i got it

nope wait 10 more secs

[manic kelp]

There’s a few things wrong with this

Most importantly, you’ve only proven this for unions consisting of two sets in curly A

[lost temple]

i know

im writing stuff xD

[manic kelp]

Okay

[lost temple]

how about this

maybe change the last term with this

[manic kelp]

No idea what that last bit means

But you still haven’t shown it’s true for all finite unions

[lost temple]

i did by not specifying A,B and considering all possible scenarios

think like this

as A,B in N

[manic kelp]

Yes, but only in the sense that it’s very easy to deduce from there

You didn’t explicitly state that any finite union can be formed using unions of these sets, and that this therefore holds true for all finite unions

Which is something I believe you ought to do in a proof

[lost temple]

why should i? The definition for curly A states that?

if i can prove they are in A then they are finite

i mean thats how i think might be wrong

[manic kelp]

I think of it like this:

When I’m asked to prove that you can freely rearrange the elements of a sum, i need to show it for any number of elements. Showing it just for a+b is very close to the solution, but it doesn’t actually state what I’m trying to prove

[lost temple]

you are right

lemme tinker a bit more

[manic kelp]

And it’s harder to show that it seems, because it’s not enough to show that you can group together these sets to finite unions and still have them be part of the family. You need to show that every possible finite union can be written this way

[manic kelp]

Which is why I would just do this

And which is why I asked how to write a finite union

Because that’s way easier than showing that curly A is closed under finite unions by showing that the union of two sets in curly A is also in curly A

[manic kelp]

Notice that in (ii), you defined C to be the union of the complements of A and B. This doesn’t show that the Union of A and B is in curly A, because C is neither the union of A and B nor the complement of the union

[manic kelp]

Going to bed now, I’ll have a look again when I wake up. Good night

[lost temple]

me too