Unfair Question!

[crazyaga !!gDYZXg02vsZ No.10534082]

This was in my test today. i had to choose only one right option. but here is the problem

I'm pretty sure option 2 cannot be proven wrong due to Godel's incompleteness theorm, and option 3 can be right or wrong depending on whether you include zero as a part of the natural numbers which purely defends on the definition you use.

I chose 2, but i think this is a very unfair question. ill be very mad if i lose points because of it.

If i'm being terribly wrong in calling this question unfair please tell me why

[crazyaga !!gDYZXg02vsZ No.10534086]

>>10534082
*which purely depends on the definition you use.

[Anonymous No.10534108]

>>10534082
You got it right, and should try giving a bijection between B and R instead of going "muh Godel" outta nowhere.

[Anonymous No.10534115]

>>10534082
What am I reading in answer 1 there? Is that an aleph zero?

>I'm pretty sure option 2 cannot be proven wrong due to Godel's incompleteness theorm,
No, that is bullshit.

>and option 3 can be right or wrong depending on whether you include zero as a part of the natural numbers which purely defends on the definition you use.
No. P(N) definitely includes the set {1, 2, 3}, which is not a real number, so (3) is definitely false.

>If i'm being terribly wrong in calling this question unfair please tell me why
I can definitely rule out answers (2), (3), and (4). I don't see anything unfair about it, assuming I have correctly interpreted answer (1).

[Anonymous No.10534120]

>>10534108
I'm dyslexic and read B=P(N) for some reason, this >>10534115 anon is correct.

[crazyaga !!gDYZXg02vsZ No.10534121]

>>10534115
>Is that an aleph zero?
yes

>I can definitely rule out answers (2), (3), and (4). I don't see anything unfair about it, assuming I have correctly interpreted answer (1).
1 is definitely not correct because the power set of the natural numbers has a cardinality which is higher than aleph zero

[Anonymous No.10534123]

>>10534121
Except that isn't the power set of the naturals, my fellow dyslexic.

[crazyaga !!gDYZXg02vsZ No.10534128]

>>10534123
It isn't, but it should have the same cardinality as the power set of naturals because An includes all the natural numbers.

Which is why option 1 cannot be correct, unless im getting it wrong

[Anonymous No.10534129]

>>10534121
>1 is definitely not correct because the power set of the natural numbers has a cardinality which is higher than aleph zero
But B is not the powerset of the natural numbers. B is in fact countable.

[Anonymous No.10534131]

>>10534128
I'll point out one obvious difference. B doesn't contain the set of all naturals. In fact, it doesn't contain a single infinite set.

[crazyaga !!gDYZXg02vsZ No.10534135]

>>10534129
How, just how is it countable?

>>10534131
I don't get why the power set of An when An includes all the natural numbers is different than the power set of all the natural numbers without zero.

What is the difference?

[Anonymous No.10534137]

>>10534086
But it’s clear what definition they used

[crazyaga !!gDYZXg02vsZ No.10534142]

>>10534137
You can deduce they included zero in the natural numbers, but they did not specify it in the question

[Anonymous No.10534143]

>>10534135
An doesn't include all natural numbers.

[Anonymous No.10534151]

>>10534082
>option 3 can be right or wrong depending on whether you include zero as a part of the natural numbers
Under what construction of the naturals and reals is the powerset of the naturals a subset of the reals?

[Anonymous No.10534153]

>>10534135
Countable union of countable (finite, at that) sets is countable. The power set of each An is finite. Question seems perfectly fine to me.

[Anonymous No.10534156]

>>10534135
>What is the difference?
For one thing, B does not contain the set of all odd numbers.

B is not the set of all subsets of the natural numbers. B is the set of all FINITE subsets of the natural numbers, which is very different.

>How, just how is it countable?
Map each set S in B to the number in binary representation with one 1 bit for each number n in S. So for S = {1, 4, 6}, map(S) = 2^1 + 2^4 + 2^6 = 82.

You can do this, because B only contains the finite sets of natural numbers, not all sets of natural numbers.

[crazyaga !!gDYZXg02vsZ No.10534157]

>>10534143
I don't get why tho. why isn't it possible to show a bijection between An and the natural numbers?

>>10534151
Oh i written it wrong. it was supposed to be a subset of B. sorry just had a typo

[Anonymous No.10534158]

>>10534135
B is set of countably infinite countable sets, but each subset of B has finite cardinality. the answer is A, basically it's a subtle version of the question "can you map N to NxN?" but the construction is a bit more nontrivial

[Anonymous No.10534162]

>>10534157
>why isn't it
It is. We ennumerate all elements of An, then all those of An+1, and so on. We can do so because all An have a finite amount of elements.
If something is in the union, it is in some An. Thus, it's eventually ennumerated.

[crazyaga !!gDYZXg02vsZ No.10534166]

>>10534158
I see. i thought An has a set which includes all the natural numbers.... :c

[Anonymous No.10534170]

>>10534142
>You can deduce they included zero in the natural numbers, but they did not specify it in the question
They also didn't define the union symbol, the powerset symbol, the N symbol, the R symbol, the = symbol, the > symbol, the aleph-0 symbol, etc. You should've been told whether or not 0 is a natural number at some point during the course

[Anonymous No.10534173]

>>10534166
[eqn]\mathbb N = \bigcup_{n\geq 1}A_n[/eqn]

[crazyaga !!gDYZXg02vsZ No.10534180]

>>10534173
It was P(An) in the question tho
But nice magic

[Anonymous No.10534186]

>>10534180
I know, I was confused as to what you meant by "i thought An has a set which includes all the natural numbers". An is a finite set for all natural n

[Anonymous No.10534192]

>>10534082
>I'm pretty sure option 2 cannot be proven wrong due to Godel's incompleteness theorm
Why? It's very easy prove

[Anonymous No.10534197]

>>10534108
>You got it right
no he didn't. the union is all finite subsets of N, not all subsets of N. that's a countable number of things.

[Anonymous No.10534225]

is this abstract algebra?

[Anonymous No.10534227]

>>10534082
The answer is 1. It's obviously not 2, brainlet. If 2 were true, then 4 must also be true. But we know only one option is true.

[Anonymous No.10534234]

>>10534225
this is just set theory

[Anonymous No.10534235]

>>10534225
No. This is set theory.

[Anonymous No.10534246]

>>10534235
>>10534234
but what class does it get taught in?

[crazyaga !!gDYZXg02vsZ No.10534248]

>>10534246
It was a discrete mathematics class

[Anonymous No.10534251]

>>10534246
Set theory.

[Anonymous No.10534252]

>>10534246
usually it's the sophomore undergrad "Fundamentals of Math" course where you learn proofs and set theory and construct the naturals and reals and dedekind cuts and stuff

[crazyaga !!gDYZXg02vsZ No.10534258]

>>10534227
No, if 2 was true it doesn't mean 4 would have to be true too, because i thought it equals to |P(N)|, not bigger than |P(N)|

However this discussion convinced me i was wrong nevertheless and the correct answer was 1. i'm not a brainlet, this was not a very trivial question, i just got it wrong

[Anonymous No.10534275]

>>10534258
>i'm not a brainlet, this was not a very trivial question, i just got it wrong
That sounds about right. Reasoning correctly about the distinction between "all subsets" and "all finite subsets" takes some practice.

[Anonymous No.10534280]

>>10534258
Did you know |P(N)| = |R| and that |P(X)| > |X| for all set X? So if |B| = |R|, then |P(B)| > |R| = |P(N)|.

[Anonymous No.10534285]

>>10534082
>I'm pretty sure option 2 cannot be proven wrong due to Godel's incompleteness theorm
You have no idea what you're talking about

[crazyaga !!gDYZXg02vsZ No.10534286]

>>10534280
You are right :|

[Anonymous No.10534289]

>>10534082
(4) seems wrong, because B is a union of finite sets and N is an infinite set. it's a bit hard to see since you have "n>=1" which kind of "suggests" infinity but it's like this. No matter what n you "take" there's always another natural number. And yet if you "take" some supposed maximum natural number, n would have to be greater than that number (a contradiction) for (4) to be true, so it's untrue no matter how you look at it.
(2) I don't think a countable set can have equal cardinality to an uncountable one
(3) is just bizarre, seems like an actual typo or something, not sure if you can really say a set of numbers is a subset of R, so it's false. i'm raising my hand and asking the professor about it. If by R he meant B then it's false for the same reason as (4)

finally i don't even know what that squiggle is in (1) but the other three are wrong so i'm hail marying (1) and hoping it's just the thing i forgot to study
i'm a nigger though -- a nigger that does well on tests, but still a nigger nonetheless, so given my low average iq you should take all this with a grain of salt
thanks for the fun

[Anonymous No.10534293]

>>10534082
The answer is 1.
Reason: countable union of finite sets is countable.
2 and 4 are both impossible as result. 3 could in principle be true if you defined the reals in a special way that nobody ever does, so we can safely rule it out for your course.

[Anonymous No.10534303]

>>10534289
>(4) seems wrong, because B is a union of finite sets and N is an infinite set
4 is wrong, but not for this reason. The union of finite sets can infinite.
>it's a bit hard to see since you have "n>=1" which kind of "suggests" infinity
It's an infinite union.
>No matter what n you "take" there's always another natural number.
Yeah, what's your point? It's a union over all natural n, not from n=1 to some other natural number
>And yet if you "take" some supposed maximum natural number, n would have to be greater than that number (a contradiction) for (4) to be true, so it's untrue no matter how you look at it.
Taking a maximum natural number is already a contradiction in of itself. You seem very dumb

[Anonymous No.10534316]

>>10534303
>Yeah, what's your point? It's a union over all natural n, not from n=1 to some other natural number
so it'd be true if it was >= instead of just >
?

>You seem very dumb
i am. average 85 IQ actually

[Anonymous No.10534324]

>>10534316
>so it'd be true if it was >= instead of just >
What? It's the same as union from n=1 to infinity

[Anonymous No.10534327]

>>10534128
By Zorns lemma, every partially order set with an upper bound contains a maximal element. An is partially ordered under the subset relation and has an upper bound, namely the first inaccessible ordinal, omega. Since every value of An is finite, the cardinality of Powerset(An) is a natural number for all values of n, hence cardinality(B) = Aleph Null.

Note: the cardinality of B should equal the cardinality of the union over n>1 An. In otherwords, the whole powerset component of the question is just their to trick you.

[crazyaga !!gDYZXg02vsZ No.10534329]

>>10534327
Thanks

[Anonymous No.10534332]

>>10534324
P(B) would be the union of the power set of all the sequences from 1 to n, and you say n goes from 1 to infinity, implying B is an infinite set
and P(N) is the power set of N
doesn't that mean P(B) = P(N)?

[Anonymous No.10534340]

>>10534332
>you say n goes from 1 to infinity, implying B is an infinite set
B is an infinite set, but not all infinite unions result in infinite sets. Do you know the definition of union?
>doesn't that mean P(B) = P(N)?
Assuming you meant |P(B)| = |P(N)|. No, B being infinite does not imply that. But it is true that |P(B)| = |P(N)|

[Anonymous No.10534343]

>>10534332
No. They have the same cardinal, but they are not the same. The elements of P(B) are sets of sets of natural numbers. P(B)'s elements are just sets whose elements are elements from P(An), which by definition of power sets are, well, sets. Meanwhile the elements of P(N) are just sets of natural numbers.

[Anonymous No.10534448]

>>10534343
Question: Does [math]P(A_n) = P(N),\ \text{with}\ n \geq 1[/math]?

[Anonymous No.10534474]

>>10534082
B is just the set of finite subsets of N.
Clearly only option 1 is correct.

[Anonymous No.10534568]

>>10534448
No

[Anonymous No.10534575]

>>10534568
How about the union of P(An) for all n? Does that equal P(N)?

[Anonymous No.10534717]

>>10534575
Yes

[Anonymous No.10534719]

>>10534575
No

[Anonymous No.10534868]

>>10534568
>>Does [math]P(A_n) = P(\mathbb{N})[/math] for [math]n \geq 1[/math]?
>No
What's the difference?

[Anonymous No.10534888]

>>10534082
>n equals or higher to one
>zero is optional
You are an idiot

[Anonymous No.10534890]

>>10534868
They differ by infinitely many elements. For example, N is in P(N) but not in P(A_n)

[Anonymous No.10534901]

bijection between cardinality of a power set and N is 2^n. so 1's true.

[/sci/ guy !!U0XrgWQLYj3 No.10535286]

>>10534082
Pretty simple question.
An = n+1\{0}
|PAn|=2^n
We count the power sets, so this is a countable union of finite sets. 1 is correct.
>defining N matters
no, if you want to use process of elimination for 4, cardinality won't change. For 3, I guess you really just need to understand the question.
>godels
If your professor hasn't explicitly written the theorem in class, just pretend it doesn't exist.

Also, 0 is in N (should probably call it omega).

[Anonymous No.10535594]

I wanna know your thought process when you came to the conclusion that Godel's incompleteness theorem implied that option 2 could not be proven wrong.

[Anonymous No.10537287]

>>10535594
probably something along the lines of this:
>me smart, me no brainlet
>number 2 hard
>heh, must be Godel
>Godel says its wrong
>I pick it anyway
Why pick an answer you think is wrong?