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Q) Let $(A_{n})_{n\geq 1}$ be the collection of non empty subsets of integer set $\mathbb{Z}$ such that $A_{m}\cap A_{n}=\varnothing$ $\forall m\neq n$. if $\mathbb{Z}=\bigcup_{n\geq 1}A_{n}$ then which of the following is/are true?

a) $A_{n}$ is finite for every integer $ n\geq 1$.

b) $A_{n}$ is finite for some integer $ n\geq 1$.

c) $A_{n}$ is infinite for some integer $ n\geq 1$.

d) $A_{n}$ is countable(finite or infinite) for every integer $ n\geq 1$.

My approach: if we choose $A_{n}$ to be singleton sets then C option is incorrect. if we choose $A_{1}=\{0,1,2,3,...\}$ and $A_{2}=\{-1,-2,-3,...\}$ then A and B options will be incorrect, and only D option is correct.

Please do check my approach and provide solution.( It was asked in CSIR NET Mathematical Sciences Dec 2023 exam)

Asaf Karagila
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Gajendra Basavaraju
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    "The" collection of all pairwise disjoint sets, or "a" collection of pairwise disjoint sets? – DonAntonio Dec 30 '23 at 11:36
  • In the exam question only what I typed in the question was there. so lets assume it is collection pair wise disjoint sets ( All sets in the collection are disjoint). – Gajendra Basavaraju Dec 30 '23 at 11:41
  • Assuming it is “a” collection, I don’t think your counterexample works, you are implicitly assuming that all A_n’s beyond A_2 are the empty set. At least that is how I am interpreting the question – Academic Dec 30 '23 at 12:10
  • It must be "a" collection, since from every such one we can construct a new different list of pairwise disjoint sets... – DonAntonio Dec 30 '23 at 12:17
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    I think one way to do B would be to use unique factorisation, let one set be the collection of primes, then other prime powers. Once those are done, take product of exactly two primes, then product of prime square and a prime etc. The issue with this is showing it can be enumerated. One natural correspondence to take the least element in a set as its distinct representative, and of course the number of such representatives will be countable – Academic Dec 30 '23 at 12:22
  • Use either \emptyset, which yields $\emptyset$, or \varnothing, which renders as $\varnothing$, rather than \phi. – Arturo Magidin Dec 30 '23 at 12:22
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    Your counterexample for C is fine. Your answer for A and B is correct but not your counterexample. You must find pairwise disjoint $A_n$s, infinite for every $n\in\Bbb N$ (not only for $n\le2$) such that their union is $\Bbb Z$. – Anne Bauval Dec 30 '23 at 12:27
  • @AnneBauval I can't see where do you deduce from that there must be countable many sets in the pairwaise family of sets... – DonAntonio Dec 30 '23 at 15:20
  • @AnneBauval From what I understand I don't think the family is indexed *necessarily@ by ALL the naturals. It is just a natural index which, as far as I can see, can end after a finite number of cases. – DonAntonio Dec 30 '23 at 15:56
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    @AnneBauval what I posted in the question was only the information that has been given in the exam, I assumed n to be finite in exam and did only D option. But now realised that union can be infinite sets. – Gajendra Basavaraju Dec 30 '23 at 17:04
  • @Academic yes I assumed union to be finite $A_{n}$'s in exam, but then realised it need not be, but still only D option is correct. – Gajendra Basavaraju Dec 30 '23 at 17:08
  • @DonAntonio at first I also assumed like you, that there need not be pairwise disjoint infinite subsets because $n\geq 1$ means we can choose any finite number , but now realised that symbol is short for $n\in \mathbb{N}$ , thats why we have to assume there will be infinitely many $ A_{n}$'s in the colllection. – Gajendra Basavaraju Dec 30 '23 at 17:24
  • Thank you for your answer to both DonAntonio and me. Can you please edit your post to clarify accordingly? – Anne Bauval Dec 30 '23 at 19:34
  • @AnneBauval No, I can't edit because this was the actual question asked in the exam, if I edit and add it is pairwise disjoint or the family of subsets are infinite then it will give hint and spoil the thinking of solver. – Gajendra Basavaraju Dec 31 '23 at 06:02
  • No, you won't give a hint nor spoil the thinking. You will just specify the question more clearly. You already specified "it is pairwise disjoint" by writing $A_{m}\cap A_{n}=\varnothing$ $\forall m\neq n$. In your comment, you clarified that the ambiguous "$\cup_{n\ge1}$" of your post meant $\cup_{n\in\Bbb N}$, i.e. that the family of subsets is infinite. I think this clear specification of the question is mandatory and should be within the post, not only in comments (which may be deleted). Future readers won't care if the "actual question asked in your exam" was not well written. – Anne Bauval Dec 31 '23 at 10:07
  • @AnneBauval there is a post on stockexchange itself which says $\cup_{n\geq 1}$ and $\cup_{n\in \mathbb{N}}$ are same. On the otherhand in this question we have to consider two cases, one where family of sets are finite, second where family of sets are infinite. – Gajendra Basavaraju Jan 03 '24 at 07:17
  • Since $\cup_{n\geq 1}$ means $\cup_{n\in\Bbb N}$, finite families of sets are off topic in this question. – Anne Bauval Jan 03 '24 at 07:58

3 Answers3

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Hints for you to understand and work out.:

  1. $\;A_1=\{0\}\,,\,\,A_2=\{z\in\Bbb Z\;/\;z\neq0\}\;,\,\,n=1,2$

  2. $\;A_1=\{z\in\Bbb Z\;/\;z=2k\;,\;\;k\in\Bbb Z\}\;,\;\;A_2=\{z\in\Bbb Z\;/\;z=2k-1\,,\,\,k\in\Bbb Z\}\;,\;\;n=1,2$

  3. $\;A_1=\{0\}\;,\;\;A_n=\{n-1\,,\,\,1-n\}\;,\;\;n=2,3...\;$

  4. $\;\forall n\in\Bbb N\;,\;\;A_n\subset \Bbb Z\;$ and $\;|\Bbb Z|=\aleph_0\;$

In case de family of pairwise disjoint sets is (must be) infinite:

(1) and (2): Take

$$A_1=\{1,3,5,....,2k-1,\ldots\}\,,\,\,A_2=\{2,6,10,14,...,4k-2,\ldots\}\;,$$

$$A_3=\{4,12,20,...,8k-4\}\,,\ldots\,\text{(try to find the general formula here)}$$

DonAntonio
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Assuming that the question requires $(A_n)_{n \geq 1}$ to be a countable collection of nonempty pairwise disjoint subsets, one can provide counterexamples for options 1 and 2 as follows:

  1. Using your suggested answer, we can modify it to the following collection of sets: $A_1 = \{0, 1, 2, ...\} = \{n \in \mathbb{Z} \ | \ n \geq 0\}$ and for all $n \geq 2$, $A_n = \{-n + 1\}$.

  2. Any positive integer can be written uniquely as $2^nm$ where $n \geq 0$ and $m$ is odd. So, one possible partition of $\mathbb{Z}$ such that each cell has infinitely many elements is given by $$ A_1 = \{0 \} \cup \{2k + 1 \ | \ k \in \mathbb{Z} \} $$ $$ A_n = \{2^{n - 1}(2k+1) \ | \ k \in \mathbb{Z}\}, \ n \geq 2. $$

Ho Wei Haw
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By "countable" we shall mean infinite countable.

Finding a counterexample to $B$ (which can then also be used to contradict $A$) amounts to find a bijection $f:\Bbb N\times\Bbb N\to\Bbb Z$ (you can then set $A_n:=\{f(n,k)\mid k\in\Bbb N\}$).

For this, simply compose your prefered bijection between $\Bbb Z$ and $\Bbb N$ and your favorite bijection between $\Bbb N$ and $\Bbb N\times\Bbb N$.

Anne Bauval
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  • This is definitely accurate, but in examhall under time pressure its very hard even to realise there is such bijection where $A_{n}$'s pairwise disjoint. – Gajendra Basavaraju Dec 30 '23 at 17:17
  • The $A_n$'s are automatically pairwise disjoint in this construction. I don't think your past exam pressure should come into play on this site. The two (families of) bijections used here, i.e. the fact that $\Bbb Z$ and $\Bbb N\times\Bbb N$ are countable, should be in your toolbox when you get prepared for such an exam. – Anne Bauval Dec 31 '23 at 10:19