Complicated Set Theory Problems: Sequence of Sets

Let A1,A2,,AnA_1, A_2, \ldots , A_n be sets such that X=i=1nAiX = \bigcup_{i=1}^n A_i . Prove that there exists a sequence of sets B1,B2,,BnB_1, B_2, \ldots , B_n such that

a) BiAiB_i \subseteq A_i for each i=1,2,,ni=1,2,\ldots ,n.

b) BiBj=B_i \cap B_j = \varnothing for iji\ne j.

c) X=i=1nBiX = \bigcup_{i=1}^n B_i .

Can you give insights on how to solve this problem? Insights is enough for me.

Note by Mharfe Micaroz
11 months, 2 weeks ago

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1 vote

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Just look at the "new" things added to XX by AiA_i. Call that BiB_i. For example B1B_1 is A1A_1, B2B_2 is A2A1A_2 \setminus A_1, etc. Can you continue from here?

Agnishom Chattopadhyay Staff - 11 months ago

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Nope. I cannot still comprehend.

Mharfe Micaroz - 10 months, 3 weeks ago

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What to do next.

Mharfe Micaroz - 10 months, 3 weeks ago

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@Mharfe Micaroz B1B_1 is A1A_1. B2B_2 is A2B1A_2 \setminus B_1. B3B_3 is A3B2A_3 \setminus B_2. B4B_4 is A4B3A_4 \setminus B_3.

Now just verify that the claims given hold for these BiB_i's

Agnishom Chattopadhyay Staff - 10 months, 3 weeks ago

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@Agnishom Chattopadhyay Can you give me the first one? I will do the rest. I just need a guide. Thanks.

Mharfe Micaroz - 10 months, 3 weeks ago

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@Mharfe Micaroz Sorry, I made some typo. Define B1B_1 to be A1A_1. For i>1i > 1, define BiB_i to be Ai(B1B2Bi1)A_i \setminus (B_1 \cup B_2 \cdots \cup B_{i-1}). The intution behind defining BiB_i is simply to take all the elements which are new, i.e, the ones that you have not seen before.

  • The first one is clear since each BiB_i is defined to be AiBi1A_i \setminus B_{i-1}. So, by definition of set difference, if something is in BiB_i, it must also be in AiA_i.
  • For the second one, just notice that BiB_i does not contain any element from BjB_j for i>ji > j. Thus, given two indices ii and jj, it must always be the case that they are disjoint.
  • If there is an element xXx \in X, then xx must be in some AiA_i. If i0i_0 is the least such value of ii, then xx must be in Bi0B_{i_0}, and hence in i=1nBi\bigcup_{i=1}^n B_i. This shows that Xi=1nBiX \subseteq \bigcup_{i=1}^n B_i. Showing the other side is easy.

Agnishom Chattopadhyay Staff - 10 months, 3 weeks ago

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@Agnishom Chattopadhyay Thank you so much. This is enough already.

Mharfe Micaroz - 10 months, 3 weeks ago

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@Mharfe Micaroz No problem

Agnishom Chattopadhyay Staff - 10 months, 3 weeks ago

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@Agnishom Chattopadhyay Do you mean the following Sir?

That for (a) I will assume first that xAk x\in A_k and do all ways to prove that xBk x\in B_k given BiAiB_i \subseteq A_i for each i=1,2,,ni=1,2,\ldots ,n?

For (b), I need to prove that Bi B_i is a disjoint set such that AiBi1AjBj1=A_{i} \setminus B_{i-1} \bigcap A_{j} \setminus B_{j-1}=\emptyset ?

Lastly, for (c), I need to prove that Bi=BkB_i=B_k which implies that AiBi1=AjBj1A_{i} \setminus B_{i-1} = A_{j} \setminus B_{j-1}?

Mharfe Micaroz - 10 months, 3 weeks ago

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