Complicated Set Theory Problems: Sequence of Sets

Let \(A_1, A_2, \ldots , A_n\) be sets such that \(X = \bigcup_{i=1}^n A_i \). Prove that there exists a sequence of sets \(B_1, B_2, \ldots , B_n\) such that

a) \(B_i \subseteq A_i \) for each \(i=1,2,\ldots ,n\).

b) \(B_i \cap B_j = \varnothing \) for \(i\ne j\).

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

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

Note by Mharfe Micaroz
1 month, 2 weeks ago

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Just look at the "new" things added to \(X\) by \(A_i\). Call that \(B_i\). For example \(B_1\) is \(A_1\), \(B_2\) is \(A_2 \setminus A_1\), etc. Can you continue from here?

Agnishom Chattopadhyay Staff - 1 month ago

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Nope. I cannot still comprehend.

Mharfe Micaroz - 3 weeks, 3 days ago

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What to do next.

Mharfe Micaroz - 3 weeks, 3 days ago

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@Mharfe Micaroz \(B_1\) is \(A_1\). \(B_2\) is \(A_2 \setminus B_1\). \(B_3\) is \(A_3 \setminus B_2\). \(B_4\) is \(A_4 \setminus B_3\).

Now just verify that the claims given hold for these \(B_i\)'s

Agnishom Chattopadhyay Staff - 3 weeks, 3 days 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 - 3 weeks, 3 days ago

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@Mharfe Micaroz Sorry, I made some typo. Define \(B_1\) to be \(A_1\). For \(i > 1\), define \(B_i\) to be \(A_i \setminus (B_1 \cup B_2 \cdots \cup B_{i-1})\). The intution behind defining \(B_i\) 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 \(B_i\) is defined to be \(A_i \setminus B_{i-1}\). So, by definition of set difference, if something is in \(B_i\), it must also be in \(A_i\).
  • For the second one, just notice that \(B_i\) does not contain any element from \(B_j\) for \(i > j\). Thus, given two indices \(i\) and \(j\), it must always be the case that they are disjoint.
  • If there is an element \(x \in X\), then \(x\) must be in some \(A_i\). If \(i_0\) is the least such value of \(i\), then \(x\) must be in \(B_{i_0}\), and hence in \(\bigcup_{i=1}^n B_i\). This shows that \(X \subseteq \bigcup_{i=1}^n B_i\). Showing the other side is easy.

Agnishom Chattopadhyay Staff - 3 weeks, 3 days ago

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

Mharfe Micaroz - 3 weeks, 2 days ago

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@Mharfe Micaroz No problem

Agnishom Chattopadhyay Staff - 3 weeks, 1 day ago

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@Agnishom Chattopadhyay Do you mean the following Sir?

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

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

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

Mharfe Micaroz - 3 weeks, 3 days ago

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