# 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?

Staff - 1 month ago

Nope. I cannot still comprehend.

- 3 weeks, 3 days ago

What to do next.

- 3 weeks, 3 days ago

$$B_1$$ 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

Staff - 3 weeks, 3 days ago

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

- 3 weeks, 3 days ago

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.

Staff - 3 weeks, 3 days ago

Thank you so much. This is enough already.

- 3 weeks, 2 days ago

No problem

Staff - 3 weeks, 1 day ago

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 $$A_{i} \setminus B_{i-1} = A_{j} \setminus B_{j-1}$$?

- 3 weeks, 3 days ago