# 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
2 years, 7 months 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?

- 2 years, 7 months ago

Nope. I cannot still comprehend.

- 2 years, 7 months ago

What to do next.

- 2 years, 7 months 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

- 2 years, 7 months ago

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

- 2 years, 7 months 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.

- 2 years, 7 months ago

Thank you so much. This is enough already.

- 2 years, 7 months ago

No problem

- 2 years, 7 months 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 $A_{i} \setminus B_{i-1} = A_{j} \setminus B_{j-1}$?

- 2 years, 7 months ago