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Problem: Let $ A_1,A_2,...,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,...,B_n$ such that

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

b) $B_i \cap B_j = \emptyset$ for $i \neq j$

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

My observation: I look at the "new things" added to $X$ by $A_i$ and call that $B_i$. For example $B_1$ is $A_1$, $B_2$ is $A_2\setminus A_1$ and $B_3$ is $A_3 \setminus A_2 \cup A_1$. In general, the sequence of sets is defined by this $B_i=A_i-\bigcup_{k=1}^{i-1} A_k$ for $k=1,2,...,n$.

Proof Attempt:

Proof (a)

Let $x\in B_i$. Since $B_i=A_i\setminus\bigcup_{k=1}^{i-1} A_k$, so by defnition of Set Difference, if something is in $B_i$, it must also be in $A_i$. Since, $x\in B_i$ implies $x\in A_i$, we conclude that $B_i\subseteq A_i$ (by definition of Subset).

Proof (b)

On which, I found an elementary approach from here.

Proof (c)

To show that $X=\bigcup_{i=1}^n B_i$, we must show that $X\subseteq\bigcup_{i=1}^n B_i$ and $\bigcup_{i=1}^n B_i\subseteq X$.

Part 1. Show that $X\subseteq\bigcup_{i=1}^n B_i$. Let $x\in X$ then $x\in\bigcup_{i=1}^n A_i$ as defined.It implies that $x$ is an element of some $A_i$'s (by definition of $\cup$).If $i_0$ is the least such value of $i$ such that $x\in A_{i_0}$ then $x\in A_{i_0}\setminus \bigcup_{k=1}^{i_0-1} A_k$ (by definition of $\setminus$). It further implies that $x\in B_{i_0}$ and hence $x\in\bigcup_{i=1}^n B_i$ by definition of $\cup$. Since $x\in X$ implies $x\in\bigcup_{i=1}^n B_i$, we conclude that $X\subseteq\bigcup_{i=1}^n B_i$ (by definition of $\subseteq$).

Part 2. Show that $\bigcup_{i=1}^n B_i\subseteq X$. Let $x\in\bigcup_{i=1}^n B_i$ then $x\in B_i$ (for some $i$). If $i_0$ is the least such value of $i$ such that $x\in B_{i_0}$ and since $B_i\subseteq A_i$ which we proved already above, it implies that $x\in A_{i_0}$ which further implies that $x\in\bigcup_{i=1}^n A_i$ (by definition of $\cup$). Hence, $x\in X$. Since, $x\in \bigcup_{i=1}^n B_i$ implies that $x\in X$, we conclude that $\bigcup_{i=1}^n B_i\subseteq X$.

Conclusion. Since, $X\subseteq\bigcup_{i=1}^n B_i$ and $\bigcup_{i=1}^n B_i\subseteq X$ then $X=\bigcup_{i=1}^n B_i$.

Note: The proof is so elementary that fits to beginners like me.

Is this right already?

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    $\begingroup$ Looks like the sets $B_i\ne \emptyset$ form a partition of the set $A$. $\endgroup$ – Wuestenfux Dec 28 '18 at 11:24
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Take $B_1=A_1,B_2=A_2\setminus A_1, \cdots, B_n=A_n\setminus (A_1 \cup A_2 \cup \cdots \cup A_{n-1})$. The idea here is if $x \in X$ then $x\in A_k$ for some $k$ and there is smallest $k$ with this property. For that $k$ we get $x \in B_k$,

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  • $\begingroup$ Thanks @kavi-rama-murthy but still unclear to me. Do I need to show my proof for each of the three conditionz given above? $\endgroup$ – Mharfe Micaroz Dec 28 '18 at 12:23
  • $\begingroup$ Yes you need to show, for your sets $B_1,\dotsc,B_n$, that a-c hold. $\endgroup$ – palmpo Dec 28 '18 at 12:27
  • $\begingroup$ Can you give just a simple oversight on how to prove each property sir? $\endgroup$ – Mharfe Micaroz Dec 28 '18 at 13:08
  • $\begingroup$ 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 $A_{i} \setminus B_{i-1} = A_{j} \setminus B_{j-1}$? $\endgroup$ – Mharfe Micaroz Dec 28 '18 at 13:56
  • $\begingroup$ No, for (a) you first assume $x\in B_i$ and show that $x\in A_i$, for any $i=1,\dotsc,n$. For (b), you should try proof by contradiction by first assuming $B_i\cap B_j\ne\emptyset$ for some $i\ne j$ and showing that this leads to a contradiction. For (c), this is just showing set equality, i.e. $\bigcup^nA_i=\bigcup^nB_i$. Include your attempt by editing your question so that others may see. $\endgroup$ – palmpo Dec 28 '18 at 15:24

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