The objective is to write a union of sets as a union of disjoint sets, that is if $A_1,A_2,......,A_n$ are subsets of $A$ then: $$\bigcup_{i=1}^{n}A_i=A_1\cup(A_1^C\cap A_2)\cup (A_1^C\cap A_2^C \cap A_3)\cup...\cup(A_1^C\cap....A_{n-1}^C\cap A_n)$$

According to me, we just need to prove that for every $i$, $A_i=(A_1^C\cap....A_{i-1}^C\cap A_i)$

For $i=2,$ it holds true.

Say, it holds true for $i=k,$ i.e.: $A_k=(A_1^C\cap....A_{k-1}^C\cap A_k)$

We need to prove this for $i=k+1$.

Since, a $A_k^C$ term will be required,

$A_k\cup A_k^C=(A_1^C\cap....A_{k-1}^C\cap A_k)\cup A_K^C$

Since, $A_k\cup A_K^C = U$, $U$ = Universal Set.

Hence, $U=[(A_1^C\cap....A_{k-1}^C\cap A_k)\cup A_K^C]$

Now, $U\cap A_{k+1}=A_{k+1}$, hence $A_{k+1}=[(A_1^C\cap....A_{k-1}^C\cap A_k)\cup A_k^C]\cap A_{k+1}$

I do not think this would make the $A_k$ term go away, What am I doing wrong ? Could anyone help ?


You've run off the rails here: "For $i=2,$ it holds true." Sadly, it doesn't, unless $A_1$ and $A_2$ are disjoint. More generally, it is quite possible that only in the $i=1$ case is $A_i$ equal to $$\left(\bigcap_{j=1}^{i-1}A_j^C\right)\cap A_i.$$

You really do have to show that the unions (rather than the components of the two unions) are equal. One nice way to do this is by double-inclusion. Since we always have $\left(\bigcap_{j=1}^{i-1}A_j^C\right)\cap A_i\subseteq A_i,$ then the disjoint union is necessarily a subset $\bigcup_{i=1}^n A_i.$ So, it suffices to take an arbitrary element of $\bigcup_{i=1}^n A_i,$ and show that it belongs to the disjoint union.

  • $\begingroup$ Thanks for your answer. So, we have established that $(\cap_{j=1}^{i-1}A_j^C)\cap A_i \subseteq A_i$, hence $\cup_{i=1}^{n}((\cap_{j=1}^{i-1}A_j^C)\cap A_i) \subseteq \cup_{i=1}^{n} A_i$ $\endgroup$ – User9523 Oct 2 '17 at 7:08
  • $\begingroup$ That's correct. $\endgroup$ – Cameron Buie Oct 2 '17 at 11:36

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