1
$\begingroup$

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 ?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.