I have a finite sequence of (not necessarily disjoint) sets $A_1, A_2, \cdots , A_n$. I have another finite sequence of sets $E_1, E_2, \cdots , E_n$ such that $E1, E2, \cdots, En$ are disjoint, and, for all $k : 1 ≤ k ≤ n$.

So essentially: $A_1 \cup A_2 \cup\cdots \cup A_k = E_1 \cup E_2 \cup\cdots\cup E_k$

So I can say:

$$\sum_i |A_i|- \sum_{i < j} |A_i \cap A_j|+ \sum_{i < j < k} |A_i \cap A_j \cap A_k|+ (-1)^{n-1} |A_1 \cap A_2 \cap\cdots \cap An| \\= |E_1| + |E_2| + |E_3| +\cdots+ |E_n|$$

Now my question is how can I represent $E_i$ in terms of $A_1 A_2 \cdots$? Am I headed in the right direction?

  • 2
    $\begingroup$ Whether or not you are heading in the wrong direction depends on pieces of information that you haven't written (namely what you are trying to accomplish). That being said, no sensible way to write the $E_i$-s in terms of the $A_i$-s comes to mind, and possibly there isn't. For instance, you can write $[0,1]=[0,\frac12)\cup [\frac12,\frac23)\cup [\frac23,1]=[0,\frac15)\cup [\frac14,\frac79]\cup (\frac{14}{27},1]$ $\endgroup$ – Saucy O'Path Aug 27 '18 at 8:04
  • $\begingroup$ So you want expressions for e.g. $|E_1|$ in the $|A_i|$? If so then impossible. Take for instance $n=2$ and $A_1=A_2=E_1\cup E_2$. $\endgroup$ – drhab Aug 27 '18 at 8:04
  • $\begingroup$ @SaucyO'Path Yes, that's what I thought, as you can never represent Ei in terms of Ai. So I basically used inverse function to find that, though that is not a good solution. $\endgroup$ – Pritam Banerjee Aug 27 '18 at 8:15
  • $\begingroup$ @PritamBanerjee What do you mean by inverse function here? I think mean by $|E_1|+|E_2|+\cdots +|E_n|$ in the right side of the equality. $\endgroup$ – tarit goswami Aug 27 '18 at 8:36

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