I have eleven sets, all of them are subsets of $X:=\{(a,b,c,d)\in[-1,1]^4: a\le b,\text{ and } c\le d\}$: $\begin{align*} A_1&:=\{(a,b,c,d)\in X: b\ge 0,\ c\le a+b+d\}\\ A_2&:=\{(a,b,c,d)\in X: b\ge 0,\ d\ge 0\}\\ A_3&:=\{(a,b,c,d)\in X: b\ge 0,\ a+b+2d\ge 0\}\\ A_4&:=\{(a,b,c,d)\in X: d\ge 0,\ a\le b+c+d\}\\ A_5&:=\{(a,b,c,d)\in X: d\ge 0,\ 2b+c+d\ge 0\}\\ A_6&:=\{(a,b,c,d)\in X: a\le b+c+d,\ c\le a+b+d\}\\ A_7&:=\{(a,b,c,d)\in X: a\le b+c+d,\ a+b+2d\ge 0\}\\ A_8&:=\{(a,b,c,d)\in X: 2b+c+d\ge 0,\ c\le a+b+d\}\\ A_9&:=\{(a,b,c,d)\in X: 2b+c+d\ge 0,\ a+b+2d\ge 0,\ c\le 2b+d\}\\ A_{10}&:=\{(a,b,c,d)\in X: 2b+c+d\ge 0,\ a+b+2d\ge 0,\ a\le b+2d\}\\ A_{11}&:=\{(a,b,c,d)\in X : 2b+c+d\ge 0,\ a+b+2d\ge 0,\ b+d\ge 0\}. \end{align*}$ My question is: Are these sets independent, i.e. is any of the sets a subset of a union of other sets?
Is any of the sets a subset of a union of other sets?
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combinatorics
analysis
inequality
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0@TCL: I was taking the elements to be integers, so$(a,b)$could be (-1,-1), (-1,0), etc up to (1,1). I see now that you were thinking reals. – 2012-02-24
1 Answers
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No. Every $A_i$ is a subset of $A_1\cup A_4$.
For $A_2$: We have $b+d\ge 0$. Since either $c\le a$ or $a\le c$, we have either $c\le a+b+d$ or $a\le b+c+d$, proving that $A_2\subset A_1\cup A_4$.
For $A_3$: If $d\ge 0$, then the point is in $A_2$ which is a subset of $A_1\cup A_4$. If $d\le 0$ then $-d\ge c$ since $c\le d\le 0$. So $a+b+d\ge -d\ge c$, proving that the point is in $A_1$.
Similar arguments apply to other $A_i$'s.