Demonstration with completes sets

[nestora]

1. We have $A\subseteq \mathcal{U}$. For $i_1, i_2 \in \{0,1\}$and $A^0 := A^c, A^1 := A$.
A is a complete set if $A\cap A_1 ^{i1} \cap A_2^{i2} \neq \emptyset$ then $A_1 ^{i1} \cap A_2^{i2} \subseteq A$

Demonstrate that $A_1, A_2$ are complete sets too. And if A is a complete set then $A^c$ is a complete set too.



2. The first part I can't get it and don't know where to begin. The second part I tried to do:
$A \cap \bigcap X \neq \emptyset \rightarrow A^c \cap (\bigcap X)^c \neq \emptyset$ with $\bigcap X = A_1^{i1} \cap A_2^{i2}$ then $A^c \subseteq (\bigcap X)^c \rightarrow A^c \subseteq (A_1^{i1})^c \cup (A_2^{i2})^c$. But when I get there I'm not sure where to go next.

Any help would be very apreciated! Thanks

 

[lippyka]

in advance.Answer: 1. Let A_1, A_2 be two subsets of U. For any i_1, i_2 in {0,1}, we have that A_1^{i1} and A_2^{i2} are also subsets of U. Since A is a complete set, we have that A∩A_1^{i1}∩A_2^{i2}≠∅. This implies that A_1^{i1}∩A_2^{i2}⊆A. Since this holds for all i_1, i_2 in {0,1}, it follows that A_1 and A_2 are both complete sets. For the second part, let A be a complete set and let X=A_1^{i1}∩A_2^{i2}. We have that A∩X≠∅, which implies that (A∩X)^c=A^c∩(X)^c≠∅. This means that A^c is also a complete set.