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This question is more of a graph problem. Suppose $A=\{1,2,..,n\}$, and we want to construct the sets $B(i)$,$1\leq i\leq n$, such that $i \not\in B(i)$ and the following constraints on the set $B(i)$ are satisfied:

1. If $j \in B(i)$ then $ i \notin B(j)$,

2. For all $i \in B(j)$, $B(j) \setminus \{i\} \not \subset B(i)$,

3. For all $i \notin B(k)$, $B(i) \not \subset {\{k\} \cup B(k)}$

Can we construct such subsets for all values of $n$? I can prove that you cannot construct such subsets for $n=4,5,6,..10$, but I can not generalize the proof for all values of $n$. I believe that such subsets cannot be constructed in general.

This question is more of a graph problem. Suppose $A=\{1,2,..,n\}$, and we want to construct the sets $B(i)$,$1\leq i\leq n$, such that $i \not\in B(i)$ and the following constraints on the set $B(i)$ are satisfied:

1. If $j \in B(i)$ then $ i \notin B(j)$,

2. For all $i \in B(j)$, $B(j) \setminus \{i\} \not \subset B(i)$,

3. For all $i \notin B(k)$, $B(i) \not \subset {\{k\} \cup B(k)}$

Can we construct such subsets for all values of $n$? I can prove that you can construct such subsets for $n=4,5,6,..10$, but I can not generalize the proof for all values of $n$. I believe that such subsets cannot be constructed in general.

This question is more of a graph problem. Suppose $A=\{1,2,..,n\}$, and we want to construct the sets $B(i)$,$1\leq i\leq n$, such that $i \not\in B(i)$ and the following constraints on the set $B(i)$ are satisfied:

1. If $j \in B(i)$ then $ i \notin B(j)$,

2. For all $i \in B(j)$, $B(j) \setminus \{i\} \not \subset B(i)$,

3. For all $i \notin B(k)$, $B(i) \not \subset {\{k\} \cup B(k)}$

Can we construct such subsets for all values of $n$? I can construct such subsets for $n=4,5,6$, but I can not generalize this method of construction for all values of $n$. I believe that such subsets can not be constructed in general.

This question is more of a graph problem. Suppose $A=\{1,2,..,n\}$, and we want to construct the sets $B(i)$,$1\leq i\leq n$, such that $i \not\in B(i)$ and the following constraints on the set $B(i)$ are satisfied:

1. If $j \in B(i)$ then $ i \notin B(j)$,

2. For all $i \in B(j)$, $B(j) \setminus \{i\} \not \subset B(i)$,

3. For all $i \notin B(k)$, $B(i) \not \subset {\{k\} \cup B(k)}$

Can we construct such subsets for all values of $n$? I can prove that you cannot construct such subsets for $n=4,5,6,..10$, but I can not generalize the proof for all values of $n$. I believe that such subsets cannot be constructed in general.

This is more of a graph problem. You are given set A={1,2,..,n}. You have to create sets B(i) for all i=1,2,..,n, such that i $\notin$ B(i). The constraint on set are the following:

1. If $j \in B(i)$ then $ i \notin B(j)$.

2. For all i $\in$ B(j), {B(j) $\setminus$ i }$\not \subset$ B(i).

3. For all i $\notin$ B(k), B(i) $\not \subset$ {k $\cup$ B(k)}

Can such subsets be created. I can prove it for n=4,5,6 but I am not able to generalise it. I have an opinion that such subsets cannot be created.

This question is more of a graph problem. Suppose $A=\{1,2,..,n\}$, and we want to construct the sets $B(i)$,$1\leq i\leq n$, such that $i \not\in B(i)$ and the following constraints on the set $B(i)$ are satisfied:

1. If $j \in B(i)$ then $ i \notin B(j)$,

2. For all $i \in B(j)$, $B(j) \setminus \{i\} \not \subset B(i)$,

3. For all $i \notin B(k)$, $B(i) \not \subset {\{k\} \cup B(k)}$

Can we construct such subsets for all values of $n$? I can construct such subsets for $n=4,5,6$, but I can not generalize this method of construction for all values of $n$. I believe that such subsets can not be constructed in general.