Question

Let A and B be two non-empty subsets of the set X such that A is not a subset of B, then
$1)$ A is always a number of the complement of B
$2)$ B is always a subset of A
$3)$ A and B are always disjoint
$4)$ A and the complement of B are always non-disjoint

Answer 1

Hint: First, to solve this problem we need to know about the concept of Set theory. The subset is the set of all elements containing from the universal set X and thus A and B are given as the subsets of the set X.
Complement means a set of elements that are not in the subset from the original subset given.
Disjoint means the two subsets intersection needs to be empty, which means no elements in the common.

Formula used:
Complement of A is ${A^1} = X - A$ and disjoint of the two subsets is $A \cap B = \phi $ (empty)

Complete step-by-step solution:
Since from the given that we have, A and B are two non-empty subsets of the set X such that A is not a subset of B.
Nonempty subsets mean the sets contain at least one element.
Let us fix some elements on the subsets A and B and universal set X
Let $X = \{ 1,2,3,4,5,6,7\} $is the universal set. From this, we make the two subsets as $A = \{ 2,5,7\} $and $B = \{ 1,5,6\} $. And from the given that A is not a subset of B which means $A \not\subset B$
Now let us check the given options correct or not
$1)$ A is always a number of the complement of B
First, we take the complement of the subset B, then we have \[{B^1} = X - B = \{ 1,2,3,4,5,6,7\} - \{ 1,5,6\} \Rightarrow \{ 2,3,4,7\} \] (which is the complements of B)
We see that A is not the complement of B, because they are not the same.
Thus option $1)$is incorrect.
$2)$ B is always a subset of A
Since A is $A = \{ 2,5,7\} $ and B is $B = \{ 1,5,6\} $
Thus, B is not the subset of A, because if its a subset then every element of B need to contain in the set A
Hence option $2)$ is incorrect, because $B \not\subset A$
$3)$ A and B are always disjoint
Since A is $A = \{ 2,5,7\} $ and B is $B = \{ 1,5,6\} $
Taking the intersection of the two sets, we get $A \cap B = \{ 2,5,7\} \cap \{ 1,5,6\} $
We see that one element in the set A and B are common, thus we get $A \cap B = \{ 2,5,7\} \cap \{ 1,5,6\} \Rightarrow 5 \ne \phi $
Hence option $(3)$ is incorrect because $A \cap B \ne \phi $ (not always disjoint)
$4)$ A and the complement of B are always non-disjoint.
Since we have A is $A = \{ 2,5,7\} $ and \[{B^1} = X - B = \{ 1,2,3,4,5,6,7\} - \{ 1,5,6\} \Rightarrow \{ 2,3,4,7\} \] (which is the complements of B)
Taking the intersection of the two sets, we get $A \cap {B^1} = \{ 2,5,7\} \cap \{ 2,3,4,7\} \Rightarrow 2,7$
Thus, we get A and the complement of B are always non-disjoint $A \cap {B^1} = 2,7 \ne \phi $
Hence option $(4)$ is correct.

Note: We are also able to solve this problem by checking the options, because in the first three statements they are always true but to check whether a statement is always true or not only if the given set is small contains elements, not in the general. Thus, non-disjoint is the most probability option we can get as the result because there is not a restriction that one element which is in the subset will not be repeated so the intersection will not be disjoint. Thus, the last option is true in general.