Question

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) \[If{\rm{ }}x \in A{\rm{ }}\,and\, {\rm{ }}A \in B,{\rm{ }}\,Then\,{\rm{ }}x \in B\]
(ii) \[If{\rm{ }}A \subset B{\rm{ }}\,and\,{\rm{ }}B \in C,{\rm{ }}\,Then\,{\rm{ }}A \in C\]
(iii) \[If{\rm{ }}A \subset B{\rm{ }}\,and\,{\rm{ }}B \subset C,{\rm{ }}\,Then\,{\rm{ }}A \subset C\]
(iv) \[If{\rm{ }}A \not\subset B{\rm{ }}and{\rm{ }}B \not\subset C,{\rm{ }}\,Then\,{\rm{ }}A \not\subset C\]
(v) \[If{\rm{ }}x \in A{\rm{ }}\,and\,{\rm{ }}A \not\subset B,{\rm{ }}\,Then\,{\rm{ }}x \in B\]
(vi) \[If{\rm{ }}A \subset B{\rm{ }}\,and\,{\rm{ }}x \notin B,{\rm{ }}\,Then\,{\rm{ }}x \notin A\]

Answer 1

Hint: Here, we must take the examples in set A, B and x, to check the different types of relation between them. We can also take example sets to illustrate this relationship for better and practical understanding of the question.

Complete step-by-step answer:
Now, let us assume : \[ \subset \to shows{\rm{ a\, symbol\, of\, subset,}}\]\[A \subset B \to shows {\rm{ all\, the\, elements\, of\, A\, are\, present\, in\, B,}}\] \[ \in \to shows\,{\rm{ the\, symbol\, of\, belongs\, to\, the\, element\, of\, a\, set\,}}\]\[ \notin \to\, shows\,{\rm{ the\, symbol\, of\, does\, not\, belongs\, to\, the\, element\, of\, a\, set,,}}\]\[ \not\subset \to \,shows{\rm{ a\, symbol\, of\, does\, not\, belongs\, to\, a\, subset}}{\rm{.}}\]
(i)\[If{\rm{ }}x \in A{\rm{ }}\,and\,{\rm{ }}A \in B,{\rm{ }}\,Then\,{\rm{ }}x \in B\]
Let us assume, \[A = \left\{ {2,3} \right\}\]
As, 2 is an element of the set.
Let, \[x = 2,2 \in \left\{ {2,3} \right\}\]
As, it is given \[A \in B\] i.e all the elements of set A must belong to set B.
Let us take \[B = \left\{ {\left\{ {2,3} \right\},4,5,6} \right\}\]
We have to prove that \[x \in B\]
\[ \Rightarrow 2 \notin \left\{ {\left\{ {2,3} \right\},4,5,6} \right\}\]
As 2 is not present in set B.
Hence, the given statement is false.

(ii)\[If{\rm{ }}A \subset B{\rm{ }}\,and\,{\rm{ }}B \in C,{\rm{ }}\,Then\,{\rm{ }}A \in C\]
Let us assume, \[A = \left\{ 3 \right\}\]
As, it is given \[A \subset B\] i.e all the elements of set A must be present in set B.
Let us take \[B = \left\{ {1,3} \right\}\]
Also, it is given \[B \in C\] i.e all the elements of set B must belong to set C.
Let us take \[C = \left\{ {0,\left\{ {1,3} \right\},4} \right\}\]
We have to prove that \[A \in C\]
\[ \Rightarrow 3 \notin \left\{ {0,\left\{ {1,3} \right\},4} \right\}\]
As 3 is not present in set C.
Hence, the given statement is false.

(iii)\[If{\rm{ }}A \subset B{\rm{ }}\,and\,{\rm{ }}B \subset C,{\rm{ }}\,Then\,{\rm{ }}A \subset C\]
Let us assume, \[A = \left\{ 3 \right\}\]
As, it is given \[A \subset B\] i.e all the elements of set A must be present in set B.
Let us take \[B = \left\{ {1,3} \right\}\]
Aslo, it is given \[B \subset C\] i.e all the elements of set B must be present in set C.
So, let us take \[C = \left\{ {1,3,4} \right\}\]
We have to prove that \[A \subset C\].
As all the elements of A are present in C.
Hence, the given statement is true.

(iv)\[If{\rm{ }}A \not\subset B{\rm{ }}\,and\,{\rm{ }}B \not\subset C,{\rm{ }}\,Then\,{\rm{ }}A \not\subset C\]
Let us assume, \[A = \left\{ 3 \right\}\]
As, it is given \[A \not\subset B\] i.e all the elements of set A must not be present in set B.
Let us take \[B = \left\{ {1,2} \right\}\]
Aslo, it is given \[B \not\subset C\] i.e all the elements of set A must not be present in set C.
So, let us take \[C = \left\{ {3,4,5} \right\}\]
We have to prove that \[A \not\subset C\].
But 3 is present in set C. \[\therefore A \subset C\]
As all the elements of A are present in C.
Hence, the given statement is false.

(v)\[If{\rm{ }}x \in A{\rm{ }}\,and\,{\rm{ }}A \not\subset B,{\rm{ }}\,Then\,{\rm{ }}x \in B\]
Let us assume, \[A = \left\{ {2,3} \right\}\]
As, 2 is an element of the set.
Let, \[x = 2,2 \in \left\{ {2,3} \right\}\]
As, it is given \[A \not\subset B\] i.e all the elements of set A must not be present in set B.
Let us take \[B = \left\{ {1,4} \right\}\]
We have to prove that \[x \in B\].
\[ \Rightarrow 2 \notin \left\{ {1,4} \right\}\]
Hence, the given statement is false.

(vi)\[If{\rm{ }}A \subset B{\rm{ }}\,and\,{\rm{ }}x \notin B,{\rm{ }}\,Then\,{\rm{ }}x \notin A\]
Let us assume, \[A = \left\{ 3 \right\}\]
As, it is given \[A \subset B\] i.e all the elements of set A must be present in set B.
Let us take \[B = \left\{ {1,3} \right\}\] and \[x = 2\]
As, it is given \[x \notin B\] i.e \[2 \notin \left\{ {1,3} \right\}\]
We have to prove that \[x \notin A\].
\[ \Rightarrow 2 \notin 3\]
Hence, the given statement is true.

Note: Here with the help of some basic examples it could be easily understandable and it will really help you to give the practically based answers whether the statement is true or false and you can verify these problems at the same time also.