State whether the given statement is true false:

Solution:

(i) True

Explanation: We have A ⊂ B since A is a subset of B then all elements of A should be in B.

Let $A=\{1,2\}$ and $B=\{1,2,3\}$

Let $x=4 \notin B$

Also we observe that $4 \notin \mathrm{A}$.

Hence, If $A \subset B$ and $x \notin B$ than $x \notin A$.

(ii) True

Explanation: We have, $A \subseteq \phi$

Now, A is a subset of : set , this implies A is also an empty set.

$\Rightarrow A=\phi$

(iii) False

Explanation: Let $A=\{a\}, B=\{\{a\}, b\}$

here, $A \in B$

Now, let $C=\{\{a\}, b, c\}$.

Since, $\{a\}, b$ is in $B$ and also in $C$ thus, $B \subset C$.

But, A ={a} and {a} is an element of C, since the element of a set cannot be a subset of a set.

Hence, $\mathrm{A} \not \subset \mathrm{C}$.

(iv) False

Explanation: Let A = {a},B = {a, b} and C = {{a, b}, c}.

Then, $A \subset B$ and $B \in C$. But, $A \notin C$ since $\{a\}$ is not an element of $C$.

(v) False

Explanation: Let $A=\{a\}, B=\{b, c\}$ and $C=\{a, c\}$.

Since $a \in A$ and $a \notin B$. Then, $A \not \subset B$

Now, $b \in B$ and $b \notin C \Rightarrow B \not \subset C$.

But, $A \subset C$ since, $a \in A$ and $a \in C$.

(vi) False

Explanation: Let $A=\{x\}, B=\{\{x\}, y\}$

Now, $\mathrm{x} \in \mathrm{A}$ and $\{\mathrm{x}\}$ is an element of $\mathrm{B} \Rightarrow \mathrm{A} \in \mathrm{B}$

But, $x$ is not an element of $B$. Thus, $x \notin B$.