Question

In each of the following, determine whether the statements is true or false. If it is true, prove it. If it is false, give an example.
$\left(i\right)$ If $x\in A$ and $A\in B$, then $x\in B$
$\left(ii\right)$ $A\subset B$ and $B\in C$, then $A\in C$
$\left(iii\right)$ $A\subset B$ and $B\subset C$, then $A\subset C$
$\left(iv\right)$ If $A\subset ̸B$ and $B\subset ̸C$, then $A\subset ̸C$
$\left(v\right)$ If $x\in A$ and $A\subset ̸B$, then $x\in B$
$\left(vi\right)$ If $A\subset B$ and $x\notin B$, then $x\notin A$

Answer 1

$\left(i\right)$ Cheacking given statement is true or false.
Let $A=\{1,2\}$
Clearly, $1\in A$
Also, $A\in B$ means whole set $A$ is an element of set $B.$
So, taking $B=\{\{1,2\},3,4,5\}$
Clearly, $1\notin \{\{1,2\},3,4,5\}$
i.e., $1$ is not an elelemnet of set B.
So, given statement is false

$\left(ii\right)$ Checking given statement is true or false.
Let $A=\left\{2\right\}$
Given, $A\subset B$
So, each element of set $A$ should be an element of set B
$\therefore$ Taking $B=\{0,2\}$
Also $B\in C$
$\therefore$ Taking set $C=\{1,\{0,2\},3\}$
Now, $\left\{2\right\}\notin C$ as $\left\{2\right\}$ is not an elelemnt of $C.$
Hence, $A\notin C$
So, given statement is false.

$\left(iii\right)$ Checking given statement is true or false
Let $x\in A$
$\Rightarrow x\in B\{\text{since}A\subset B\}$
Also, it is given that $B\subset C$
$\Rightarrow x\in C$
So, $x\in A\Rightarrow x\in C$
$\therefore A\subset C$
So, given statement is true

$\left(iv\right)$ Checking given statement is true or false.
Let $A=\{1,2\}$
$\because A\subset ̸B$ so, each element of set $A$ should not be an element of set $B.$
$\therefore$ Taking $B=\{0,6,8\}$
Also, $B\subset ̸C$ so, each element of set $B$ should not be an element of set $C.$ But each element of set $A$ can be in set $C.$
$\therefore$ Set A is a subset of set C.
So, there is a chance that if $A\subset ̸B$ and $B\subset ̸C$ then $A\subset C$
So, given statement is false.

$\left(v\right)$ Checking given statement is true or false.
Let $A=\{3,5,7\}$
Let $x=5,5\in A$
$\because A\subset ̸B$, each element of set $A$ should not be an element of set $B.$
$\therefore$ Taking $B=\{0,2\}$
Also, $5\notin B$ i.e., $x\notin B$
So, given statement is false.

$\left(vi\right)$ Checking given statement is true or false.
Given $A\subset B$
$\Rightarrow$ All elements of $A$ are also in $B$
This means if $x\in A$, then $x\in B$
Or, if $x\notin B$, then $x\notin A$
Thus, given statement is true.