Question

Show that the following four conditions are equivalent:
(i) $A\subset B$ (ii) $A-B=\varphi$ (iii) $A\cup B=B$ (iv) $A\cap B=A$

Answer 1

First, we have to show that (i) $\iff$ (ii).

$\left(i\right)\Rightarrow$ (ii).
Let $A\subset B$
To show: $A-B=\varphi$
If possible, suppose $A-B\ne \varphi$
This means that there exists $x\in A,x\notin B$, which is not possible as $A\subset B$.
$\therefore A-B=\varphi$
$\therefore A\subset B\Rightarrow A-B=\varphi$

$\left(ii\right)\Rightarrow$ (i).
Let $A-B=\varphi$
To show: $A\subset B$
Let $xϵA$
Clearly, $xϵB$ because if $x\text{/}ϵB$, then $A-B\ne \varphi$
$\therefore A-B=\varphi \Rightarrow A\subset B$

Hence, $\left(ii\right)\iff \left(i\right)$

$\left(i\right)\Rightarrow$ (iii).

Let $A\subset B$
To show: $A\cup B=B$
Clearly, $B\subset A\cup B$
Let $xϵA\cup B$
$\Rightarrow xϵA$ or $xϵB$
Case I: $xϵA$
$\Rightarrow xϵB$ $[\because A\subset B]$
$\therefore A\cup B\subset B$
Case II: $xϵB$

Then $A\cup B\subset B$
So, $A\cup B=B$

$\left(iii\right)\Rightarrow$ (i).

Conversely, let $A\cup B=B$

To show : $A\subset B$
Let $xϵA$
$\Rightarrow xϵA\cup B$ $[\because A\subset A\cup B]$
$\Rightarrow xϵB$ $[\because A\cup B=B]$
$\therefore A\subset B$

Hence, $\left(iii\right)\iff \left(i\right)$

Now, we have to show that $\left(i\right)\iff \left(iv\right)$.
Let $A\subset B$
Clearly $A\cap B\subset A$
Let $xϵA$
We have to show that $xϵA\cap B$
As $A\subset B,xϵB$
$\therefore xϵA\cap B$
$\therefore A\subset A\cap B$
Hence, $A=A\cap B$
Conversely, suppose $A\cap B=A$
Let $xϵA$
$\Rightarrow xϵA\cap B$
$\Rightarrow xϵA$ and $xϵB$
$\Rightarrow xϵB$
$\therefore A\subset B$
Hence, $\left(i\right)\iff \left(iv\right)$.