Question

Show that the following four conditions are equivalent: $(i) A ⊂ B (ii) A – B = \phi (iii) A ∪ B = B (iv) A ∩ B = A$

Answer 1

First, we have to show that $(i) ⇔ (ii).$
Let $A ⊂ B$
To show: $A – B \ne\phi$
If possible, suppose $A – B \ne\phi$
This means that there exists $x ∈ A, x ≠ B$, which is not possible as $A ⊂ B.$
$∴ A – B = \phi$
$∴ A ⊂ B ⇒ A – B = \phi$
Let $A – B = \phi$
To show: $A ⊂ B$
Let $x ∈ A$
Clearly, $x ∈ B$ because if $x ∉ B$, then $A – B ≠ \phi$
$∴ A – B = \phi ⇒ A ⊂ B$
$∴ (i) ⇔ (ii)$
Let $A ⊂ B$
To show: $A ∪ B = B$
Clearly, $B ⊂ A ∪ B$
Let $x ∈ A ∪ B$
$⇒ x ∈ A \space or x ∈ B$

Case I : $x ∈ A$
$⇒ x ∈ B [∴ A ⊂ B]$
$∴ A ∪ B ⊂ B$

Case II : $x ∈ B$
Then, $A ∪ B = B$
Conversely, let $A ∪ B = B$
Let $x ∈ A$
$⇒ x ∈ A ∪ B$ $[∴ A ⊂ A ∪ B]$
$⇒ x ∈ B$ $[ ∴ A ∪ B = B]$
$∴ A ⊂ B$

Hence, $(i) ⇔ (iii)$
Now, we have to show that $(i) ⇔ (iv).$
Let $A ⊂ B$
Clearly $A ∩ B ⊂ A$
Let $x ∈ A$
We have to show that $x ∈ A ∩ B$
As $A ⊂ B, x ∈ B$
$∴ x ∈ A ∩ B$
$∴ A ⊂ A ∩ B$
Hence, $A = A ∩ B$
Conversely, suppose. $A ∩ B = A$
Let $x ∈ A$
$⇒ x ∈ A ∩ B$
$⇒ x ∈ A$ and $x ∈ B$
$⇒ x ∈ B$
$∴ A ⊂ B$
Hence, $(i) ⇔ (iv).$