Question
Show that for any sets A and B,

$\text{A}\cup(\text{B}-\text{A})=(\text{A}\cup\text{B})$

Answer

Let $\text{x}\in\text{A}\cup\text{(B - A)}$

$\Rightarrow\text{x}\in\text{A or x}\in(\text{B - A})$

$\Rightarrow\text{x}\in\text{A or x}\in\text{B and x}\not\in\text{A}$

$\Rightarrow\text{x}\in\text{A or x}\in\text{B}$

$\Rightarrow\text{x}\in\text{(A}\cup\text{B})$

$\therefore\text{A}\cup\text{(B - A)}\subset\text{(A}\cup\text{B}).....\text{(i)}$

Let and $\text{x}\in\text{(A}\cup\text{B})$

$\Rightarrow\text{x}\in\text{A or x}\in\text{B}$

$\Rightarrow\text{x}\in\text{A or x}\in\text{B and x}\not\in\text{A}$

$\Rightarrow\text{x}\in\text{A or x}\in\text{(B - A)}$

$\Rightarrow\text{x}\in\text{A}\cup\text{(B - A)}$

$\therefore(\text{A}\cup\text{B})\subset\text{A}\cup\text{(B - A)}.....\text{(ii)}$

From (i) and (ii), we get

$\text{A}\cup\text{(B - A)}=\text{A}\cup\text{B.}$

 

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