For any two sets A and B, prove that :

A' - B' = B- A

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Solution

To show A' - B' = B- A

We show that $A^{'} - B^{'} \subseteq B - A$ and vice versa

Let $x ϵ A^{'} - B^{'}$

$\Rightarrow x ϵ A^{'} a n d x / ϵ B^{'}$

$\Rightarrow x ϵ A a n d x ϵ B$

[ $∵ A \cup A^{'} = ϕ a n d B \cap B^{'} = ϕ$ ]

$\Rightarrow x ϵ B - A$

This is true for all $x ϵ A^{'} - B^{'}$

Hence $A^{'} - B^{'} \subseteq B - A$

Conversely,

Let, $x ϵ B - A$

$\Rightarrow x ϵ B a n d Z x / ϵ A$

$\Rightarrow x / ϵ B^{'} a n d x ϵ A^{'}$

[ $∵ B \cap B^{'} = ϕ a n d A \cap A^{'} = ϕ$ ]

$\Rightarrow x ϵ A^{'} a n d x / ϵ B^{'}$

$\Rightarrow x ϵ A^{'} - B^{'}$

This is true for all $x ϵ B - A$

Hence $B - A \subseteq A^{'} - B^{'}$

$∴ A^{'} - B^{'} = B - A$ Proved.

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Similar questions

For any two sets A and B , prove that :

$A \cap B = ϕ \Rightarrow A \subseteq B^{'}$ .

B^{'} \subset A^{'} \Rightarrow A \subset B

For any two sets of A and B, prove that :

(i) $A^{'} \cup B = U \Rightarrow A \subset B$

(ii) $B^{'} \subset A^{'} \Rightarrow A \subset B$

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