Subjective Type

If $$ \xi=\{x: x \in W, x \leq 10\}, A .=\{x: x \geq 5\} $$ and $$ B=\{x: 3 \leq x<8\}, $$ then verify that: $$ B-A=B \cap A^{\prime} $$

Solution

$$ \xi=\{0,1,2,3,4,5,6,7,8,9,10\} $$
$$ A=\{5,6,7,8,9,10\} $$
$$ B=\{3,4,5,6,7\} $$
$$ \mathrm{B}-\mathrm{A}=\mathrm{B} \cap \mathrm{A}^{\prime} $$
Consider the $$ \mathrm{LHS}=\mathrm{B}-\mathrm{A} $$
$$ \mathrm{B}-\mathrm{A}=\{3,4,5,6,7\}-\{5,6,7,8,9,10\} $$
$$ B-A=\{3,4\} $$
Therefore, $$ L H S=B-A=\{3,4\} $$
Now, consider $$ \mathrm{RHS}=\mathrm{B} \cap \mathrm{A}^{\prime} $$
$$ \mathrm{B} \cap \mathrm{A}^{\prime}=\{3,4,5,6,7\} \cap\{5,6,7,8,9,10\} $$
$$ B \cap A^{\prime}=\{3,4\} $$
Therefore, $$ \mathrm{RHS}=\mathrm{B} \cap \mathrm{A}^{\prime}=\{3,4\} $$
By comparing the $$ L H S $$ and $$ R H S, B-A=B \cap A^{\prime} $$

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