in Set Theory & Algebra
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1 vote
1 vote
Want to verify

let set $\left | A \right |=n$ and $\left | B \right |=m$

Then $max(m,n)\leq \left | A\cup B \right |\leq (m+n)$

$0\leq \left | A\cap B \right |\leq min(m,n)$

$0\leq \left | A- B \right |\leq \left | n \right |$

$max(m,n)\leq \left | A\bigoplus B \right |\leq (m+n)$ here $\bigoplus$ is symmetric difference

$0\leq \left | \overline{A} \right |\leq U$ here $\overline{A}$ is compliment of A and U is universal Set
in Set Theory & Algebra
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3 Comments

u can take suitable example and verify it,,,,,
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Just want to confirm that i didn't miss any case
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ya all are correct,,,
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1 Answer

1 vote
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Best answer

1. |A∪B| -> max-  (m+n) when A∩B is ∅                                          

min-   max(m,n) when one set is a subset of another. 

2. |A∩B| -> max- min(m,n) (cardinality of smaller set)                      

min- 0   when sets are disjoint.

3.|A−B| ->  max- |n| i.e cardinality of A . When sets are disjoint      

min. - 0 .When sets are equal.

4.|A⨁B| -> max- (m+n). When sets are disjoint                               

min. - 0 when A= B

5. $|\overline{A}|$ -> max - U when A is an empty set                    

min. - 0 when A=U

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@Tesla! minimum symmetric difference can be. 0 when A=B
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