Question

For any two sets A and B, show that the following statements are equivalent:

(i) \[A \subset B\] 

(ii) \[A \subset B\]=ϕ 

(iii) \[A \cup B = B\]

(iv) \[A \cap B = A .\] 

Answer 1

We have that the following statements are equivalent: 

(i) \[A \subset B\] 

(ii) \[A \subset B\] 

(iii) \[A \cup B = B\] 

(iv) \[A \cap B = A\] 

Proof: 

\[\text{ Let } A \subset B\]
\[\text{ Let } x \text{ be an arbitary element of } (A - B) . \]
\[\text{ Now }, \]
\[x \text{ in } (A - B)\]
\[ \Rightarrow x \text{ in A & x } \text{ not }\text{ in } B (\text{ Which is contradictory }) \]
\[\text{ Also }, \]
\[ \text{ because } A \text{ subset } B\]
\[ \Rightarrow A - B \text{ subseteq } \phi . . . (1) \]
\[\text{ We know that null sets are the subsets of every set } . \]
\[ \text{ therefore } \text{ phi } \text{ subseteq } A - B . . . (2)\]
\[\text{ From } (1) \text{ & } (2), \text{ we get } \]
\[(A - B) = \phi\]
\[ \text{ therefore } (i) = (ii)\]
\[\text{ Now, we have }, \]
\[(A - B) = \text{ phi }\]
\[\text{ That means that there is no element in A that does not belong to B } . \]
\[\text{ Now }, \]
\[ A \text{ cup } B = B\]
\[ \text{ therefore } (ii) = (iii) \]
\[\text{ We have }, \]
\[A \text{ cup } B = B\]
\[ \Rightarrow A \text{ subset } B\]
\[ \Rightarrow A \text{ cap } B = A\]
\[ \text{ therefore } (iii) = (iv)\]
\[\text{ We have }, \]
\[ A \text{ cap } B = A\]
\[\text{ It should be possible if A } \text{ subset B } . \]
\[\text{ Now }, \]
\[A \text{ subset } B\]
\[ \text{ therefore } (iv) = (i)\]
\[\text{ We have }, \]
\[(i) = (ii) = (iii) = (iv) \]
\[\text{ Therefore, we can say that all statements are equivalent } . \]