Let A , B and C be sets such that 

\phi \neq A\cap B\subseteq C.  Then which of the following 

statements is not true ? 

  • Option 1)

    B\cap C\neq \phi

  • Option 2)

    If\: \: (A-B)\subseteq C, then\: \: A\subseteq C

  • Option 3)

    (C\cup A)\cap(C\cup B)=C

  • Option 4)

    If\: \: (A-C)\subseteq B, then\: \: A\subseteq B

 

Answers (1)

As (A\cap B)\subseteq C

=> (A\cap B)\subseteq (B\cap C)

as (A\cap B)\neq \phi => (B\cap C)\neq \phi

So, option (1) is true

let x\epsilon A \: \: and\: \: x\notin B => x\epsilon (A- B)=> x\epsilon C

let x\epsilon A \: \: and\: \: x\epsilon B => x\epsilon (A\cap B)=> x\epsilon C

Hence x\epsilon A \: \: and\: \:  x\epsilon C  => A\subseteq C

So, option (2) is  true.

Let x\in C , x\in(C\cup A)\cap(C\cup B))

=>x\epsilon (C\cup A)\: and\: x\epsilon (C\cup B))

\left (x\in C\;or\;x\in A \right )\;\;and\;\;\left ( x\in C\;or\;x\in B \right )

=>x\epsilon C\: or\: x\epsilon (A\cap B)

=>x\epsilon C\;\;\;As,\;A\cup B\subseteq C

(C\cup A)\cap(C\cup B)\subseteq C................(1)

Now,

=>x\epsilon C

=>x\epsilon (C\cup A)\cap(C\cup B)

=>C\subseteq (C\cup A)\cap(C\cup B)..............(2)

From (1) and (2)

(C\cup A)\cap(C\cup B)=C

=> option (3) is  correct.

For A = C , A-C=\phi

               =>\phi\subseteq B  but  =>A\nsubseteq B

So, option (4) is not true.

 


Option 1)

B\cap C\neq \phi

Option 2)

If\: \: (A-B)\subseteq C, then\: \: A\subseteq C

Option 3)

(C\cup A)\cap(C\cup B)=C

Option 4)

If\: \: (A-C)\subseteq B, then\: \: A\subseteq B

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