Dhaka Higher Secondary 2011/3

[BdMO]

If $A = \{\Phi\}$, $A \cup B = P(A)$ and $A \cap B = \Phi$, find $B$. $\Phi$ represents empty set.

[Zzzz]

Hint:

A set can be element of a set

[TTowsif]

I think the ans is empty set, isn't it?

[Hasib]

$B=\{\phi\}$ isn't it?

[Moon]

Remember set of empty set can be inside a set.

[Zzzz]

Ah.. guys are getting closer. One said $B=\Phi$, another said $B=\{\Phi\}$. What's next?

You should explain your answers ...

[Hasib]

Oh shit! I haven't saw the braccet around $\phi$ i saw $A=\phi$ ok ok okz! Now, trying again.

[leonardo shawon]

here $A= \{\phi\}$
$P(A)=\{\phi,empty set\}$
$A \cup B=\{\phi,empty set\}$ ,
$A \cap B=\phi$
if we put the value of set "A" then we can see that set "B" is an empty set . That means. . .
$\{\phi,empty set\} \cap empty set = \phi$
so,,$ B= \phi $


Did i do any mistake....?

*** EDITED

[Zzzz]

Hey . $\Phi $ is a set. $A=\{\Phi\}$ is not an empty set. It has an element which is $\Phi$. If they said $A=\Phi$ then we could say $A$ is an empty set.

[Hasib]

$A=\{\emptyset\}$
$A \cup B=P(A)$
now, $P(A)=\{\{\emptyset\},\emptyset\}$
so,
$\underline{\{\emptyset\}} \cup B=\{\{\emptyset\},\underline{\emptyset\}}$

the underlined are same. And $A \cap B=\emptyset$
So, $B=\{\{\emptyset\}\}$

ri8?