P(A) is the power set of A. Show that-
(i) P(AUB) = P(A) ∩ P(B) is always true, but
(ii) P(AUB) = P(A) U P(B) may not be always true.

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SalmonWorldTopper SalmonWorldTopper · Answer 1 · 2021-10-24T13:29:05+02:00

[tex]\large\underline{\sf{Solution-}}[/tex]

Let assume that,

[tex]\rm \longmapsto\:A \: = \:\{x\} \: [/tex]

and

[tex]\rm \longmapsto\:B \: = \:\{y\} \: [/tex]

[tex]\rm\implies \:A \: \cap \: B = \phi [/tex]

and

[tex]\rm \longmapsto\:A\cup B = \{x, \: y\}[/tex]

We know,

Power set of a set A is defined as set of subsets of A.

So,

[tex]\rm \longmapsto\:P(A) = \{ \phi ,\: x\}[/tex]

and

[tex]\rm \longmapsto\:P(B) = \{ \phi ,\: y\}[/tex]

and

[tex]\rm\implies \:P(A \: \cap \: B) =\{ \phi \}[/tex]

and

[tex]\rm \longmapsto\:P(A\cup B) = \{\phi ,x,y,xy\}[/tex]

Now, Consider

[tex]\rm \longmapsto\:P(A)\cup P(B) = \{\phi ,x,y\}[/tex]

and

[tex]\rm \longmapsto\:P(A\cup B) = \{\phi ,x,y,xy\}[/tex]

[tex]\rm\implies \:\boxed{\tt{ \: P(A)\cup P(B) \: \ne \: P(A\cup B) \: }} \\ [/tex]

Now, Consider

[tex]\rm \longmapsto\:P(A\cap B) = \{\phi \}[/tex]

and

[tex]\rm \longmapsto\:P(A) \: \cap \: P( B) = \{\phi \}[/tex]

[tex]\rm\implies \:\boxed{\tt{ \: P(A)\cap P(B) \: = \: P(A\cap B) \: }} \\ [/tex]