for any sets A and B . show that
P ( A intersection B ) = P ( A ) intersection P (B)

Hi Akshya,
Please find below the solution to the asked query:

L.H.S.=PABLetxPABxABxA and xB ; StatementiR.H.S.=PAPBLet xPAPBxA and xB   ; StatementiiBy statementi and statementiiL.H.S.=R.H.S.PAB=PAPB  Hence proved

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  • 25
Let X belongs to P (AnB).then
XcAnB nB.so, XcA and XcB.
; X blngs to P (A)and X blngs to P (B) which implies X blngs to P(A) nP (B).This gives P (AnB) c P (A)n P(B).
Let Y blngs to P ( A) n P ( B).
then Y blngs to P (A) and Y blngs to P (B). so, Y c A and Y c B.Therefore , Y can, which implies Y blngs to P (AnB).
This gives P (A) n P (B) c P(AnB)
hence , P (AnB) equal to P (A) nP (B).
hint : c :subset
n:intersection
  • 6
or it can be done as ;
Take a set X which belongs to
P(AnB) implies x blngs to AnB
ie, x blngs to A and X blngs to B
so , set X blngs to P (A) and set X blngs to P (B).
Therefore , set X belongs to
P (A) nP (B).
  • 8
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