Respuesta :
Answer:
We have to prove that,
|P(A)| = [tex]2^n[/tex] , if |A| = n for n = 0, 1, 2, 3.
For n = 0,
A = {}
P(A) = { {} } = [tex]2^0[/tex] = 1
For n = 1,
A = { a } ( suppose )
P(A) = { {}, a } = [tex]2^{1}[/tex] = 2,
For n = 2,
A = { a, b }
P(A) = { {}, {a}, {b}, {a, b} = [tex]2^2[/tex] = 4,
For n = 3,
A = { a, b, c },
P(A) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } = [tex]2^3[/tex] = 8
Thus, it is verified for n = 0, 1, 2, 3.
Now, suppose it is valid for a set B having k elements,
That is, |P(B)| = [tex]2^k[/tex]
Also, there is a set A,
Such that, A = B ∪ {x}
Since, after including the element x in set B,
The element x will be come with every element of set B in the power set of B,
i.e. P(A) = [tex]2^k+2^k[/tex] = [tex]2^k(1+1)[/tex] = [tex]2^{k}.2[/tex] = [tex]2^{k+1}[/tex]
Hence, by the induction it has been proved,
|P(A)| = [tex]2^n[/tex] , if |A| = n, Where, n∈ N ( set of natural numbers )
