I have the following question:
Let A = {a,b,c,d},
Let B = {c,d,e}.
Compute P(A) and P(A (intersection) B).
Here's my answer:
P(A) = {{a,b,c,d}}
A(intersection)B = {c,d}
P(A(intersection)B) = {{c,d}}
Is my answer correct?
one way to understand "why" |P(A)| = $\displaystyle 2^{|A|}$ is to think of all possible functions:
f:A→{0,1}.
for example, one such function is:
f(a) = 1
f(b) = 1
f(c) = 0
f(d) = 1
which corresponds to the set {a,b,d}. see what i did? i just picked a subset of A, in this case, {a,b,d}, and defined f(x) = 1, if x is in {a,b,d}, f(x) = 0, otherwise.
there are $\displaystyle 2^{|A|}$ such functions (since we have 2 choices for every element f(x), and |A| elements).
if you like, you can think of this as a truth-table for the proposition: "is x in this subset X of A?" a value of 1, means "yes"(true), and a value of 0 means "no"(false).
if you have a set of 3 elements, its power set will have 8 elements. and there are precisely 8 functions from {x,y,z} to {0,1}.