I am confused about what is the answer of P(P(P(Ø)))?
I have tried to solve it as follows
P(Ø) = ?
Ø has no elements therefore P(Ø) will contain 2^0=1 element, that is
P(Ø) = {Ø}
What will be the power set of the above set?
P(P(Ø)) = ?
P(Ø) has one element therefore P(P(Ø)) will contain 2^1=2 elements, that is
P(P(Ø)) = {Ø, {Ø}}
What will be the power set of the above set?
P(P(P(Ø))) = ?
P(P(Ø)) has two elements therefore P(P(P(Ø))) will contain 2^2=4 elements, that is
P(P(P(Ø))) = {Ø, {Ø}, {{Ø}}, {Ø, {Ø}}}
.....
Please tell me is the last power set correct? How?
Thanks!
Yes.
>
> What will be the power set of the above set?
>
> P(P(Ø)) = ?
>
> P(Ø) has one element therefore P(P(Ø)) will contain 2^1=2 elements, that is
> P(P(Ø)) = {Ø, {Ø}}
Yes.
>
> What will be the power set of the above set?
>
> P(P(P(Ø))) = ?
>
> P(P(Ø)) has two elements therefore P(P(P(Ø))) will contain 2^2=4 elements, that is
> P(P(P(Ø))) = {Ø, {Ø}, {{Ø}}, {Ø, {Ø}}}
Yes.
>
> .....
>
> Please tell me is the last power set correct? How?
"How" what?
Look, take your previous set, P(P(emptyset)). Call it A. A has two
elements; call them a and b (in "reality", "a' is emptyset, and "b" is
{emptyset}).
If A = {a,b}, then P(A) = { emptyset, {a}, {b}, {a,b} }. No problem,
right?
Well, now substitute the "real" values of A, a, and b to convince
yourself you got the correct answer.
Doesn't matter what you call the elements, after all.
--
Arturo Magidin
> Please tell me is the last power set correct? How?
One way to approach the problem of computing the powerset of a finite
set A is to first list the zero-element subsets (i.e. the empty set),
then the one-element subsets (singletons of elements of A), then the
two-element subsets, ... finally collecting them all together and
removing duplicates.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus