Assignment 1 : Problem 1

[Aaron Albers]

How many elements are in P(P(A)) where A is a set containing 1 element.

A = {a}

P(A) = { 0, {a}}

P(P(A)) = {0, {0}, {a}, {0, a}}

P(P(A)) contains 5 elements and 4 sets.

[gage saber]

Hey, I missed class yesterday. Can anyone please post the assignment
for me so I can get started on the problems and contribute to the
discussion?

[Anne S]

I think, looking at the answer to pause 3.a on page 42, that {a,b} is
a single element of P(P(A)). So P(P(A)) has four elements.

I also think the sets follow the rules of permutations, which would
mean there are 2^4 (16) subsets possible from the 4 elements.

On Sep 9, 2:41 pm, Aaron Albers <aaroncalb...@gmail.com> wrote:

[alex nguyen]

Anne is right, it should have four elements. hence, this is the
answer :)

{{}, {{}}, (a), {{}, {a}}}

that answer looks ugly.

[Aaron Albers]

I agree. That was the answer that I posted earlier.

[Diane]

I think per the definition of the power set, it can only contains sets.  so the elements of P(P(A)) are the 4 sets as follows:

1.  the empty set   0
2. the set containing the empty set   {0}
3. the set containing a   {a}
4 the set containing both the empty set and a  {a, 0}

so I came up with the same thing you did, but with slightly different terminology.

[alex]

Yeah I just posted my answer in response to anne.  Sorry i didn't take a look at what you wrote  :P