# Thread: Computing Powers of Sets

1. ## Computing Powers of Sets

I have the following question:

Let A = {a,b,c,d},
Let B = {c,d,e}.

Compute P(A) and P(A (intersection) B).

P(A) = {{a,b,c,d}}
A(intersection)B = {c,d}
P(A(intersection)B) = {{c,d}}

2. ## Re: Computing Powers of Sets

no. the power set of a set, contains ALL subsets of the original set.

for example, if C = {a,b}, then P(C) = {Ø,{a},{b},{a}U{b}} = {Ø,{a},{b},{a,b}} = {Ø, {a},{b},C}

the power set is "bigger" than just {C}.

3. ## Re: Computing Powers of Sets

Originally Posted by maclunian
I have the following question:
Let A = {a,b,c,d},
Let B = {c,d,e}.
Compute P(A) and P(A (intersection) B).
P(A) = {{a,b,c,d}}
A(intersection)B = {c,d}
P(A(intersection)B) = {{c,d}}
No, not even close to being correct.

$\mathcal{P}(A)$ is a set of all subsets of A.
It is true that $A\in\mathcal{P}(A)$ and $\emptyset\in\mathcal{P}(A)$.
But there are fourteen more sets in $\mathcal{P}(A)$.

4. ## Re: Computing Powers of Sets

Originally Posted by Plato
No, not even close to being correct.

$\mathcal{P}(A)$ is a set of all subsets of A.
It is true that $A\in\mathcal{P}(A)$ and $\emptyset\in\mathcal{P}(A)$.
But there are fourteen more sets in $\mathcal{P}(A)$.
How do you know that there are fourteen more sets in $\mathcal{P}(A)$?

5. ## Re: Computing Powers of Sets

Originally Posted by maclunian
How do you know that there are fourteen more sets in $\mathcal{P}(A)$?
If a set, $A$ contains $n$ elements the $\mathcal{P}(A)$ contains $2^n$ elements.

6. ## Re: Computing Powers of Sets

Originally Posted by Deveno
no. the power set of a set, contains ALL subsets of the original set.

for example, if C = {a,b}, then P(C) = {Ø,{a},{b},{a}U{b}} = {Ø,{a},{b},{a,b}} = {Ø, {a},{b},C}

the power set is "bigger" than just {C}.
So, if A = {a,b,c,d}, then P(A) = {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}} = {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}

Is this correct?

7. ## Re: Computing Powers of Sets

Originally Posted by maclunian
So, if A = {a,b,c,d}, then P(A) = {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}} = {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}
Is this correct?
This much is correct: P(A) = {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}}

8. ## Re: Computing Powers of Sets

P(A) = {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}

9. ## Re: Computing Powers of Sets

Originally Posted by maclunian
= {Ø,{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c}{b,d},{c ,d},{a,b,c},{a,b,d},{a,c,d},{b,c,d},A}
They are identical. You just renamed one set.

10. ## Re: Computing Powers of Sets

one way to understand "why" |P(A)| = $2^{|A|}$ 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 $2^{|A|}$ 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}.