Thread: 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).

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?

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}.

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).
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?

No, not even close to being correct.

is a set of all subsets of A.
It is true that and .
But there are fourteen more sets in .

Originally Posted by Plato

No, not even close to being correct.

is a set of all subsets of A.
It is true that and .
But there are fourteen more sets in .

How do you know that there are fourteen more sets in ?

Originally Posted by maclunian

How do you know that there are fourteen more sets in ?

If a set, contains elements the contains elements.

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?

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}}

What about the second bit?

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}

Originally Posted by maclunian

What about the second bit?
= {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.

one way to understand "why" |P(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 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}.