power sets equality

shahar

shahar

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A and B are sets.
I know that: |A| = |B|
How I prove that the "size" of the power sets of them is equal (one to each other)?
Or: |P(A)| = |P(B)|

Given: |A| = |B|
Prove: |P(A)| = |P(B)|
 
shahar said:
A and B are sets.
I know that: |A| = |B|
How I prove that the "size" of the power sets of them is equal (one to each other)?
Or: |P(A)| = |P(B)|

Given: |A| = |B|
Prove: |P(A)| = |P(B)|
Click to expand...
Sorry but I use [imath]\|A\|[/imath] for the number of elements.
To prove that [imath]\|\mathcal{P}(A)\|=\|\mathcal{P}(B)\|[/imath] show that there is a bijection between the sets.
Give that a try and post your efforts.
 
null set in A -> null set in B
a_1 in A -> b_1 in B
a_i in A -> b_i in B
a_(i+1) in A -> b_(i+1) in B
a_(i+j) in A -> b_(i+j) in B
...
That what we have to prove.
 
shahar said:
null set in A -> null set in B
a_1 in A -> b_1 in B
a_i in A -> b_i in B
a_(i+1) in A -> b_(i+1) in B
a_(i+j) in A -> b_(i+j) in B
...
That what we have to prove.
Click to expand...
I cannot agree with this statement. What are elements of [imath]\mathcal P (A)[/imath] and [imath]\mathcal P(B)[/imath]?
 
Do you know what it means to even say that two set are equinumerous: i.e. that [imath]\|A\|=\|B\|[/imath]
If you do not know the basics the how do you expect to do proofs???
The so-called proof you posted looks like an proof by induction. Those are only valid for countable sets,
B.T.W. [imath]\mathcal{P}(B)=\{T: T\subseteq B\}[/imath]
Please answer my question: what does it mean that two set are equinumerous and how is it proved?
 
As aka has stated, if |A| = |B|, then there is a bijection between the two sets. This bijection just states a different way to write the elements in set A with elements in set B.

BTW, are your sets finite?
 
pka said:
Do you know what it means to even say that two set are equinumerous: i.e. that [imath]\|A\|=\|B\|[/imath]
If you do not know the basics the how do you expect to do proofs???
The so-called proof you posted looks like an proof by induction. Those are only valid for countable sets,
B.T.W. [imath]\mathcal{P}(B)=\{T: T\subseteq B\}[/imath]
Please answer my question: what does it mean that two set are equinumerous and how is it proved?
Click to expand...
Powerset of A is the set A, the null element and the sets that consist from partial elements of A.
 
shahar said:
Powerset of A is the set A, the null element and the sets that consist from partial elements of A.
Click to expand...
Please write a simple sentence stating what it means to say that two sets are equinumerous,
[imath]\|X\|=\|Y\|[/imath], and how we would prove that [imath]\|\mathcal{P}(X)\|=\|\mathcal{P}(Y)\|[/imath].
Please do that to show that there is some reason to keep this going.
 
pka said:
Please write a simple sentence stating what it means to say that two sets are equinumerous,
[imath]\|X\|=\|Y\|[/imath], and how we would prove that [imath]\|\mathcal{P}(X)\|=\|\mathcal{P}(Y)\|[/imath].
Please do that to show that there is some reason to keep this going.
Click to expand...
There is a bijection function between Set A to Set B, so there is a bijection function between Powersets A to Powersets B.
O.K. I am stopping here.
I have a question:
How can I write this in the symbols of sets theory the sentence I wrote? (I think it will help to understand - there is a way to write it in sets theory symbol???)
 
shahar said:
There is a bijection function between Set A to Set B, so there is a bijection function between Powersets A to Powersets B.
Click to expand...
I think you actually have to prove this statement. I don't know how to avoid using plain English in such proof, but it will necessarily contain some symbols from the set theory.
 
Finally, thank you. Given [imath]\|A\|=\|B\|[/imath] we know that there is a bijection [imath]g:A\leftrightarrow B[/imath]
Notation: If [imath]J\subseteq A\text{ then }\overrightarrow g (J) = \left\{ {g(x):x \in J} \right\}[/imath]
Likewise, If [imath]S\subseteq B\text{ then }\overleftarrow g (S) = \left\{ {x\in A: g(x) \in S} \right\}[/imath]
In other words [imath]\overrightarrow g (\mathcal{P}(A))\to \mathcal{P}(B)~\&~\overleftarrow g (\mathcal{P}(B))\to \mathcal{P}(A)[/imath]
As part of your background course in set-theory we prove that if [imath]g \text{ is a bijection then so is }\overleftarrow g[/imath][imath][/imath] and we are done.
 
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