Proving |2^{A}| = 2^|A| Using Mathematical Induction - Step By Step Guide

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The discussion focuses on proving the equation |2^{A}| = 2^{|A|} for a finite set A using mathematical induction. The initial approach involves defining the powersets P(S) and P(S') for a set S with n elements and its extension S' with n+1 elements. Key steps include showing that P(S') can be expressed as the union of P(S) and a new set A, ensuring they are disjoint, and confirming that |A| equals 2^n. An alternative proof method is also suggested, utilizing characteristic maps to demonstrate the relationship between the number of subsets and the elements in set A. The discussion emphasizes the importance of establishing these foundational steps for a successful proof.
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Hey all,

As I said in one of my earlier posts I'm not that good with proofs, hence why I have another post.

Here is the one I'm on now...

Prove using mathematical induction that |2^{A}| = 2^{|A|} where A is a finite set.

..Here is what I have so far

http://img293.imageshack.us/img293/1944/proofdu9.jpg

Any help is greatly appreciated.

Thanks!
 
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bigrodey77 said:
Hey all,

As I said in one of my earlier posts I'm not that good with proofs, hence why I have another post.

Here is the one I'm on now...

Prove using mathematical induction that |2^{A}| = 2^{|A|} where A is a finite set.

..Here is what I have so far

http://img293.imageshack.us/img293/1944/proofdu9.jpg

Any help is greatly appreciated.

Thanks!

You might look at a proof this way.

Let's let |S| = n and S' = S U {a} (where a is not in S). Clearly, |S'| = n+1.
Let P(S) and P(S') denote their powersets.
Let A = {B U {a} | B is in P(S)}.

In the induction step, the induction hypothesis says |P(S)| = 2^n.
So if you can show the following

1. P(S') = P(S) U A,
2. P(S) and A are disjoint,
3. |A| = 2^n,

you'll be home.

Of course, induction isn't the only approach.
You might think about a proof based on characteristic maps.
 
2A means, of course, the collection of all subsets of A. |A| is the number of elements in A and |2A| is the number of elements in 2A, i.e. the number of subsets of A. And, of course, we are assuming A is a finite set.

Assume, as you did in the induction, that |2A|= 2|A| for all sets with k members: |A|= k. Now let B be a set with k+1 members: |B|= k+1. Let x be a specific member of B (since |B|= k+1> 0, there exist such a member). Every subset of B either includes x or it does not. Those subsets that do not include x are subsets of B\{x} which has k members. How many of them are there? Those subsets that do include x can be thought of as C\cup {x} where C is a subset of B that does not include x. How of them are there? How many subsets does B have?
 
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I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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