# Discrete Math Proof With Power-sets [duplicate]

The question I am working on is:

Can you conclude that $A=B$ if $A$ and $B$ are two sets with the same power set?

At first, I thought of doing a proof by contraposition; however, that didn't appear (ostensibly, at least) it would prove effective. Although, you of more experience, if you happen to know a way of using that proof method, I would doubtlessly be interested.

So, I then thought of doing a proof by contradiction. Here is my attempt, if you'd be so kind as to look it over and evaluate it, I would appreciate it:

PS I'll have a few questions in bold, sprinkled throughout the proof.

If $A$ and $B$ are two sets with the same power-set, then $A=B$.

Let's assume the opposite is true, that if $A$ and $B$ are two sets with the same power-set, then $A$ and $B$ must be different sets ($A \ne B$).

If $A \ne B$, then either $A$ or $B$ will have at least one element that isn't in the other set. Say $A$ has the extra element, denoted by $e$. (Although you could have $B$ contain the extra element, the argument will work either way, it is independent of the choice).

Let $A= \{a_1,a_2, a_3, ...,e,...\}$, and $B=\{b_1,b_2,b_3,...\}$, such that $B$ doesn't contain $e$ (How would I write that little bit in set notation? Also, do I have to explicitly write that $a_1 =b_1$, $a_2=b_2$,and so on?)

So, $\mathcal P(A)=\biggr \{ \{a_1\},\{a_2\},\{a_3\},...\{e\},...\{a_1,a_2,a_3,...,e,...\} \biggr \}$

and $\mathcal P(A)=\biggr \{ \{b_1\},\{b_2\},\{b_3\},...\biggr \}$

As you can see, the power-set of $A$ contains at least one element that the power-set of $B$ doesn't have (although, the differ by more than just one). This leads to a contradiction, because we assumed that the power-sets were equal.

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## marked as duplicate by Asaf Karagila, 5pm, Jim, Cameron Buie, Chris EagleFeb 27 '13 at 17:17

This looks fine to me, if slightly muddled in the wording. Be careful with statement's such as "Let's assume the opposite is true, that if $A$ and $B$ are two sets with the same power-set, then $A$ and $B$ must be different sets ($A \ne B$). " The nagation of the statement is only that $P(A)=P(B)$ does not imply $A=B$. This is a different statement than $P(A)=P(B)$ implies that $A\neq B$. The important point I think, is that you understand an element in $A$ but not in $B$ will be in an element of $P(A)$ but not in an element of $P(B)$ which contradicts the fact that $P(A)=P(B)$. – Dan Rust Feb 27 '13 at 16:24

Let $A= \{a_1,a_2, a_3, ...,e,...\}$, and $B=\{b_1,b_2,b_3,...\}$, such that $B$ doesn't contain $e$ (How would I write that little bit in set notation? Also, do I have to explicitly write that $a_1 =b_1$, $a_2=b_2$,and so on?)

Rather than worrying about stating that each of $a_i = b_i$, which you would need to do otherwise, why not list:

$A= \{x_1,x_2, x_3, ...,e,...\}$, and $B=\{x_1,x_2,x_3,...\}$, or better yet, $$A= \{x_1,x_2, x_3, ...,e,...\},\;\;B = A\setminus \{e\}.$$

Therefore, if $\mathcal P(A) = \mathcal P(B) \implies A = B$.

Then you could proceed a bit more easily with your expansion of the power sets of $A$ and $B$.

It might help to conclude that clearly, while $\mathcal{P}(B) \subseteq \mathcal{P}(A)$, $\mathcal{P}(A) \not\subseteq \mathcal{P}(B)$. Hence $\mathcal{P}(A) \neq \mathcal{P}(B)$: contradition!

Indeed, the only element of importance here, is the element $e$, provided $e \in A$, $e\notin B$, hence $\{e\} \in \mathcal{P}(A),\;\land \; \{e\} \notin \mathcal P (B)$...Hence $\mathcal P (A) \not\subset \mathcal P (B),\;$ hence $\mathcal P(A) \neq \mathcal P(B)$, which is a contradiction.

(Alternatively, it follows by contrapositive from $A \neq B \implies \mathcal P(A) \neq \mathcal P(B)$, if you start from the mere premise that "Suppose $A \neq B$...without loss of generality, we can assume there is some element $e\in A, e \notin B$...)

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Oh, yes. Those are some excellent suggestions. Thank you, amWhy! – Mack Feb 27 '13 at 16:28
Let me know when you've "wrapped it all up": your proof! – amWhy Feb 27 '13 at 17:26
Oh, so by taking the contrapositive, we would end up having to do a proof by contradiction? – Mack Feb 27 '13 at 17:40
No, you can prove the contrapositive simply by starting with the single premise: suppose $A \neq B$, then....then....thus $P(A) \neq P(B)$. Thus, by contraposition, since $A \neq B \implies P(A) \neq P(B) \iff P(A) = P(B) \implies A = B$. Proof by contradiction assumes both the truth of the premise: $P(A) = P(B)$ and the supposition $A \neq B$...from which we derive a contradiction. Strictly speaking, these approaches are intimately connected. My last paragraph, in parentheses, would be how to approach pretty much the same proof but by contraposition... – amWhy Feb 27 '13 at 18:01
But I think your approach is just fine, with some of the modifications I suggested initially. – amWhy Feb 27 '13 at 18:14

Two simple proofs:

(1) Assume $P(A)=P(B)$. Since $A\in P(A)$, we have $A\in P(B)$, which means $A\subseteq B$. Similarly, $B\subseteq A$. Therefore $A=B$.

(2) Every set $X$ is the union of all the members of $P(X)$. So, if $P(A)=P(B)$, apply to both sides of this equation the operation (often denoted by $\bigcup$) "union of all the elements of" to get $A=B$.

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(3) The maximal element (with respect to $\subseteq$) of $P(A)$ is $A$. +1 for simplicity. – dtldarek Feb 27 '13 at 17:05

I'll post what I think is a fairly model answer. Notice how I don't need to make reference to any of the other elements in $A$ or $B$, only a single one.

Suppose $P(A)=P(B)$ but that $A\neq B$. Without loss of generality, the statement that $A\neq B$ implies that there exists an element $e\in A$ such that $e\notin B$. It follows that $\{e\}\in P(A)$ and $\{e\}\notin P(B)$. But $P(A)=P(B)$ implies that if $\{e\}\in P(A)$ then $\{e\}\in P(B)$. This is a contradiction and so our assumption that $A\neq B$ is false. Hence $P(A)=P(B)$ implies that $A=B$. QED

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