If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$?

Question

Prove, disprove, or give a counterexample:

If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$.

Assume $\mathcal{P}(A)=\mathcal{P}(B)$. Since we know $A \subseteq A$, we know $A \in \mathcal{P}(A)$.

Since $\mathcal{P}(A)=\mathcal{P}(B)$, we know that $A \in \mathcal{P}(B)$.

Therefore, $A \subseteq B$ and $A=B$.

Is this proof okay?

Edit: I should note this isn't probability, $\mathcal{P}$ is the power set.

It's missing the $B \subseteq A$ bit, but the idea is right. — Karolis Juodelė, Mar 15 '14 at 19:52

score 17 · Accepted Answer · answered Mar 15 '14 at 20:32

17

You can prove: $$\bigcup \mathcal{P}(A)=A$$$$\bigcup\mathcal{P}(B)=B$$ and by hypothesis you have $\mathcal{P}(A)=\mathcal{P}(B)$ therefore $$B=\bigcup\mathcal{P}(B)=\bigcup\mathcal{P}(A)=A$$

answered Mar 15 '14 at 20:32

mle

2,287

score 11 · Answer 2 · answered Mar 15 '14 at 20:09

Everything in your answer is correct until the implication "Therefore $A \subseteq B$ and $A=B$". The second implication (i.e. that $A=B$ is not yet justified). So keep only the first, i.e. that $$A \subseteq B \tag 1$$ Reason in exactly the same way to deduce that $$B \subseteq A \tag2$$ Now combining $(1)$ and $(2)$ yields the result $$A=B$$

score 8 · Answer 3 · answered Mar 15 '14 at 19:54

8

The proof is almost okay. You need to argue why $B\subseteq A$. But it follows from the same argument.

answered Mar 15 '14 at 19:54

Asaf Karagila

393,674

I should use the same logic that showed A$\subseteq$B to show B$\subseteq$A? And thus, A=B? – Vincent Mar 15 '14 at 20:03
1

@CharlesNosbig Just say something like, "By symmetry, $B \subseteq A$ as well." – Caleb Stanford Mar 15 '14 at 20:11
@Charles: Yes, you should. – Asaf Karagila Mar 15 '14 at 20:11

Asinomás · Answer 4 · 2014-12-13T17:26:15.317

1

$A\in \mathcal P (A)\rightarrow A \in \mathcal P (B)\rightarrow A\subset B$

$B\in \mathcal P (B)\rightarrow B \in \mathcal P(A)\rightarrow B\subset A$

Using the conclusions of each line $A=B$

edited Dec 13 '14 at 17:26

answered Mar 15 '14 at 20:39

Asinomás

105,651

Darius · Answer 5 · 2014-04-05T22:08:02.790

-1

If we know $A$ we can determine unequivocally $\mathcal{P}(A)$ and if we know $\mathcal{P}(A)$ we can determine unequivocally A, as element of $\mathcal{P}(A)$ which includes any subset , so if $A=B$ then $\mathcal{P}(A)=\mathcal{P}(B)$ and if $A\neq B$ then $\mathcal{P}(A)\neq\mathcal{P}(B)$ what completes the proof of your theorem.

Another way of proving: $$(\mathcal{P}(A)=\mathcal{P}(B))\implies [(A\in\mathcal{P}(B)) \wedge (B\in\mathcal{P}(A))]\implies [(A \subseteq B) \wedge (B \subseteq A)]\\ [(A \subseteq B) \wedge (B \subseteq A)] \implies [(A\cup B=B) \wedge (B\cup A=A)]\implies (A=B)$$

PS. -1 as rating of my answer is more against reputation of math.stackexchange.com , than mine.

edited Apr 05 '14 at 22:08

answered Mar 15 '14 at 20:15

Darius

1,334

This construction only works for finite sets. We should take the element that is maximal under inclusion. – Andrew Dudzik Mar 15 '14 at 20:18
Injection, not bijection. – egreg Mar 15 '14 at 20:25
... and exactly the assertation that $A\mapsto P(A)$ (not $P(A)$ which is a set, not a function) is an injection is what needs to be proved here. – Johannes Hahn Mar 15 '14 at 20:26
Or take the union of all elements i.e sets in $\mathcal{P}(A)$. – Uri Brezner Mar 15 '14 at 20:27
@egreg it can be also bijection. – Darius Mar 15 '14 at 21:01
@JohannesHahn it was proved when you were reading my answer. – Darius Mar 15 '14 at 21:03
@egreg injective function, be accurate, if you require it from me. – Darius Mar 15 '14 at 21:05
@UriBrezner if set of the sets is not the power set , it doesn't work. – Darius Mar 15 '14 at 21:07
3

@Darius You were using the term “bijection”. – egreg Mar 15 '14 at 21:20
@egreg injective function may be also bijection – Darius Mar 15 '14 at 21:21
There is no reason for hating correct answer. It's against reputation of math.stackexchange.com – Darius Mar 15 '14 at 22:05
@JohannesHahn You have written: "...... (not P$(A)$ which is a set, not a function)...", if understand you $\sin(x)$ is a number , not a function? – Darius Mar 16 '14 at 05:06
@Darius: Exactly. sin is the function, sin(x) is the value of that function at the point x. – Johannes Hahn Mar 16 '14 at 13:45
@JohannesHahn but there is a custom using term , function $sin(x)$ , and there is no rational reason for persecuting such a people, or using sanction against them, because it can lead to defeat and scarcity , isn't it Johann? – Darius Mar 16 '14 at 16:32
1

Whoa calm down there. Nobody said anything about persecuting or sanctioning. – Johannes Hahn Mar 16 '14 at 16:50

If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$?

5 Answers5