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How to prove that $a=a',b' Rightarrow a=a'=b' $?

1

$begingroup$

I'm using Jean-Louis Krivine's book "Theorie des ensembles" as a reference book so the notations might be a little different from english literature, but i guess the question is easy to follow.

Let $mathcal U$ be a universe (a collection of objects that is supposedly not empty, the objects are called sets). We define a binary relation on this universe that we note "$in$".

Axiom of extensionality: $forall x forall y[forall z(zin xLeftrightarrow z in y)Rightarrow x=y]$.

Axiom of pairing: $forall x forall y exists zforall t[tin zLeftrightarrow (t=x lor t=y)]$.

My question is, assuming only these two axioms, how to prove that if we had sets $a,a' and b'$ satisfying $a=a',b'$ then $ a=a'=b' $?

elementary-set-theory logic

share|cite|improve this question

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

$endgroup$

add a comment | 

1

$begingroup$

I'm using Jean-Louis Krivine's book "Theorie des ensembles" as a reference book so the notations might be a little different from english literature, but i guess the question is easy to follow.

Let $mathcal U$ be a universe (a collection of objects that is supposedly not empty, the objects are called sets). We define a binary relation on this universe that we note "$in$".

Axiom of extensionality: $forall x forall y[forall z(zin xLeftrightarrow z in y)Rightarrow x=y]$.

Axiom of pairing: $forall x forall y exists zforall t[tin zLeftrightarrow (t=x lor t=y)]$.

My question is, assuming only these two axioms, how to prove that if we had sets $a,a' and b'$ satisfying $a=a',b'$ then $ a=a'=b' $?

elementary-set-theory logic

share|cite|improve this question

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

$endgroup$

add a comment | 

$begingroup$

I'm using Jean-Louis Krivine's book "Theorie des ensembles" as a reference book so the notations might be a little different from english literature, but i guess the question is easy to follow.

Let $mathcal U$ be a universe (a collection of objects that is supposedly not empty, the objects are called sets). We define a binary relation on this universe that we note "$in$".

Axiom of extensionality: $forall x forall y[forall z(zin xLeftrightarrow z in y)Rightarrow x=y]$.

Axiom of pairing: $forall x forall y exists zforall t[tin zLeftrightarrow (t=x lor t=y)]$.

My question is, assuming only these two axioms, how to prove that if we had sets $a,a' and b'$ satisfying $a=a',b'$ then $ a=a'=b' $?

elementary-set-theory logic

share|cite|improve this question

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

$endgroup$

share|cite|improve this question

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

share|cite|improve this question

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

edited Mar 26 at 16:52

Mauro ALLEGRANZA

68.1k449117

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

asked Mar 26 at 16:43

Matsukazi IciMatsukazi Ici

397

2 Answers
2

active

oldest

votes

4

$begingroup$

We have that $a' in a',b′ = a $ and $b' in a',b′ = a $.

But $a' in a $ means that $a'=a$, and the same with $b'$.

Thus, from transitivity of equality:

$a'=a=b'$.

share|cite|improve this answer

edited Mar 26 at 19:14

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My bad, i thought that $a=a′,b′$ means that $ a' in a,$ which is false, and instead we have $ a' in a$, and $a$ being the only set containig $a$ and just $a$ gives $a'=a$. thank you so much for your answer.
  $endgroup$
  – Matsukazi Ici
  Mar 26 at 20:07
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: If you don't like the direct route you can go by contradiction: Suppose $a neq a'$, for example. Then use pairing to obtain the sets $ a $ and $ a',b' $. Now use the contrapositive of extensionality; what do you get?

share|cite|improve this answer

answered Mar 26 at 18:08

MacRanceMacRance

1826

$endgroup$

add a comment | 

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4

$begingroup$

We have that $a' in a',b′ = a $ and $b' in a',b′ = a $.

But $a' in a $ means that $a'=a$, and the same with $b'$.

Thus, from transitivity of equality:

$a'=a=b'$.

share|cite|improve this answer

edited Mar 26 at 19:14

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My bad, i thought that $a=a′,b′$ means that $ a' in a,$ which is false, and instead we have $ a' in a$, and $a$ being the only set containig $a$ and just $a$ gives $a'=a$. thank you so much for your answer.
  $endgroup$
  – Matsukazi Ici
  Mar 26 at 20:07
  
  

  
  
  
  

add a comment | 

4

$begingroup$

We have that $a' in a',b′ = a $ and $b' in a',b′ = a $.

But $a' in a $ means that $a'=a$, and the same with $b'$.

Thus, from transitivity of equality:

$a'=a=b'$.

share|cite|improve this answer

edited Mar 26 at 19:14

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My bad, i thought that $a=a′,b′$ means that $ a' in a,$ which is false, and instead we have $ a' in a$, and $a$ being the only set containig $a$ and just $a$ gives $a'=a$. thank you so much for your answer.
  $endgroup$
  – Matsukazi Ici
  Mar 26 at 20:07
  
  

  
  
  
  

add a comment | 

$begingroup$

We have that $a' in a',b′ = a $ and $b' in a',b′ = a $.

But $a' in a $ means that $a'=a$, and the same with $b'$.

Thus, from transitivity of equality:

$a'=a=b'$.

share|cite|improve this answer

edited Mar 26 at 19:14

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

$endgroup$

share|cite|improve this answer

edited Mar 26 at 19:14

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

answered Mar 26 at 16:51

Mauro ALLEGRANZAMauro ALLEGRANZA

68.1k449117

1

$begingroup$

Hint: If you don't like the direct route you can go by contradiction: Suppose $a neq a'$, for example. Then use pairing to obtain the sets $ a $ and $ a',b' $. Now use the contrapositive of extensionality; what do you get?

share|cite|improve this answer

answered Mar 26 at 18:08

MacRanceMacRance

1826

$endgroup$

add a comment | 

1

$begingroup$

Hint: If you don't like the direct route you can go by contradiction: Suppose $a neq a'$, for example. Then use pairing to obtain the sets $ a $ and $ a',b' $. Now use the contrapositive of extensionality; what do you get?

share|cite|improve this answer

answered Mar 26 at 18:08

MacRanceMacRance

1826

$endgroup$

add a comment | 

$begingroup$

Hint: If you don't like the direct route you can go by contradiction: Suppose $a neq a'$, for example. Then use pairing to obtain the sets $ a $ and $ a',b' $. Now use the contrapositive of extensionality; what do you get?

share|cite|improve this answer

answered Mar 26 at 18:08

MacRanceMacRance

1826

$endgroup$

share|cite|improve this answer

answered Mar 26 at 18:08

MacRanceMacRance

1826

answered Mar 26 at 18:08

MacRanceMacRance

1826

answered Mar 26 at 18:08

MacRanceMacRance

1826