How to prove that $a=a',b' Rightarrow a=a'=b' $? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Set theory like first-order theory of ordered pairsSome questions regarding set theory.A question regarding the validity of the Union AxiomComprehension and ImpredicativityAxiom of pairing finite structureThe natural numbers as the intersection of all inductive setsA question about collections that agree with logical statements in $ZF$Equivalent statement for theAxiom of PairingAre notations in ZF conservative?How to prove statements in formal set theory? Substitution, the empty set and an example.
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How to prove that $a=a',b' Rightarrow a=a'=b' $?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Set theory like first-order theory of ordered pairsSome questions regarding set theory.A question regarding the validity of the Union AxiomComprehension and ImpredicativityAxiom of pairing finite structureThe natural numbers as the intersection of all inductive setsA question about collections that agree with logical statements in $ZF$Equivalent statement for theAxiom of PairingAre notations in ZF conservative?How to prove statements in formal set theory? Substitution, the empty set and an example.
$begingroup$
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
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
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
edited Mar 26 at 16:52
Mauro ALLEGRANZA
68.1k449117
asked Mar 26 at 16:43
Matsukazi IciMatsukazi Ici
397
$endgroup$
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
elementary-set-theory logic
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
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
edited Mar 26 at 16:52
Mauro ALLEGRANZA
68.1k449117
68.1k449117
asked Mar 26 at 16:43
Matsukazi IciMatsukazi Ici
397
asked Mar 26 at 16:43
Matsukazi IciMatsukazi Ici
397
asked Mar 26 at 16:43
Matsukazi IciMatsukazi Ici
397
397
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
$a'=a=b'$.
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 |
$begingroup$
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?
answered Mar 26 at 18:08
MacRanceMacRance
1826
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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:
$a'=a=b'$.
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:
$a'=a=b'$.
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:
$a'=a=b'$.
answered Mar 26 at 16:51
Mauro ALLEGRANZAMauro ALLEGRANZA
68.1k449117
$endgroup$
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:
$a'=a=b'$.
answered Mar 26 at 16:51
Mauro ALLEGRANZAMauro ALLEGRANZA
68.1k449117
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
68.1k449117
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
$begingroup$
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?
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?
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?
answered Mar 26 at 18:08
MacRanceMacRance
1826
$endgroup$
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?
answered Mar 26 at 18:08
MacRanceMacRance
1826
answered Mar 26 at 18:08
MacRanceMacRance
1826
answered Mar 26 at 18:08
MacRanceMacRance
1826
answered Mar 26 at 18:08
MacRanceMacRance
1826
1826
add a comment |
add a comment |
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