If $A$ and $B$ are sets,does $AcupA=BcupB$ imply $A=B$? [closed]

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Please help.

I have read the prove of the existence of the natural numbers in Analysis I written by Amann and Escher .It uses the claim that the function $mathcalV:mathbbNlongrightarrowmathbbN$, $zlongmapsto zcupz$ is injective.Here $mathbbN$ is a set we construct to be the set of natural numbers.I don't know how to prove the claim.

My English is not good.If you don't understand what I said,please tell me.

analysis elementary-set-theory

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edited Jul 14 at 14:52

Andrés E. Caicedo

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asked Jul 14 at 13:20

Djx

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closed as off-topic by Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, user223391 Jul 14 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

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up vote
2
down vote

favorite

Please help.

I have read the prove of the existence of the natural numbers in Analysis I written by Amann and Escher .It uses the claim that the function $mathcalV:mathbbNlongrightarrowmathbbN$, $zlongmapsto zcupz$ is injective.Here $mathbbN$ is a set we construct to be the set of natural numbers.I don't know how to prove the claim.

My English is not good.If you don't understand what I said,please tell me.

analysis elementary-set-theory

share|cite|improve this question

edited Jul 14 at 14:52

Andrés E. Caicedo

63.2k7151236

asked Jul 14 at 13:20

Djx

132

closed as off-topic by Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, user223391 Jul 14 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Please help.

I have read the prove of the existence of the natural numbers in Analysis I written by Amann and Escher .It uses the claim that the function $mathcalV:mathbbNlongrightarrowmathbbN$, $zlongmapsto zcupz$ is injective.Here $mathbbN$ is a set we construct to be the set of natural numbers.I don't know how to prove the claim.

My English is not good.If you don't understand what I said,please tell me.

analysis elementary-set-theory

share|cite|improve this question

edited Jul 14 at 14:52

Andrés E. Caicedo

63.2k7151236

asked Jul 14 at 13:20

Djx

132

Please help.

I have read the prove of the existence of the natural numbers in Analysis I written by Amann and Escher .It uses the claim that the function $mathcalV:mathbbNlongrightarrowmathbbN$, $zlongmapsto zcupz$ is injective.Here $mathbbN$ is a set we construct to be the set of natural numbers.I don't know how to prove the claim.

My English is not good.If you don't understand what I said,please tell me.

analysis elementary-set-theory

share|cite|improve this question

edited Jul 14 at 14:52

Andrés E. Caicedo

63.2k7151236

asked Jul 14 at 13:20

Djx

132

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 14:52

Andrés E. Caicedo

63.2k7151236

edited Jul 14 at 14:52

Andrés E. Caicedo

63.2k7151236

edited Jul 14 at 14:52

Andrés E. Caicedo

63.2k7151236

63.2k7151236

asked Jul 14 at 13:20

Djx

132

asked Jul 14 at 13:20

Djx

132

asked Jul 14 at 13:20

Djx

132

132

closed as off-topic by Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, user223391 Jul 14 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, user223391 Jul 14 at 22:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, Shaun, Chris Custer, Isaac Browne, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

1 Answer
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up vote
3
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accepted

The answer to the question in the title is yes, assuming the axiom of foundation.

Indeed $A$ is the only set that satisfies "for all $xin Acup A$, $xin y$ or $x=y$, and $yin Acup A$".

Indeed if $y$ satisfies it, then $y=A$ or $yin A$. But if $yin A$, then either $Ain y$ or $A=y$, which is absurd (by the axiom of foundation). Thus $y=A$

However, I believe (I'm not sure about this part) it is consistent to have $A= B$ and $B=A$, $Aneq B$ without the axiom of foundation. In such an event, clearly the answer is no.

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answered Jul 14 at 13:30

Max

10.5k1836

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is consistent. But for the OP's use you don't need that.
  – Asaf Karagila♦
  Jul 14 at 13:47
  
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

The answer to the question in the title is yes, assuming the axiom of foundation.

Indeed $A$ is the only set that satisfies "for all $xin Acup A$, $xin y$ or $x=y$, and $yin Acup A$".

Indeed if $y$ satisfies it, then $y=A$ or $yin A$. But if $yin A$, then either $Ain y$ or $A=y$, which is absurd (by the axiom of foundation). Thus $y=A$

However, I believe (I'm not sure about this part) it is consistent to have $A= B$ and $B=A$, $Aneq B$ without the axiom of foundation. In such an event, clearly the answer is no.

share|cite|improve this answer

answered Jul 14 at 13:30

Max

10.5k1836

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is consistent. But for the OP's use you don't need that.
  – Asaf Karagila♦
  Jul 14 at 13:47
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

The answer to the question in the title is yes, assuming the axiom of foundation.

Indeed $A$ is the only set that satisfies "for all $xin Acup A$, $xin y$ or $x=y$, and $yin Acup A$".

Indeed if $y$ satisfies it, then $y=A$ or $yin A$. But if $yin A$, then either $Ain y$ or $A=y$, which is absurd (by the axiom of foundation). Thus $y=A$

However, I believe (I'm not sure about this part) it is consistent to have $A= B$ and $B=A$, $Aneq B$ without the axiom of foundation. In such an event, clearly the answer is no.

share|cite|improve this answer

answered Jul 14 at 13:30

Max

10.5k1836

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is consistent. But for the OP's use you don't need that.
  – Asaf Karagila♦
  Jul 14 at 13:47
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

The answer to the question in the title is yes, assuming the axiom of foundation.

Indeed $A$ is the only set that satisfies "for all $xin Acup A$, $xin y$ or $x=y$, and $yin Acup A$".

Indeed if $y$ satisfies it, then $y=A$ or $yin A$. But if $yin A$, then either $Ain y$ or $A=y$, which is absurd (by the axiom of foundation). Thus $y=A$

However, I believe (I'm not sure about this part) it is consistent to have $A= B$ and $B=A$, $Aneq B$ without the axiom of foundation. In such an event, clearly the answer is no.

share|cite|improve this answer

answered Jul 14 at 13:30

Max

10.5k1836

The answer to the question in the title is yes, assuming the axiom of foundation.

Indeed $A$ is the only set that satisfies "for all $xin Acup A$, $xin y$ or $x=y$, and $yin Acup A$".

Indeed if $y$ satisfies it, then $y=A$ or $yin A$. But if $yin A$, then either $Ain y$ or $A=y$, which is absurd (by the axiom of foundation). Thus $y=A$

However, I believe (I'm not sure about this part) it is consistent to have $A= B$ and $B=A$, $Aneq B$ without the axiom of foundation. In such an event, clearly the answer is no.

share|cite|improve this answer

answered Jul 14 at 13:30

Max

10.5k1836

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 13:30

Max

10.5k1836

answered Jul 14 at 13:30

Max

10.5k1836

answered Jul 14 at 13:30

Max

10.5k1836

10.5k1836

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is consistent. But for the OP's use you don't need that.
  – Asaf Karagila♦
  Jul 14 at 13:47
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it is consistent. But for the OP's use you don't need that.
  – Asaf Karagila♦
  Jul 14 at 13:47
  
  

  
  
  
  

Yes, it is consistent. But for the OP's use you don't need that.
– Asaf Karagila♦
Jul 14 at 13:47

Yes, it is consistent. But for the OP's use you don't need that.
– Asaf Karagila♦
Jul 14 at 13:47

add a comment |