Is the structure $(mathbb{R}, +, *)$ rigid? [duplicate]

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  Is an automorphism of the field of real numbers the identity map?
  
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Does the structure $(mathbb{R}, + ,*)$ have only the trivial automorphism?

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asked Jan 30 at 17:34

user107952user107952

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marked as duplicate by stressed out, Mike Earnest, Wojowu, verret, Gibbs Jan 31 at 1:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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0

$begingroup$

This question already has an answer here:

• 
  Is an automorphism of the field of real numbers the identity map?
  
  6 answers
  
  

Does the structure $(mathbb{R}, + ,*)$ have only the trivial automorphism?

model-theory

share|cite|improve this question

asked Jan 30 at 17:34

user107952user107952

2,71931032

$endgroup$

marked as duplicate by stressed out, Mike Earnest, Wojowu, verret, Gibbs Jan 31 at 1:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Is an automorphism of the field of real numbers the identity map?
  
  6 answers
  
  

Does the structure $(mathbb{R}, + ,*)$ have only the trivial automorphism?

model-theory

share|cite|improve this question

asked Jan 30 at 17:34

user107952user107952

2,71931032

$endgroup$

This question already has an answer here:

• 
  Is an automorphism of the field of real numbers the identity map?
  
  6 answers
  
  

Does the structure $(mathbb{R}, + ,*)$ have only the trivial automorphism?

This question already has an answer here:

• 
  Is an automorphism of the field of real numbers the identity map?
  
  6 answers
  
  

model-theory

model-theory

share|cite|improve this question

asked Jan 30 at 17:34

user107952user107952

2,71931032

share|cite|improve this question

asked Jan 30 at 17:34

user107952user107952

2,71931032

share|cite|improve this question

share|cite|improve this question

asked Jan 30 at 17:34

user107952user107952

2,71931032

asked Jan 30 at 17:34

user107952user107952

2,71931032

asked Jan 30 at 17:34

user107952user107952

2,71931032

2,71931032

marked as duplicate by stressed out, Mike Earnest, Wojowu, verret, Gibbs Jan 31 at 1:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by stressed out, Mike Earnest, Wojowu, verret, Gibbs Jan 31 at 1:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

add a comment | 

1 Answer
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Suppose $phiintext{Aut}(mathbb{R})$.
$$phi(1)^2 = phi(1) implies phi(1) = 1 $$
since $phi$ is an automorphism. It's easy to see that for $xinmathbb{Q}$, $phi(x) = x.$ For $x> 0$
$$phi(x) = phi(sqrt{x})^2> 0. $$
So $$ x>y implies phi(x)>phi(y).$$
If $x$ is irrational, there exists $2$ sequences, one increasing $x_n$, and one decreasing $y_n$, both are sequences of rational numbers, and converge to $x$.
$$y_n = phi(y_n)geq phi(x)geq phi(x_n) = x_n $$
By taking limit, $phi(x) =x$ for all $x$.

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answered Jan 30 at 17:46

JakobianJakobian

2,733721

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout There's no continuity involved
  $endgroup$
  – Jakobian
  Jan 30 at 17:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
  $endgroup$
  – Wojowu
  Jan 30 at 17:49
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Suppose $phiintext{Aut}(mathbb{R})$.
$$phi(1)^2 = phi(1) implies phi(1) = 1 $$
since $phi$ is an automorphism. It's easy to see that for $xinmathbb{Q}$, $phi(x) = x.$ For $x> 0$
$$phi(x) = phi(sqrt{x})^2> 0. $$
So $$ x>y implies phi(x)>phi(y).$$
If $x$ is irrational, there exists $2$ sequences, one increasing $x_n$, and one decreasing $y_n$, both are sequences of rational numbers, and converge to $x$.
$$y_n = phi(y_n)geq phi(x)geq phi(x_n) = x_n $$
By taking limit, $phi(x) =x$ for all $x$.

share|cite|improve this answer

answered Jan 30 at 17:46

JakobianJakobian

2,733721

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout There's no continuity involved
  $endgroup$
  – Jakobian
  Jan 30 at 17:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
  $endgroup$
  – Wojowu
  Jan 30 at 17:49
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose $phiintext{Aut}(mathbb{R})$.
$$phi(1)^2 = phi(1) implies phi(1) = 1 $$
since $phi$ is an automorphism. It's easy to see that for $xinmathbb{Q}$, $phi(x) = x.$ For $x> 0$
$$phi(x) = phi(sqrt{x})^2> 0. $$
So $$ x>y implies phi(x)>phi(y).$$
If $x$ is irrational, there exists $2$ sequences, one increasing $x_n$, and one decreasing $y_n$, both are sequences of rational numbers, and converge to $x$.
$$y_n = phi(y_n)geq phi(x)geq phi(x_n) = x_n $$
By taking limit, $phi(x) =x$ for all $x$.

share|cite|improve this answer

answered Jan 30 at 17:46

JakobianJakobian

2,733721

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout There's no continuity involved
  $endgroup$
  – Jakobian
  Jan 30 at 17:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
  $endgroup$
  – Wojowu
  Jan 30 at 17:49
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Suppose $phiintext{Aut}(mathbb{R})$.
$$phi(1)^2 = phi(1) implies phi(1) = 1 $$
since $phi$ is an automorphism. It's easy to see that for $xinmathbb{Q}$, $phi(x) = x.$ For $x> 0$
$$phi(x) = phi(sqrt{x})^2> 0. $$
So $$ x>y implies phi(x)>phi(y).$$
If $x$ is irrational, there exists $2$ sequences, one increasing $x_n$, and one decreasing $y_n$, both are sequences of rational numbers, and converge to $x$.
$$y_n = phi(y_n)geq phi(x)geq phi(x_n) = x_n $$
By taking limit, $phi(x) =x$ for all $x$.

share|cite|improve this answer

answered Jan 30 at 17:46

JakobianJakobian

2,733721

$endgroup$

Suppose $phiintext{Aut}(mathbb{R})$.
$$phi(1)^2 = phi(1) implies phi(1) = 1 $$
since $phi$ is an automorphism. It's easy to see that for $xinmathbb{Q}$, $phi(x) = x.$ For $x> 0$
$$phi(x) = phi(sqrt{x})^2> 0. $$
So $$ x>y implies phi(x)>phi(y).$$
If $x$ is irrational, there exists $2$ sequences, one increasing $x_n$, and one decreasing $y_n$, both are sequences of rational numbers, and converge to $x$.
$$y_n = phi(y_n)geq phi(x)geq phi(x_n) = x_n $$
By taking limit, $phi(x) =x$ for all $x$.

share|cite|improve this answer

answered Jan 30 at 17:46

JakobianJakobian

2,733721

share|cite|improve this answer

share|cite|improve this answer

answered Jan 30 at 17:46

JakobianJakobian

2,733721

answered Jan 30 at 17:46

JakobianJakobian

2,733721

answered Jan 30 at 17:46

JakobianJakobian

2,733721

2,733721

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout There's no continuity involved
  $endgroup$
  – Jakobian
  Jan 30 at 17:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
  $endgroup$
  – Wojowu
  Jan 30 at 17:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout There's no continuity involved
  $endgroup$
  – Jakobian
  Jan 30 at 17:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
  $endgroup$
  – Wojowu
  Jan 30 at 17:49
  

  
  
  
  

1

1

$begingroup$
@stressedout There's no continuity involved
$endgroup$
– Jakobian
Jan 30 at 17:49

$begingroup$
@stressedout There's no continuity involved
$endgroup$
– Jakobian
Jan 30 at 17:49

$begingroup$
@stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
$endgroup$
– Wojowu
Jan 30 at 17:49

$begingroup$
@stressedout You only apply limits to $y_ngeqphi(x)geq x_n$.
$endgroup$
– Wojowu
Jan 30 at 17:49

add a comment | 

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