Very specyfic linear order not isomorphic with $langlemathbb N,lerangle$ [closed]

0

$begingroup$

Give an example of a numerable linear order such that each element has a successor, there is the smallest element, each element except the smallest one has a predecessor, but the order is not isomorphic with $langlemathbb N,lerangle$.

I understand it all, but I can't come up with anything.

discrete-mathematics logic relations

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edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

asked Jan 27 at 11:03

KarolKarol

205

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closed as off-topic by user21820, Holo, Did, TheSimpliFire, Lord_Farin Jan 27 at 17:04

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, TheSimpliFire, Lord_Farin

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

add a comment | 

0

$begingroup$

Give an example of a numerable linear order such that each element has a successor, there is the smallest element, each element except the smallest one has a predecessor, but the order is not isomorphic with $langlemathbb N,lerangle$.

I understand it all, but I can't come up with anything.

discrete-mathematics logic relations

share|cite|improve this question

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

asked Jan 27 at 11:03

KarolKarol

205

$endgroup$

closed as off-topic by user21820, Holo, Did, TheSimpliFire, Lord_Farin Jan 27 at 17:04

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, TheSimpliFire, Lord_Farin

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

add a comment | 

0

0

0

$begingroup$

Give an example of a numerable linear order such that each element has a successor, there is the smallest element, each element except the smallest one has a predecessor, but the order is not isomorphic with $langlemathbb N,lerangle$.

I understand it all, but I can't come up with anything.

discrete-mathematics logic relations

share|cite|improve this question

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

asked Jan 27 at 11:03

KarolKarol

205

$endgroup$

Give an example of a numerable linear order such that each element has a successor, there is the smallest element, each element except the smallest one has a predecessor, but the order is not isomorphic with $langlemathbb N,lerangle$.

I understand it all, but I can't come up with anything.

discrete-mathematics logic relations

discrete-mathematics logic relations

share|cite|improve this question

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

asked Jan 27 at 11:03

KarolKarol

205

share|cite|improve this question

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

asked Jan 27 at 11:03

KarolKarol

205

share|cite|improve this question

share|cite|improve this question

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

edited Jan 27 at 15:13

Andrés E. Caicedo

65.8k8160251

65.8k8160251

asked Jan 27 at 11:03

KarolKarol

205

asked Jan 27 at 11:03

KarolKarol

205

asked Jan 27 at 11:03

KarolKarol

205

205

closed as off-topic by user21820, Holo, Did, TheSimpliFire, Lord_Farin Jan 27 at 17:04

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, TheSimpliFire, Lord_Farin

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

closed as off-topic by user21820, Holo, Did, TheSimpliFire, Lord_Farin Jan 27 at 17:04

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, TheSimpliFire, Lord_Farin

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
1

active

oldest

votes

5

$begingroup$

Consider $Bbb N$ followed by $Bbb Z$, i.e., ${0}times Bbb Ncup {1}times Bbb Z$ with lexicographic order. Or if you prefer it embedded into the real line: ${,arctan nmid ninBbb N,}cup{4+arctan kmid kinBbb Z,}$.

share|cite|improve this answer

edited Jan 27 at 14:53

Jishin Noben

1072

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
  $endgroup$
  – Karol
  Jan 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
  $endgroup$
  – Holo
  Jan 27 at 12:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I wouldn't think of it myself.
  $endgroup$
  – Karol
  Jan 27 at 18:59
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

5

$begingroup$

Consider $Bbb N$ followed by $Bbb Z$, i.e., ${0}times Bbb Ncup {1}times Bbb Z$ with lexicographic order. Or if you prefer it embedded into the real line: ${,arctan nmid ninBbb N,}cup{4+arctan kmid kinBbb Z,}$.

share|cite|improve this answer

edited Jan 27 at 14:53

Jishin Noben

1072

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
  $endgroup$
  – Karol
  Jan 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
  $endgroup$
  – Holo
  Jan 27 at 12:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I wouldn't think of it myself.
  $endgroup$
  – Karol
  Jan 27 at 18:59
  

  
  
  
  

add a comment | 

5

$begingroup$

Consider $Bbb N$ followed by $Bbb Z$, i.e., ${0}times Bbb Ncup {1}times Bbb Z$ with lexicographic order. Or if you prefer it embedded into the real line: ${,arctan nmid ninBbb N,}cup{4+arctan kmid kinBbb Z,}$.

share|cite|improve this answer

edited Jan 27 at 14:53

Jishin Noben

1072

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
  $endgroup$
  – Karol
  Jan 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
  $endgroup$
  – Holo
  Jan 27 at 12:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I wouldn't think of it myself.
  $endgroup$
  – Karol
  Jan 27 at 18:59
  

  
  
  
  

add a comment | 

5

5

5

$begingroup$

Consider $Bbb N$ followed by $Bbb Z$, i.e., ${0}times Bbb Ncup {1}times Bbb Z$ with lexicographic order. Or if you prefer it embedded into the real line: ${,arctan nmid ninBbb N,}cup{4+arctan kmid kinBbb Z,}$.

share|cite|improve this answer

edited Jan 27 at 14:53

Jishin Noben

1072

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

$endgroup$

Consider $Bbb N$ followed by $Bbb Z$, i.e., ${0}times Bbb Ncup {1}times Bbb Z$ with lexicographic order. Or if you prefer it embedded into the real line: ${,arctan nmid ninBbb N,}cup{4+arctan kmid kinBbb Z,}$.

share|cite|improve this answer

edited Jan 27 at 14:53

Jishin Noben

1072

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

share|cite|improve this answer

share|cite|improve this answer

edited Jan 27 at 14:53

Jishin Noben

1072

edited Jan 27 at 14:53

Jishin Noben

1072

edited Jan 27 at 14:53

Jishin Noben

1072

1072

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

answered Jan 27 at 11:20

Hagen von EitzenHagen von Eitzen

283k23272507

283k23272507

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
  $endgroup$
  – Karol
  Jan 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
  $endgroup$
  – Holo
  Jan 27 at 12:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I wouldn't think of it myself.
  $endgroup$
  – Karol
  Jan 27 at 18:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
  $endgroup$
  – Karol
  Jan 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
  $endgroup$
  – Holo
  Jan 27 at 12:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I wouldn't think of it myself.
  $endgroup$
  – Karol
  Jan 27 at 18:59
  

  
  
  
  

$begingroup$
Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
$endgroup$
– Karol
Jan 27 at 11:26

$begingroup$
Hi, thanks for answering, I have a question, You have $Bbb Ncup {1}cup Bbb Z$. Isn't it just $Bbb Z$?
$endgroup$
– Karol
Jan 27 at 11:26

1

1

$begingroup$
@Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
$endgroup$
– Holo
Jan 27 at 12:08

$begingroup$
@Karol I think it should be $({0}×mathbb N)∪({1}×mathbb Z)$ with the Lexicographical order: $(a,a')le (b,b')$ if and only if $a<b$ or ($a=b$ and $a'≤ b'$). Where $a=0,1$ and if $a=0$ then $a'∈Bbb N$, if $a=1$ then $a'∈Bbb Z$(and the same for $b$)
$endgroup$
– Holo
Jan 27 at 12:08

$begingroup$
Thank you very much. I wouldn't think of it myself.
$endgroup$
– Karol
Jan 27 at 18:59

$begingroup$
Thank you very much. I wouldn't think of it myself.
$endgroup$
– Karol
Jan 27 at 18:59

add a comment | 

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