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Let $emptyset ≠A⊆F$. $exists max(A) implies exists sup(A)$ and $sup(A)=max(A)$.

1

$begingroup$

Let $∅≠A⊆F$. Prove that if $max(A)$ exists, then $sup(A)$ exists and $sup(A)=max(A)$.

How do I prove such statement?

real-analysis supremum-and-infimum

share|cite|improve this question

edited Mar 31 at 14:16

Felipe Duda

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you checked what the definition of $max(A)$ is?
  $endgroup$
  – blub
  Mar 31 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, my problem is not with the concept itself, but with the proof.
  $endgroup$
  – Felipe Duda
  Mar 31 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, what is your $F$?
  $endgroup$
  – blub
  Mar 31 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  F is a general ordered field.
  $endgroup$
  – Felipe Duda
  Mar 31 at 13:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say $emptyset = A$, twice. Surely that should be $emptysetne A$???
  $endgroup$
  – David C. Ullrich
  Mar 31 at 14:13
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

Let $∅≠A⊆F$. Prove that if $max(A)$ exists, then $sup(A)$ exists and $sup(A)=max(A)$.

How do I prove such statement?

real-analysis supremum-and-infimum

share|cite|improve this question

edited Mar 31 at 14:16

Felipe Duda

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you checked what the definition of $max(A)$ is?
  $endgroup$
  – blub
  Mar 31 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, my problem is not with the concept itself, but with the proof.
  $endgroup$
  – Felipe Duda
  Mar 31 at 13:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, what is your $F$?
  $endgroup$
  – blub
  Mar 31 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  F is a general ordered field.
  $endgroup$
  – Felipe Duda
  Mar 31 at 13:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say $emptyset = A$, twice. Surely that should be $emptysetne A$???
  $endgroup$
  – David C. Ullrich
  Mar 31 at 14:13
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Let $∅≠A⊆F$. Prove that if $max(A)$ exists, then $sup(A)$ exists and $sup(A)=max(A)$.

How do I prove such statement?

real-analysis supremum-and-infimum

share|cite|improve this question

edited Mar 31 at 14:16

Felipe Duda

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

$endgroup$

share|cite|improve this question

edited Mar 31 at 14:16

Felipe Duda

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

share|cite|improve this question

edited Mar 31 at 14:16

Felipe Duda

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

asked Mar 31 at 13:42

Felipe DudaFelipe Duda

62

2 Answers
2

active

oldest

votes

0

$begingroup$

The supremum must exist, and in this case having a maximum implies the set is bounded above, so the supremum is finite. Suppose by contradiction that the max and sup are different. Then it must be the case that $sup(A)>max(A)$ since $max(A)in A$. Can you derive a contradiction?

share|cite|improve this answer

answered Mar 31 at 14:25

DaveDave

9,22211033

$endgroup$

add a comment | 

0

$begingroup$

This is straightforward enough, if you recall the definitions of a supremum of a partially ordered set and the maximum. The condition that $F$ is partially ordered is important in these definitions. Specifically:

Maximal(Maximum) : Let $(F,leq)$ be a partially ordered set and $ Asubseteq F$. Then $min A$ is a maximal element of $A$ if $forall ain A$ it is : $m leq a implies m = a$.

Supremum : An upper bound $b$ of a partially ordered set $(F,leq)$ will be called a supremum of $A$ if for all upper bounds $u_b$ of $A$ in $F$, $u_b geq b$.

Now, since $max(A)$ exists, that means that there exists $m in A$ such that for all $a in A$ it is $$ m leq a implies m=a qquad (1)$$

Suppose it is $max(A) neq sup(A)$. Since $max(A) in A$ and for all upper bounds $u_b$ of $A$ we have $u_b geq b$, it should be $sup(A) > max(A)$. But we have arrived at a contradiction, since $(1)$ holds. Thus $max(A) = sup(A)$.

share|cite|improve this answer

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250

$endgroup$

add a comment | 

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0

$begingroup$

The supremum must exist, and in this case having a maximum implies the set is bounded above, so the supremum is finite. Suppose by contradiction that the max and sup are different. Then it must be the case that $sup(A)>max(A)$ since $max(A)in A$. Can you derive a contradiction?

share|cite|improve this answer

answered Mar 31 at 14:25

DaveDave

9,22211033

$endgroup$

add a comment | 

0

$begingroup$

The supremum must exist, and in this case having a maximum implies the set is bounded above, so the supremum is finite. Suppose by contradiction that the max and sup are different. Then it must be the case that $sup(A)>max(A)$ since $max(A)in A$. Can you derive a contradiction?

share|cite|improve this answer

answered Mar 31 at 14:25

DaveDave

9,22211033

$endgroup$

add a comment | 

$begingroup$

The supremum must exist, and in this case having a maximum implies the set is bounded above, so the supremum is finite. Suppose by contradiction that the max and sup are different. Then it must be the case that $sup(A)>max(A)$ since $max(A)in A$. Can you derive a contradiction?

share|cite|improve this answer

answered Mar 31 at 14:25

DaveDave

9,22211033

$endgroup$

share|cite|improve this answer

answered Mar 31 at 14:25

DaveDave

9,22211033

answered Mar 31 at 14:25

DaveDave

9,22211033

answered Mar 31 at 14:25

DaveDave

9,22211033

0

$begingroup$

This is straightforward enough, if you recall the definitions of a supremum of a partially ordered set and the maximum. The condition that $F$ is partially ordered is important in these definitions. Specifically:

Maximal(Maximum) : Let $(F,leq)$ be a partially ordered set and $ Asubseteq F$. Then $min A$ is a maximal element of $A$ if $forall ain A$ it is : $m leq a implies m = a$.

Supremum : An upper bound $b$ of a partially ordered set $(F,leq)$ will be called a supremum of $A$ if for all upper bounds $u_b$ of $A$ in $F$, $u_b geq b$.

Now, since $max(A)$ exists, that means that there exists $m in A$ such that for all $a in A$ it is $$ m leq a implies m=a qquad (1)$$

Suppose it is $max(A) neq sup(A)$. Since $max(A) in A$ and for all upper bounds $u_b$ of $A$ we have $u_b geq b$, it should be $sup(A) > max(A)$. But we have arrived at a contradiction, since $(1)$ holds. Thus $max(A) = sup(A)$.

share|cite|improve this answer

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250

$endgroup$

add a comment | 

0

$begingroup$

This is straightforward enough, if you recall the definitions of a supremum of a partially ordered set and the maximum. The condition that $F$ is partially ordered is important in these definitions. Specifically:

Maximal(Maximum) : Let $(F,leq)$ be a partially ordered set and $ Asubseteq F$. Then $min A$ is a maximal element of $A$ if $forall ain A$ it is : $m leq a implies m = a$.

Supremum : An upper bound $b$ of a partially ordered set $(F,leq)$ will be called a supremum of $A$ if for all upper bounds $u_b$ of $A$ in $F$, $u_b geq b$.

Now, since $max(A)$ exists, that means that there exists $m in A$ such that for all $a in A$ it is $$ m leq a implies m=a qquad (1)$$

Suppose it is $max(A) neq sup(A)$. Since $max(A) in A$ and for all upper bounds $u_b$ of $A$ we have $u_b geq b$, it should be $sup(A) > max(A)$. But we have arrived at a contradiction, since $(1)$ holds. Thus $max(A) = sup(A)$.

share|cite|improve this answer

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250

$endgroup$

add a comment | 

$begingroup$

This is straightforward enough, if you recall the definitions of a supremum of a partially ordered set and the maximum. The condition that $F$ is partially ordered is important in these definitions. Specifically:

Maximal(Maximum) : Let $(F,leq)$ be a partially ordered set and $ Asubseteq F$. Then $min A$ is a maximal element of $A$ if $forall ain A$ it is : $m leq a implies m = a$.

Supremum : An upper bound $b$ of a partially ordered set $(F,leq)$ will be called a supremum of $A$ if for all upper bounds $u_b$ of $A$ in $F$, $u_b geq b$.

Now, since $max(A)$ exists, that means that there exists $m in A$ such that for all $a in A$ it is $$ m leq a implies m=a qquad (1)$$

Suppose it is $max(A) neq sup(A)$. Since $max(A) in A$ and for all upper bounds $u_b$ of $A$ we have $u_b geq b$, it should be $sup(A) > max(A)$. But we have arrived at a contradiction, since $(1)$ holds. Thus $max(A) = sup(A)$.

share|cite|improve this answer

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250

$endgroup$

share|cite|improve this answer

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250

answered Mar 31 at 14:30

RebellosRebellos

15.7k31250