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Definitions of supremum

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2

1

$begingroup$

This question is for me to better understand the beginning of a real analysis course.

We are provided with two definitions of supremum as follows:

Def 1 : Let $S$ be a set in $mathbbR$ be bounded above, then $m$ is called the least upper bound (supremum) if $m ge sspace, forall sin S $ and if $m'$ is some other upper bound, then $m < m'$

Def 2: Let $S$ be a set in $mathbbR$ be bounded above, then $m$ is a supremum if for some arbitrary $epsilon>0$ $exists s in S, m-epsilon < s$

I understand how both statements are true, however, would it be possible to prove the Def 2 based on Def 1?

Any hint is appreciated!

real-analysis real-numbers supremum-and-infimum

share|cite|improve this question

edited Mar 22 at 0:43

qwr

6,68242755

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ epsilon := m' - m $
  $endgroup$
  – ninjaaa
  Jan 22 '17 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how would that work? When we take the $epsilon$ we intend that it can be infinitly small, implying that $m- epsilon$ leaves no space for any m' between m and all of s in S. How would your argument using $epsilon := m'-m$ go? Would we assume $m'$ some arbitrary upper bound making $epsilon$ also arbitrary?
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 19:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry now I can see it isn't true - consider the set $ S := 1, 2 $ and $epsilon = 1/2$ .
  $endgroup$
  – ninjaaa
  Jan 22 '17 at 19:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @rannoudanames "When we take the ϵϵ we intend that it can be infinitly small," No no no. No such thing as "infinitely small" in the real numbers. Arbitrarily small is how we express it. Huge difference.
  $endgroup$
  – user4894
  Jan 22 '17 at 20:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There are no infinitely small real numbers. The idea of replacing the vague idea of "infinitely small" with the precise idea of arbitrarily small is the key breakthrough in the modern formalization of the real numbers. Note that $epsilon$ is always a positive real number.
  $endgroup$
  – user4894
  Jan 22 '17 at 21:13
  
  

  
  
  
  

 | 
show 4 more comments

2

1

$begingroup$

This question is for me to better understand the beginning of a real analysis course.

We are provided with two definitions of supremum as follows:

Def 1 : Let $S$ be a set in $mathbbR$ be bounded above, then $m$ is called the least upper bound (supremum) if $m ge sspace, forall sin S $ and if $m'$ is some other upper bound, then $m < m'$

Def 2: Let $S$ be a set in $mathbbR$ be bounded above, then $m$ is a supremum if for some arbitrary $epsilon>0$ $exists s in S, m-epsilon < s$

I understand how both statements are true, however, would it be possible to prove the Def 2 based on Def 1?

Any hint is appreciated!

real-analysis real-numbers supremum-and-infimum

share|cite|improve this question

edited Mar 22 at 0:43

qwr

6,68242755

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ epsilon := m' - m $
  $endgroup$
  – ninjaaa
  Jan 22 '17 at 19:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how would that work? When we take the $epsilon$ we intend that it can be infinitly small, implying that $m- epsilon$ leaves no space for any m' between m and all of s in S. How would your argument using $epsilon := m'-m$ go? Would we assume $m'$ some arbitrary upper bound making $epsilon$ also arbitrary?
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 19:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry now I can see it isn't true - consider the set $ S := 1, 2 $ and $epsilon = 1/2$ .
  $endgroup$
  – ninjaaa
  Jan 22 '17 at 19:59
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @rannoudanames "When we take the ϵϵ we intend that it can be infinitly small," No no no. No such thing as "infinitely small" in the real numbers. Arbitrarily small is how we express it. Huge difference.
  $endgroup$
  – user4894
  Jan 22 '17 at 20:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There are no infinitely small real numbers. The idea of replacing the vague idea of "infinitely small" with the precise idea of arbitrarily small is the key breakthrough in the modern formalization of the real numbers. Note that $epsilon$ is always a positive real number.
  $endgroup$
  – user4894
  Jan 22 '17 at 21:13
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

This question is for me to better understand the beginning of a real analysis course.

We are provided with two definitions of supremum as follows:

Def 1 : Let $S$ be a set in $mathbbR$ be bounded above, then $m$ is called the least upper bound (supremum) if $m ge sspace, forall sin S $ and if $m'$ is some other upper bound, then $m < m'$

Def 2: Let $S$ be a set in $mathbbR$ be bounded above, then $m$ is a supremum if for some arbitrary $epsilon>0$ $exists s in S, m-epsilon < s$

I understand how both statements are true, however, would it be possible to prove the Def 2 based on Def 1?

Any hint is appreciated!

real-analysis real-numbers supremum-and-infimum

share|cite|improve this question

edited Mar 22 at 0:43

qwr

6,68242755

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

$endgroup$

share|cite|improve this question

edited Mar 22 at 0:43

qwr

6,68242755

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

share|cite|improve this question

edited Mar 22 at 0:43

qwr

6,68242755

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

edited Mar 22 at 0:43

qwr

6,68242755

edited Mar 22 at 0:43

qwr

6,68242755

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

asked Jan 22 '17 at 19:39

rannoudanamesrannoudanames

565717

1 Answer
1

active

oldest

votes

1

$begingroup$

Suppose $m$ is a supremum by definition 1. Suppose $exists$ $epsilon>0$ such that $m-epsilongeq s$ $forall sin S$, then $m'=m-epsilon$ is another upper bound so it must be $m<m'=m-epsilon$ which is impossible. So we must have $forall epsilon>0$, $m-epsilon <s$ $exists sin S$.

This proves definition 2 in terms of definition 1.

share|cite|improve this answer

edited Jan 22 '17 at 19:55

answered Jan 22 '17 at 19:49

mathmamathma

577217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see you are doing a proof by contradiction, and it makes sense to me. m cannot be inferior to some number smaller than m.
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 20:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also let me add that definition 2 implies definition 1. If $m$ is a supremum by definition 2 then $mgeq s$ $forall sin S$. Now suppose $m'$ is another upper bound and $m'<m$ then $m-m'>0$. Take $0<epsilon <m-m'$ so by def 2 $exists sin S$ s.t. $m'<m-epsilon<s$ so $m'$ is not an upper bound, so we get again a contradiction.
  $endgroup$
  – mathma
  Jan 22 '17 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  makes sense! (hypothetical +1)
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 20:29
  

  
  
  
  

add a comment | 

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1

$begingroup$

Suppose $m$ is a supremum by definition 1. Suppose $exists$ $epsilon>0$ such that $m-epsilongeq s$ $forall sin S$, then $m'=m-epsilon$ is another upper bound so it must be $m<m'=m-epsilon$ which is impossible. So we must have $forall epsilon>0$, $m-epsilon <s$ $exists sin S$.

This proves definition 2 in terms of definition 1.

share|cite|improve this answer

edited Jan 22 '17 at 19:55

answered Jan 22 '17 at 19:49

mathmamathma

577217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see you are doing a proof by contradiction, and it makes sense to me. m cannot be inferior to some number smaller than m.
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 20:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also let me add that definition 2 implies definition 1. If $m$ is a supremum by definition 2 then $mgeq s$ $forall sin S$. Now suppose $m'$ is another upper bound and $m'<m$ then $m-m'>0$. Take $0<epsilon <m-m'$ so by def 2 $exists sin S$ s.t. $m'<m-epsilon<s$ so $m'$ is not an upper bound, so we get again a contradiction.
  $endgroup$
  – mathma
  Jan 22 '17 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  makes sense! (hypothetical +1)
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 20:29
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose $m$ is a supremum by definition 1. Suppose $exists$ $epsilon>0$ such that $m-epsilongeq s$ $forall sin S$, then $m'=m-epsilon$ is another upper bound so it must be $m<m'=m-epsilon$ which is impossible. So we must have $forall epsilon>0$, $m-epsilon <s$ $exists sin S$.

This proves definition 2 in terms of definition 1.

share|cite|improve this answer

edited Jan 22 '17 at 19:55

answered Jan 22 '17 at 19:49

mathmamathma

577217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see you are doing a proof by contradiction, and it makes sense to me. m cannot be inferior to some number smaller than m.
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 20:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also let me add that definition 2 implies definition 1. If $m$ is a supremum by definition 2 then $mgeq s$ $forall sin S$. Now suppose $m'$ is another upper bound and $m'<m$ then $m-m'>0$. Take $0<epsilon <m-m'$ so by def 2 $exists sin S$ s.t. $m'<m-epsilon<s$ so $m'$ is not an upper bound, so we get again a contradiction.
  $endgroup$
  – mathma
  Jan 22 '17 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  makes sense! (hypothetical +1)
  $endgroup$
  – rannoudanames
  Jan 22 '17 at 20:29
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $m$ is a supremum by definition 1. Suppose $exists$ $epsilon>0$ such that $m-epsilongeq s$ $forall sin S$, then $m'=m-epsilon$ is another upper bound so it must be $m<m'=m-epsilon$ which is impossible. So we must have $forall epsilon>0$, $m-epsilon <s$ $exists sin S$.

This proves definition 2 in terms of definition 1.

share|cite|improve this answer

edited Jan 22 '17 at 19:55

answered Jan 22 '17 at 19:49

mathmamathma

577217

$endgroup$

share|cite|improve this answer

edited Jan 22 '17 at 19:55

answered Jan 22 '17 at 19:49

mathmamathma

577217

answered Jan 22 '17 at 19:49

mathmamathma

577217

answered Jan 22 '17 at 19:49

mathmamathma

577217