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Using Archimedian property to prove that infimum of set is $0$

Proof that the infimum of a given set $E$ is uniqueQuick question about Archimedian propertyFinding the infimum of a set of real numbersInfimum of $S = fracx2^y $ using archimedean propertyRigorously proving infimum of a set.Approximation Property for InfimumIf $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.Proof of least upper bound and greatest lower bound property of an ordered setProving an infimum using the archimedean property of RTo prove that $ cap_n=1^infty (0 , frac1n) = emptyset $

1

$begingroup$

Given set A = [ $frac1n | n in mathbbN$]

It seems to me i have to prove two things :

1. $frac1n geq 0 , forall n in mathbbN$

2, Assuming $b$ be another lower bound , and so $b leq 0$

Work :

1 Now $frac1n geq 0 , forall n in N$

Also $n geq 1$

So $frac1n leq 1$ . But $frac1n in R^+$. So, $0 < frac1n leq
1$

2. Assume that $b > 0$
   

so by Archimedian property $exists n in mathbbN$
such that $frac1n < b$ which is contradiction to the fact that b is lower bound for the set

Is this correct ?

real-analysis proof-verification self-learning

share|cite|improve this question

edited Mar 22 at 9:04

J. Deff

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

$endgroup$

add a comment | 

1

$begingroup$

Given set A = [ $frac1n | n in mathbbN$]

It seems to me i have to prove two things :

1. $frac1n geq 0 , forall n in mathbbN$

2, Assuming $b$ be another lower bound , and so $b leq 0$

Work :

1 Now $frac1n geq 0 , forall n in N$

Also $n geq 1$

So $frac1n leq 1$ . But $frac1n in R^+$. So, $0 < frac1n leq
1$

2. Assume that $b > 0$
   

so by Archimedian property $exists n in mathbbN$
such that $frac1n < b$ which is contradiction to the fact that b is lower bound for the set

Is this correct ?

real-analysis proof-verification self-learning

share|cite|improve this question

edited Mar 22 at 9:04

J. Deff

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

$endgroup$

add a comment | 

$begingroup$

Given set A = [ $frac1n | n in mathbbN$]

It seems to me i have to prove two things :

1. $frac1n geq 0 , forall n in mathbbN$

2, Assuming $b$ be another lower bound , and so $b leq 0$

Work :

1 Now $frac1n geq 0 , forall n in N$

Also $n geq 1$

So $frac1n leq 1$ . But $frac1n in R^+$. So, $0 < frac1n leq
1$

2. Assume that $b > 0$
   

so by Archimedian property $exists n in mathbbN$
such that $frac1n < b$ which is contradiction to the fact that b is lower bound for the set

Is this correct ?

real-analysis proof-verification self-learning

share|cite|improve this question

edited Mar 22 at 9:04

J. Deff

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

$endgroup$

share|cite|improve this question

edited Mar 22 at 9:04

J. Deff

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

share|cite|improve this question

edited Mar 22 at 9:04

J. Deff

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

asked Mar 22 at 8:04

J. DeffJ. Deff

717519

2 Answers
2

active

oldest

votes

1

$begingroup$

$0$ is a lower bound, $ 1/n >0$, $n in mathbbZ^+$.

Assume $b >0$, real, is a lower bound.

Archimedean principle:

There is a $n_0 in mathbbZ^+$ s.t.

$n_0 >1/b$, i.e.

$b > 1/n_0$, a contradiction(Why?).

Hence $inf (A)=0$.

share|cite|improve this answer

edited Mar 22 at 8:43

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please check my attempt
  $endgroup$
  – J. Deff
  Mar 22 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  J.Deff.I 1/n >0, 0 is a lower bound.Ok?To show that 0 is the greatest lower bound you assume a lower bound b that is bigger. 0< 1/n, and assuming it is not the greatesower bound you pick b >0 , and contradict.You choose b<0, a smaller lower bound. b <0 <1/n automatically.Let me know.
  $endgroup$
  – Peter Szilas
  Mar 22 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check my proof now please and let me know
  $endgroup$
  – J. Deff
  Mar 22 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Deff.You got it. I would put in an extra line, Archimedean principle: There is a n >1/b ( real).Then b >1/n, i.e. a little more explicit on the Arch.principle.
  $endgroup$
  – Peter Szilas
  Mar 22 at 9:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that 0 is a lower bound of $frac1n$
Archimedian property- if $x > 0 $ then there exists $n epsilon mathbbR $ such that $nx > y; x,y epsilon mathbbR$. Set $x = epsilon$ and $y=1$ $$ epsilon > frac1n forall epsilon > 0$$ . Hence infimum of $frac1n $ is $0$ .

share|cite|improve this answer

edited Mar 22 at 8:42

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont understand this version of archimedian property.I have following version which is :...Given there is any rela number b then there exists some natural number n such that $n > b$
  $endgroup$
  – J. Deff
  Mar 22 at 8:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just search archimedian property for real nos. on Google. You'll get this version of archimedian property.
  $endgroup$
  – Prakhar Neema
  Mar 22 at 8:45
  

  
  
  
  

add a comment | 

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1

$begingroup$

$0$ is a lower bound, $ 1/n >0$, $n in mathbbZ^+$.

Assume $b >0$, real, is a lower bound.

Archimedean principle:

There is a $n_0 in mathbbZ^+$ s.t.

$n_0 >1/b$, i.e.

$b > 1/n_0$, a contradiction(Why?).

Hence $inf (A)=0$.

share|cite|improve this answer

edited Mar 22 at 8:43

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please check my attempt
  $endgroup$
  – J. Deff
  Mar 22 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  J.Deff.I 1/n >0, 0 is a lower bound.Ok?To show that 0 is the greatest lower bound you assume a lower bound b that is bigger. 0< 1/n, and assuming it is not the greatesower bound you pick b >0 , and contradict.You choose b<0, a smaller lower bound. b <0 <1/n automatically.Let me know.
  $endgroup$
  – Peter Szilas
  Mar 22 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check my proof now please and let me know
  $endgroup$
  – J. Deff
  Mar 22 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Deff.You got it. I would put in an extra line, Archimedean principle: There is a n >1/b ( real).Then b >1/n, i.e. a little more explicit on the Arch.principle.
  $endgroup$
  – Peter Szilas
  Mar 22 at 9:16
  

  
  
  
  

add a comment | 

1

$begingroup$

$0$ is a lower bound, $ 1/n >0$, $n in mathbbZ^+$.

Assume $b >0$, real, is a lower bound.

Archimedean principle:

There is a $n_0 in mathbbZ^+$ s.t.

$n_0 >1/b$, i.e.

$b > 1/n_0$, a contradiction(Why?).

Hence $inf (A)=0$.

share|cite|improve this answer

edited Mar 22 at 8:43

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please check my attempt
  $endgroup$
  – J. Deff
  Mar 22 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  J.Deff.I 1/n >0, 0 is a lower bound.Ok?To show that 0 is the greatest lower bound you assume a lower bound b that is bigger. 0< 1/n, and assuming it is not the greatesower bound you pick b >0 , and contradict.You choose b<0, a smaller lower bound. b <0 <1/n automatically.Let me know.
  $endgroup$
  – Peter Szilas
  Mar 22 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check my proof now please and let me know
  $endgroup$
  – J. Deff
  Mar 22 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Deff.You got it. I would put in an extra line, Archimedean principle: There is a n >1/b ( real).Then b >1/n, i.e. a little more explicit on the Arch.principle.
  $endgroup$
  – Peter Szilas
  Mar 22 at 9:16
  

  
  
  
  

add a comment | 

$begingroup$

$0$ is a lower bound, $ 1/n >0$, $n in mathbbZ^+$.

Assume $b >0$, real, is a lower bound.

Archimedean principle:

There is a $n_0 in mathbbZ^+$ s.t.

$n_0 >1/b$, i.e.

$b > 1/n_0$, a contradiction(Why?).

Hence $inf (A)=0$.

share|cite|improve this answer

edited Mar 22 at 8:43

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

$endgroup$

share|cite|improve this answer

edited Mar 22 at 8:43

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

answered Mar 22 at 8:37

Peter SzilasPeter Szilas

11.8k2822

1

$begingroup$

Observe that 0 is a lower bound of $frac1n$
Archimedian property- if $x > 0 $ then there exists $n epsilon mathbbR $ such that $nx > y; x,y epsilon mathbbR$. Set $x = epsilon$ and $y=1$ $$ epsilon > frac1n forall epsilon > 0$$ . Hence infimum of $frac1n $ is $0$ .

share|cite|improve this answer

edited Mar 22 at 8:42

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont understand this version of archimedian property.I have following version which is :...Given there is any rela number b then there exists some natural number n such that $n > b$
  $endgroup$
  – J. Deff
  Mar 22 at 8:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just search archimedian property for real nos. on Google. You'll get this version of archimedian property.
  $endgroup$
  – Prakhar Neema
  Mar 22 at 8:45
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that 0 is a lower bound of $frac1n$
Archimedian property- if $x > 0 $ then there exists $n epsilon mathbbR $ such that $nx > y; x,y epsilon mathbbR$. Set $x = epsilon$ and $y=1$ $$ epsilon > frac1n forall epsilon > 0$$ . Hence infimum of $frac1n $ is $0$ .

share|cite|improve this answer

edited Mar 22 at 8:42

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont understand this version of archimedian property.I have following version which is :...Given there is any rela number b then there exists some natural number n such that $n > b$
  $endgroup$
  – J. Deff
  Mar 22 at 8:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just search archimedian property for real nos. on Google. You'll get this version of archimedian property.
  $endgroup$
  – Prakhar Neema
  Mar 22 at 8:45
  

  
  
  
  

add a comment | 

$begingroup$

Observe that 0 is a lower bound of $frac1n$
Archimedian property- if $x > 0 $ then there exists $n epsilon mathbbR $ such that $nx > y; x,y epsilon mathbbR$. Set $x = epsilon$ and $y=1$ $$ epsilon > frac1n forall epsilon > 0$$ . Hence infimum of $frac1n $ is $0$ .

share|cite|improve this answer

edited Mar 22 at 8:42

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174

$endgroup$

share|cite|improve this answer

edited Mar 22 at 8:42

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174

answered Mar 22 at 8:21

Prakhar NeemaPrakhar Neema

1174