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For every $epsilongt 0$, $|a-b|

I need to prove the continuity of $f(x)=log x$ using a $epsilon-delta$ proofIf a is an element of R, such that $0leq a<epsilon$ for every $epsilon>0$,then a = 0If $P(A_n) ge epsilon>0$ for large $n$, then $P(A_n i.o.) ge epsilon$Proving: if $|a|<epsilon forall epsilon>0$ then $a=0$ using a direct proofHelp with composing an $epsilon-N$ proof of mineEpsilon-Delta Proof StructureProve every convergent sequence is Cauchy.Show that for every $epsilon > 0$ there exists an $M>0$ such that $|f(x)−f(y)|<epsilon$Quick epsilon-delta questionReal Analysis: Proving Continuity ($epsilon$ $delta$ argument)

0

$begingroup$

I have done a proof by myself but not sure about it

proof: $|b-a|<epsilon $
=$a-epsilon $

real-analysis calculus measure-theory proof-writing

share|cite|improve this question

edited Mar 18 at 15:06

Somos

14.7k11337

asked Mar 18 at 12:43

nancynancy

165

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try epsilon or varepsilon to get $epsilon$ or $varepsilon$ respectively.
  $endgroup$
  – Randall
  Mar 18 at 12:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use dollar signs to wrap up math, like $epsilongt 0$ for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.
  $endgroup$
  – learner
  Mar 18 at 12:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
  $endgroup$
  – learner
  Mar 18 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for telling how to write epsilon .it worked
  $endgroup$
  – nancy
  Mar 18 at 13:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  learner will you please elaborate it more that how to choose suitable $ epsilon $
  $endgroup$
  – nancy
  Mar 18 at 13:10
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I have done a proof by myself but not sure about it

proof: $|b-a|<epsilon $
=$a-epsilon $

real-analysis calculus measure-theory proof-writing

share|cite|improve this question

edited Mar 18 at 15:06

Somos

14.7k11337

asked Mar 18 at 12:43

nancynancy

165

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try epsilon or varepsilon to get $epsilon$ or $varepsilon$ respectively.
  $endgroup$
  – Randall
  Mar 18 at 12:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use dollar signs to wrap up math, like $epsilongt 0$ for $epsilongt 0$. Read this for an introduction on how to format your content properly on this website. You can also view how others type math on this site by right clicking on the math -> show math as -> TeX commands.
  $endgroup$
  – learner
  Mar 18 at 12:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here's a hint to get you started: suppose $|a-b|=kgt 0$, derive a contradiction by choosing suitable $epsilon$; hence conclude that $|a-b|=0iff a-b=0iff a=b$
  $endgroup$
  – learner
  Mar 18 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for telling how to write epsilon .it worked
  $endgroup$
  – nancy
  Mar 18 at 13:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  learner will you please elaborate it more that how to choose suitable $ epsilon $
  $endgroup$
  – nancy
  Mar 18 at 13:10
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I have done a proof by myself but not sure about it

proof: $|b-a|<epsilon $
=$a-epsilon $

real-analysis calculus measure-theory proof-writing

share|cite|improve this question

edited Mar 18 at 15:06

Somos

14.7k11337

asked Mar 18 at 12:43

nancynancy

165

$endgroup$

share|cite|improve this question

edited Mar 18 at 15:06

Somos

14.7k11337

asked Mar 18 at 12:43

nancynancy

165

share|cite|improve this question

edited Mar 18 at 15:06

Somos

14.7k11337

asked Mar 18 at 12:43

nancynancy

165

edited Mar 18 at 15:06

Somos

14.7k11337

edited Mar 18 at 15:06

Somos

14.7k11337

asked Mar 18 at 12:43

nancynancy

165

asked Mar 18 at 12:43

nancynancy

165

1 Answer
1

active

oldest

votes

5

$begingroup$

We have $forall epsilon > 0,quad |a-b| < epsilon$

If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac{|a-b|}{2} > 0$

Then $|a-b| < dfrac{|a-b|}{2} implies 1 < dfrac{1}{2}$, contradiction!

Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$

share|cite|improve this answer

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanku for such quick answer @LAGRIDA
  $endgroup$
  – nancy
  Mar 18 at 13:58
  

  
  
  
  

add a comment | 

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5

$begingroup$

We have $forall epsilon > 0,quad |a-b| < epsilon$

If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac{|a-b|}{2} > 0$

Then $|a-b| < dfrac{|a-b|}{2} implies 1 < dfrac{1}{2}$, contradiction!

Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$

share|cite|improve this answer

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanku for such quick answer @LAGRIDA
  $endgroup$
  – nancy
  Mar 18 at 13:58
  

  
  
  
  

add a comment | 

5

$begingroup$

We have $forall epsilon > 0,quad |a-b| < epsilon$

If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac{|a-b|}{2} > 0$

Then $|a-b| < dfrac{|a-b|}{2} implies 1 < dfrac{1}{2}$, contradiction!

Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$

share|cite|improve this answer

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanku for such quick answer @LAGRIDA
  $endgroup$
  – nancy
  Mar 18 at 13:58
  

  
  
  
  

add a comment | 

$begingroup$

We have $forall epsilon > 0,quad |a-b| < epsilon$

If we suppose that $a neq b$ then $|a-b| neq 0$, we choose $epsilon = dfrac{|a-b|}{2} > 0$

Then $|a-b| < dfrac{|a-b|}{2} implies 1 < dfrac{1}{2}$, contradiction!

Conclusion : $forall epsilon > 0,quad |a-b| < epsilon implies a =b$

share|cite|improve this answer

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111

$endgroup$

share|cite|improve this answer

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111

answered Mar 18 at 13:37

LAGRIDALAGRIDA

320111