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5

1

$begingroup$

Perhaps a trivial question, but something I never completely understood. If we have shown that $a-b < epsilon$ for all $epsilon > 0$, then does that imply that $a-b le 0$?

I"m interested in it in the context of this question: Proving that $ f: [a,b] to BbbR $ is Riemann-integrable using an $ epsilon $-$ delta $ definition.

and in page 172 of these notes: http://math.uga.edu/~pete/2400full.pdf

real-analysis calculus epsilon-delta

share|cite|improve this question

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

asked May 19 at 20:12

nundonundo

18018

$endgroup$

add a comment | 

5

1

$begingroup$

Perhaps a trivial question, but something I never completely understood. If we have shown that $a-b < epsilon$ for all $epsilon > 0$, then does that imply that $a-b le 0$?

I"m interested in it in the context of this question: Proving that $ f: [a,b] to BbbR $ is Riemann-integrable using an $ epsilon $-$ delta $ definition.

and in page 172 of these notes: http://math.uga.edu/~pete/2400full.pdf

real-analysis calculus epsilon-delta

share|cite|improve this question

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

asked May 19 at 20:12

nundonundo

18018

$endgroup$

add a comment | 

$begingroup$

Perhaps a trivial question, but something I never completely understood. If we have shown that $a-b < epsilon$ for all $epsilon > 0$, then does that imply that $a-b le 0$?

I"m interested in it in the context of this question: Proving that $ f: [a,b] to BbbR $ is Riemann-integrable using an $ epsilon $-$ delta $ definition.

and in page 172 of these notes: http://math.uga.edu/~pete/2400full.pdf

real-analysis calculus epsilon-delta

share|cite|improve this question

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

asked May 19 at 20:12

nundonundo

18018

$endgroup$

share|cite|improve this question

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

asked May 19 at 20:12

nundonundo

18018

share|cite|improve this question

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

asked May 19 at 20:12

nundonundo

18018

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

edited May 20 at 7:48

Asaf Karagila♦

311k33445777

asked May 19 at 20:12

nundonundo

18018

asked May 19 at 20:12

nundonundo

18018

1 Answer
1

active

oldest

votes

13

$begingroup$

Yes, it does.

Suppose not. Then $a-b > 0$ and in particular we can take $epsilon = a-b$. The statement then says that $a-b < a-b$, which is absurd.

share|cite|improve this answer

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  To add to this answer, a common proof technique is to conclude that $a = 0$ from $|a| < epsilon$ for all $epsilon > 0$.
  $endgroup$
  – Charles Hudgins
  May 19 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a constructive proof of this? I.e. one that doesn't use proof by contradiction or excluded middle?
  $endgroup$
  – Michael Bächtold
  May 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can formulate the same proof with contraposition. Is that good enough?
  $endgroup$
  – EpsilonDelta
  May 20 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EpsilonDelta not sure what you mean. Btw. my question was not directly addressed at you. Maybe I should ask it as a separate question.
  $endgroup$
  – Michael Bächtold
  May 20 at 10:50
  

  
  
  
  

add a comment | 

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13

$begingroup$

Yes, it does.

Suppose not. Then $a-b > 0$ and in particular we can take $epsilon = a-b$. The statement then says that $a-b < a-b$, which is absurd.

share|cite|improve this answer

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  To add to this answer, a common proof technique is to conclude that $a = 0$ from $|a| < epsilon$ for all $epsilon > 0$.
  $endgroup$
  – Charles Hudgins
  May 19 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a constructive proof of this? I.e. one that doesn't use proof by contradiction or excluded middle?
  $endgroup$
  – Michael Bächtold
  May 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can formulate the same proof with contraposition. Is that good enough?
  $endgroup$
  – EpsilonDelta
  May 20 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EpsilonDelta not sure what you mean. Btw. my question was not directly addressed at you. Maybe I should ask it as a separate question.
  $endgroup$
  – Michael Bächtold
  May 20 at 10:50
  

  
  
  
  

add a comment | 

13

$begingroup$

Yes, it does.

Suppose not. Then $a-b > 0$ and in particular we can take $epsilon = a-b$. The statement then says that $a-b < a-b$, which is absurd.

share|cite|improve this answer

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  To add to this answer, a common proof technique is to conclude that $a = 0$ from $|a| < epsilon$ for all $epsilon > 0$.
  $endgroup$
  – Charles Hudgins
  May 19 at 20:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a constructive proof of this? I.e. one that doesn't use proof by contradiction or excluded middle?
  $endgroup$
  – Michael Bächtold
  May 20 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can formulate the same proof with contraposition. Is that good enough?
  $endgroup$
  – EpsilonDelta
  May 20 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EpsilonDelta not sure what you mean. Btw. my question was not directly addressed at you. Maybe I should ask it as a separate question.
  $endgroup$
  – Michael Bächtold
  May 20 at 10:50
  

  
  
  
  

add a comment | 

$begingroup$

Yes, it does.

Suppose not. Then $a-b > 0$ and in particular we can take $epsilon = a-b$. The statement then says that $a-b < a-b$, which is absurd.

share|cite|improve this answer

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819

$endgroup$

share|cite|improve this answer

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819

answered May 19 at 20:15

EpsilonDeltaEpsilonDelta

39819