Need of $1$ in this proof?

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What is the need of a $1$ inside the $mathbbmax$ in this proof?

real-analysis proof-explanation

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asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is really important to ask these questions by the way. Over time, these details will make more and more sense.
  – Elliot G
  Aug 6 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ElliotG, Thanks for encouragement. I was really concerned that asking these minute details would evoke criticism.
  – Aditya Agarwal
  Aug 6 at 18:33
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

What is the need of a $1$ inside the $mathbbmax$ in this proof?

real-analysis proof-explanation

share|cite|improve this question

asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is really important to ask these questions by the way. Over time, these details will make more and more sense.
  – Elliot G
  Aug 6 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ElliotG, Thanks for encouragement. I was really concerned that asking these minute details would evoke criticism.
  – Aditya Agarwal
  Aug 6 at 18:33
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

What is the need of a $1$ inside the $mathbbmax$ in this proof?

real-analysis proof-explanation

share|cite|improve this question

asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

What is the need of a $1$ inside the $mathbbmax$ in this proof?

real-analysis proof-explanation

share|cite|improve this question

asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

asked Aug 6 at 18:22

Aditya Agarwal

2,93111536

2,93111536

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is really important to ask these questions by the way. Over time, these details will make more and more sense.
  – Elliot G
  Aug 6 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ElliotG, Thanks for encouragement. I was really concerned that asking these minute details would evoke criticism.
  – Aditya Agarwal
  Aug 6 at 18:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It is really important to ask these questions by the way. Over time, these details will make more and more sense.
  – Elliot G
  Aug 6 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ElliotG, Thanks for encouragement. I was really concerned that asking these minute details would evoke criticism.
  – Aditya Agarwal
  Aug 6 at 18:33
  

  
  
  
  

2

2

It is really important to ask these questions by the way. Over time, these details will make more and more sense.
– Elliot G
Aug 6 at 18:31

It is really important to ask these questions by the way. Over time, these details will make more and more sense.
– Elliot G
Aug 6 at 18:31

@ElliotG, Thanks for encouragement. I was really concerned that asking these minute details would evoke criticism.
– Aditya Agarwal
Aug 6 at 18:33

@ElliotG, Thanks for encouragement. I was really concerned that asking these minute details would evoke criticism.
– Aditya Agarwal
Aug 6 at 18:33

add a comment | 

4 Answers
4

active

oldest

votes

up vote
2
down vote



accepted

I don't think it's actually needed. The only thing I could of is to ensure that no division by zero takes place, but the condition $-m'<a(n)<M'$ means that not both $m'$ and $M'$ can be zero, so it's actually unnecessary.

share|cite|improve this answer

answered Aug 6 at 18:36

saulspatz

10.7k21323

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
  – Clayton
  Aug 6 at 18:38
  

  
  
  
  

add a comment | 

up vote
1
down vote

It ensures that $M$ is not zero, which is problematic as you divide by $M$.

Edit: However, by strictness in $m' lt a(n) lt M'$ we have $m' neq M'$ and else $M$ could not be zero even without a $1$ there. I guess the $1$ makes it clearer that we do not divide by zero?

share|cite|improve this answer

edited Aug 6 at 18:38

answered Aug 6 at 18:34

Keba

1,246518

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
  – Clayton
  Aug 6 at 18:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for raising the idea of division by $0$.
  – Aditya Agarwal
  Aug 6 at 18:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clayton: Yes, I also realized that and added the edit.
  – Keba
  Aug 6 at 18:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

The $1$ is redundant. We know that $M$ cannot equal zero as $-m'<a(n)<M'$ are strict inequalities.

Writing $M=M'$ would be sufficient for the proof.

share|cite|improve this answer

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

add a comment | 

up vote
1
down vote

The $1$ is there so that we have $$fracvarepsilon2Mleqfracvarepsilon2.$$ This way, we have each sequence being with $fracvarepsilon2$ of its limit.

share|cite|improve this answer

edited Aug 6 at 18:39

answered Aug 6 at 18:30

Clayton

17.9k22882

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  When did we use that? We just used that $mathbbL_2leq mathbbM$??
  – Aditya Agarwal
  Aug 6 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
  – Clayton
  Aug 6 at 18:34
  

  
  
  
  

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

I don't think it's actually needed. The only thing I could of is to ensure that no division by zero takes place, but the condition $-m'<a(n)<M'$ means that not both $m'$ and $M'$ can be zero, so it's actually unnecessary.

share|cite|improve this answer

answered Aug 6 at 18:36

saulspatz

10.7k21323

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
  – Clayton
  Aug 6 at 18:38
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

I don't think it's actually needed. The only thing I could of is to ensure that no division by zero takes place, but the condition $-m'<a(n)<M'$ means that not both $m'$ and $M'$ can be zero, so it's actually unnecessary.

share|cite|improve this answer

answered Aug 6 at 18:36

saulspatz

10.7k21323

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
  – Clayton
  Aug 6 at 18:38
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

I don't think it's actually needed. The only thing I could of is to ensure that no division by zero takes place, but the condition $-m'<a(n)<M'$ means that not both $m'$ and $M'$ can be zero, so it's actually unnecessary.

share|cite|improve this answer

answered Aug 6 at 18:36

saulspatz

10.7k21323

I don't think it's actually needed. The only thing I could of is to ensure that no division by zero takes place, but the condition $-m'<a(n)<M'$ means that not both $m'$ and $M'$ can be zero, so it's actually unnecessary.

share|cite|improve this answer

answered Aug 6 at 18:36

saulspatz

10.7k21323

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 18:36

saulspatz

10.7k21323

answered Aug 6 at 18:36

saulspatz

10.7k21323

answered Aug 6 at 18:36

saulspatz

10.7k21323

10.7k21323

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
  – Clayton
  Aug 6 at 18:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
  – Clayton
  Aug 6 at 18:38
  

  
  
  
  

1

1

I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
– Clayton
Aug 6 at 18:38

I agree with this answer. I think it's just a stylistic approach so that $a(n)$ and $b(n)$ are within $fracvarepsilon2$ of their limits (without relying on the bounds of the sequence).
– Clayton
Aug 6 at 18:38

add a comment | 

up vote
1
down vote

It ensures that $M$ is not zero, which is problematic as you divide by $M$.

Edit: However, by strictness in $m' lt a(n) lt M'$ we have $m' neq M'$ and else $M$ could not be zero even without a $1$ there. I guess the $1$ makes it clearer that we do not divide by zero?

share|cite|improve this answer

edited Aug 6 at 18:38

answered Aug 6 at 18:34

Keba

1,246518

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
  – Clayton
  Aug 6 at 18:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for raising the idea of division by $0$.
  – Aditya Agarwal
  Aug 6 at 18:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clayton: Yes, I also realized that and added the edit.
  – Keba
  Aug 6 at 18:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

It ensures that $M$ is not zero, which is problematic as you divide by $M$.

Edit: However, by strictness in $m' lt a(n) lt M'$ we have $m' neq M'$ and else $M$ could not be zero even without a $1$ there. I guess the $1$ makes it clearer that we do not divide by zero?

share|cite|improve this answer

edited Aug 6 at 18:38

answered Aug 6 at 18:34

Keba

1,246518

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
  – Clayton
  Aug 6 at 18:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for raising the idea of division by $0$.
  – Aditya Agarwal
  Aug 6 at 18:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clayton: Yes, I also realized that and added the edit.
  – Keba
  Aug 6 at 18:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

It ensures that $M$ is not zero, which is problematic as you divide by $M$.

Edit: However, by strictness in $m' lt a(n) lt M'$ we have $m' neq M'$ and else $M$ could not be zero even without a $1$ there. I guess the $1$ makes it clearer that we do not divide by zero?

share|cite|improve this answer

edited Aug 6 at 18:38

answered Aug 6 at 18:34

Keba

1,246518

It ensures that $M$ is not zero, which is problematic as you divide by $M$.

Edit: However, by strictness in $m' lt a(n) lt M'$ we have $m' neq M'$ and else $M$ could not be zero even without a $1$ there. I guess the $1$ makes it clearer that we do not divide by zero?

share|cite|improve this answer

edited Aug 6 at 18:38

answered Aug 6 at 18:34

Keba

1,246518

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 18:38

edited Aug 6 at 18:38

edited Aug 6 at 18:38

answered Aug 6 at 18:34

Keba

1,246518

answered Aug 6 at 18:34

Keba

1,246518

answered Aug 6 at 18:34

Keba

1,246518

1,246518

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
  – Clayton
  Aug 6 at 18:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for raising the idea of division by $0$.
  – Aditya Agarwal
  Aug 6 at 18:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clayton: Yes, I also realized that and added the edit.
  – Keba
  Aug 6 at 18:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
  – Clayton
  Aug 6 at 18:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for raising the idea of division by $0$.
  – Aditya Agarwal
  Aug 6 at 18:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clayton: Yes, I also realized that and added the edit.
  – Keba
  Aug 6 at 18:39
  

  
  
  
  

2

2

Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
– Clayton
Aug 6 at 18:36

Because of the strict inequalities, I don't think $M$ could be $0$ (in particular, $-m'<M'$, so both can't be $0$).
– Clayton
Aug 6 at 18:36

+1 for raising the idea of division by $0$.
– Aditya Agarwal
Aug 6 at 18:37

+1 for raising the idea of division by $0$.
– Aditya Agarwal
Aug 6 at 18:37

@Clayton: Yes, I also realized that and added the edit.
– Keba
Aug 6 at 18:39

@Clayton: Yes, I also realized that and added the edit.
– Keba
Aug 6 at 18:39

add a comment | 

up vote
1
down vote

The $1$ is redundant. We know that $M$ cannot equal zero as $-m'<a(n)<M'$ are strict inequalities.

Writing $M=M'$ would be sufficient for the proof.

share|cite|improve this answer

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

add a comment | 

up vote
1
down vote

The $1$ is redundant. We know that $M$ cannot equal zero as $-m'<a(n)<M'$ are strict inequalities.

Writing $M=M'$ would be sufficient for the proof.

share|cite|improve this answer

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

add a comment | 

up vote
1
down vote

up vote
1
down vote

The $1$ is redundant. We know that $M$ cannot equal zero as $-m'<a(n)<M'$ are strict inequalities.

Writing $M=M'$ would be sufficient for the proof.

share|cite|improve this answer

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

The $1$ is redundant. We know that $M$ cannot equal zero as $-m'<a(n)<M'$ are strict inequalities.

Writing $M=M'$ would be sufficient for the proof.

share|cite|improve this answer

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

answered Aug 6 at 18:38

TheSimpliFire

9,69261952

9,69261952

add a comment | 

add a comment | 

up vote
1
down vote

The $1$ is there so that we have $$fracvarepsilon2Mleqfracvarepsilon2.$$ This way, we have each sequence being with $fracvarepsilon2$ of its limit.

share|cite|improve this answer

edited Aug 6 at 18:39

answered Aug 6 at 18:30

Clayton

17.9k22882

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  When did we use that? We just used that $mathbbL_2leq mathbbM$??
  – Aditya Agarwal
  Aug 6 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
  – Clayton
  Aug 6 at 18:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

The $1$ is there so that we have $$fracvarepsilon2Mleqfracvarepsilon2.$$ This way, we have each sequence being with $fracvarepsilon2$ of its limit.

share|cite|improve this answer

edited Aug 6 at 18:39

answered Aug 6 at 18:30

Clayton

17.9k22882

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  When did we use that? We just used that $mathbbL_2leq mathbbM$??
  – Aditya Agarwal
  Aug 6 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
  – Clayton
  Aug 6 at 18:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

The $1$ is there so that we have $$fracvarepsilon2Mleqfracvarepsilon2.$$ This way, we have each sequence being with $fracvarepsilon2$ of its limit.

share|cite|improve this answer

edited Aug 6 at 18:39

answered Aug 6 at 18:30

Clayton

17.9k22882

The $1$ is there so that we have $$fracvarepsilon2Mleqfracvarepsilon2.$$ This way, we have each sequence being with $fracvarepsilon2$ of its limit.

share|cite|improve this answer

edited Aug 6 at 18:39

answered Aug 6 at 18:30

Clayton

17.9k22882

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 18:39

edited Aug 6 at 18:39

edited Aug 6 at 18:39

answered Aug 6 at 18:30

Clayton

17.9k22882

answered Aug 6 at 18:30

Clayton

17.9k22882

answered Aug 6 at 18:30

Clayton

17.9k22882

17.9k22882

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  When did we use that? We just used that $mathbbL_2leq mathbbM$??
  – Aditya Agarwal
  Aug 6 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
  – Clayton
  Aug 6 at 18:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  When did we use that? We just used that $mathbbL_2leq mathbbM$??
  – Aditya Agarwal
  Aug 6 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
  – Clayton
  Aug 6 at 18:34
  

  
  
  
  

2

2

When did we use that? We just used that $mathbbL_2leq mathbbM$??
– Aditya Agarwal
Aug 6 at 18:32

When did we use that? We just used that $mathbbL_2leq mathbbM$??
– Aditya Agarwal
Aug 6 at 18:32

@AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
– Clayton
Aug 6 at 18:34

@AdityaAgarwal: The proof doesn't explicitly use it. I think it is so that you have $|a(n)-L_1|<fracvarepsilon2$ and $|b(n)-L_2|<fracvarepsilon2$.
– Clayton
Aug 6 at 18:34

add a comment | 

 

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Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email