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If $x_n$ is an increasing sequence and $lim_ntoinftyx_n=L$, then $L$ is an upper bound of $x_n$

Least Upper Bound of 2 SetsNon-increasing Monotone Sequence Convergence ProofCheck the proof of an upper bound must be greater than or equal to the a lower boundProof verification for the uniqueness of the least upper bound of a set.Upper and lower limitShrinking upper bound in limitUpper bound of a recursive sequence for a fixed $n$Showing the set with a supremum has an increasing sequence converging to that supremum.Upper limit less than M condition proofProof Verification: A monotonically increasing sequence that is bounded above always has a LUB

9

1

$begingroup$

I was wondering if somebody could critique my proof -- I feel I have the general idea but my solution lacks elegance. I appreciate your help!

Proposition:

An increasing sequence $x_n$ has a limit $L$. Then $L$ is an upper bound for $x_n$.

Indirect Proof:

Assume that $L$ is not an upper bound. Then for some value $n=N$, we have:
$$x_N>L.$$

Since $ x_n $ is increasing, we have:

$$x_n> Ltext for all n geq N.$$

However, because $L$ is the limit of $x_n$, we also have, given $epsilon>0$,:

$$L-epsilon<x_n<L+epsilon,$$
for sufficiently large $n$.

Because the above equation must hold for all $epsilon >0$, I can select a value of $epsilon$ such that:

$$ 0<epsilon<x_N-L.$$

It follows that:
$$L+epsilon<x_N.$$

And since $x_n$ is an increasing sequence, I can extend the above equation to:

$$L+epsilon<x_n$$ for sufficiently large $n$.

This contradicts our earlier definition of $L$ as the limit of $x_n$, which required $x_n<L+epsilon$ for sufficiently large $n$.

Therefore, our original assumption is false, and we conclude that $L$ is an upper bound.

real-analysis proof-verification

share|cite|improve this question

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

asked Jun 1 at 16:57

WillWill

483

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'd go just straight to definition of an upper bound.
  $endgroup$
  – Jakobian
  Jun 1 at 16:59
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I think that this proof is perfect.
  $endgroup$
  – Crostul
  Jun 1 at 17:01
  

  
  
  
  

add a comment | 

9

1

$begingroup$

I was wondering if somebody could critique my proof -- I feel I have the general idea but my solution lacks elegance. I appreciate your help!

Proposition:

An increasing sequence $x_n$ has a limit $L$. Then $L$ is an upper bound for $x_n$.

Indirect Proof:

Assume that $L$ is not an upper bound. Then for some value $n=N$, we have:
$$x_N>L.$$

Since $ x_n $ is increasing, we have:

$$x_n> Ltext for all n geq N.$$

However, because $L$ is the limit of $x_n$, we also have, given $epsilon>0$,:

$$L-epsilon<x_n<L+epsilon,$$
for sufficiently large $n$.

Because the above equation must hold for all $epsilon >0$, I can select a value of $epsilon$ such that:

$$ 0<epsilon<x_N-L.$$

It follows that:
$$L+epsilon<x_N.$$

And since $x_n$ is an increasing sequence, I can extend the above equation to:

$$L+epsilon<x_n$$ for sufficiently large $n$.

This contradicts our earlier definition of $L$ as the limit of $x_n$, which required $x_n<L+epsilon$ for sufficiently large $n$.

Therefore, our original assumption is false, and we conclude that $L$ is an upper bound.

real-analysis proof-verification

share|cite|improve this question

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

asked Jun 1 at 16:57

WillWill

483

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'd go just straight to definition of an upper bound.
  $endgroup$
  – Jakobian
  Jun 1 at 16:59
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I think that this proof is perfect.
  $endgroup$
  – Crostul
  Jun 1 at 17:01
  

  
  
  
  

add a comment | 

$begingroup$

I was wondering if somebody could critique my proof -- I feel I have the general idea but my solution lacks elegance. I appreciate your help!

Proposition:

An increasing sequence $x_n$ has a limit $L$. Then $L$ is an upper bound for $x_n$.

Indirect Proof:

Assume that $L$ is not an upper bound. Then for some value $n=N$, we have:
$$x_N>L.$$

Since $ x_n $ is increasing, we have:

$$x_n> Ltext for all n geq N.$$

However, because $L$ is the limit of $x_n$, we also have, given $epsilon>0$,:

$$L-epsilon<x_n<L+epsilon,$$
for sufficiently large $n$.

Because the above equation must hold for all $epsilon >0$, I can select a value of $epsilon$ such that:

$$ 0<epsilon<x_N-L.$$

It follows that:
$$L+epsilon<x_N.$$

And since $x_n$ is an increasing sequence, I can extend the above equation to:

$$L+epsilon<x_n$$ for sufficiently large $n$.

This contradicts our earlier definition of $L$ as the limit of $x_n$, which required $x_n<L+epsilon$ for sufficiently large $n$.

Therefore, our original assumption is false, and we conclude that $L$ is an upper bound.

real-analysis proof-verification

share|cite|improve this question

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

asked Jun 1 at 16:57

WillWill

483

$endgroup$

share|cite|improve this question

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

asked Jun 1 at 16:57

WillWill

483

share|cite|improve this question

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

asked Jun 1 at 16:57

WillWill

483

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

edited Jun 2 at 12:50

Asaf Karagila♦

312k33446780

asked Jun 1 at 16:57

WillWill

483

asked Jun 1 at 16:57

WillWill

483

3 Answers
3

active

oldest

votes

9

$begingroup$

This is a very clear proof, easy and natural to follow. The only 'improvement' that I could see would be that it's possible to compactify it a lot, as in squeeze the information into a paragraph or less, with the upside being space advantage and the downside being reduced readability. All in all though, this proof is clear and concise which is pretty much what everybody should aim for.

share|cite|improve this answer

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for the feedback -- I am relatively new to proofs and appreciate the insight.
  $endgroup$
  – Will
  Jun 1 at 17:07
  

  
  
  
  

add a comment | 

6

$begingroup$

Your proof is fine.

Slightly different:

Given: $x_n$ is increasing, convergent to $L$.

Need to show that $x_n le L$,

Assume there is a $n_0 in mathbbN$ s.t.

$L < x_n_0$.

For $n ge n_0$ :

$x_n_0 le x_n$ since $x_n$ is increasing.

But then

$L <x_n_0 le lim_ n rightarrow inftyx_n = L$,

a contradiction.

share|cite|improve this answer

edited Jun 2 at 6:12

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

$endgroup$

add a comment | 

0

$begingroup$

By the definition of the limit,

$$forall epsilon>0:exists N:forall nge N:L-epsilon<x_n<L+epsilon.$$

Then assume we found some $x_m>L$. Taking $epsilon:=x_m-L>0$, we can write

$$exists N:forall nge N:x_n<x_m.$$

This contradicts that the sequence is growing (because we can have $n>m$).

share|cite|improve this answer

edited Jun 2 at 9:28

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237

$endgroup$

add a comment | 

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9

$begingroup$

This is a very clear proof, easy and natural to follow. The only 'improvement' that I could see would be that it's possible to compactify it a lot, as in squeeze the information into a paragraph or less, with the upside being space advantage and the downside being reduced readability. All in all though, this proof is clear and concise which is pretty much what everybody should aim for.

share|cite|improve this answer

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for the feedback -- I am relatively new to proofs and appreciate the insight.
  $endgroup$
  – Will
  Jun 1 at 17:07
  

  
  
  
  

add a comment | 

9

$begingroup$

This is a very clear proof, easy and natural to follow. The only 'improvement' that I could see would be that it's possible to compactify it a lot, as in squeeze the information into a paragraph or less, with the upside being space advantage and the downside being reduced readability. All in all though, this proof is clear and concise which is pretty much what everybody should aim for.

share|cite|improve this answer

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for the feedback -- I am relatively new to proofs and appreciate the insight.
  $endgroup$
  – Will
  Jun 1 at 17:07
  

  
  
  
  

add a comment | 

$begingroup$

This is a very clear proof, easy and natural to follow. The only 'improvement' that I could see would be that it's possible to compactify it a lot, as in squeeze the information into a paragraph or less, with the upside being space advantage and the downside being reduced readability. All in all though, this proof is clear and concise which is pretty much what everybody should aim for.

share|cite|improve this answer

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

$endgroup$

share|cite|improve this answer

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

answered Jun 1 at 17:01

auscryptauscrypt

7,416614

6

$begingroup$

Your proof is fine.

Slightly different:

Given: $x_n$ is increasing, convergent to $L$.

Need to show that $x_n le L$,

Assume there is a $n_0 in mathbbN$ s.t.

$L < x_n_0$.

For $n ge n_0$ :

$x_n_0 le x_n$ since $x_n$ is increasing.

But then

$L <x_n_0 le lim_ n rightarrow inftyx_n = L$,

a contradiction.

share|cite|improve this answer

edited Jun 2 at 6:12

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

$endgroup$

add a comment | 

6

$begingroup$

Your proof is fine.

Slightly different:

Given: $x_n$ is increasing, convergent to $L$.

Need to show that $x_n le L$,

Assume there is a $n_0 in mathbbN$ s.t.

$L < x_n_0$.

For $n ge n_0$ :

$x_n_0 le x_n$ since $x_n$ is increasing.

But then

$L <x_n_0 le lim_ n rightarrow inftyx_n = L$,

a contradiction.

share|cite|improve this answer

edited Jun 2 at 6:12

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

$endgroup$

add a comment | 

$begingroup$

Your proof is fine.

Slightly different:

Given: $x_n$ is increasing, convergent to $L$.

Need to show that $x_n le L$,

Assume there is a $n_0 in mathbbN$ s.t.

$L < x_n_0$.

For $n ge n_0$ :

$x_n_0 le x_n$ since $x_n$ is increasing.

But then

$L <x_n_0 le lim_ n rightarrow inftyx_n = L$,

a contradiction.

share|cite|improve this answer

edited Jun 2 at 6:12

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

$endgroup$

share|cite|improve this answer

edited Jun 2 at 6:12

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

answered Jun 1 at 17:31

Peter SzilasPeter Szilas

12.6k2822

0

$begingroup$

By the definition of the limit,

$$forall epsilon>0:exists N:forall nge N:L-epsilon<x_n<L+epsilon.$$

Then assume we found some $x_m>L$. Taking $epsilon:=x_m-L>0$, we can write

$$exists N:forall nge N:x_n<x_m.$$

This contradicts that the sequence is growing (because we can have $n>m$).

share|cite|improve this answer

edited Jun 2 at 9:28

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237

$endgroup$

add a comment | 

0

$begingroup$

By the definition of the limit,

$$forall epsilon>0:exists N:forall nge N:L-epsilon<x_n<L+epsilon.$$

Then assume we found some $x_m>L$. Taking $epsilon:=x_m-L>0$, we can write

$$exists N:forall nge N:x_n<x_m.$$

This contradicts that the sequence is growing (because we can have $n>m$).

share|cite|improve this answer

edited Jun 2 at 9:28

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237

$endgroup$

add a comment | 

$begingroup$

By the definition of the limit,

$$forall epsilon>0:exists N:forall nge N:L-epsilon<x_n<L+epsilon.$$

Then assume we found some $x_m>L$. Taking $epsilon:=x_m-L>0$, we can write

$$exists N:forall nge N:x_n<x_m.$$

This contradicts that the sequence is growing (because we can have $n>m$).

share|cite|improve this answer

edited Jun 2 at 9:28

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237

$endgroup$

share|cite|improve this answer

edited Jun 2 at 9:28

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237

answered Jun 2 at 9:17

Yves DaoustYves Daoust

139k878237