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June 5th, 2013, 01:04 PM   #1
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need help on the proof of the inf on a subset!!

Hi, I tried one exercise question from my text book!! Please correct me if I did something wrong!
The questions is about the greatest lower bound!! It is presented right after the Completeness Axiom!!
Some parts of the proof is written in the book but there were two parts left as exercise!!

Corollary.
Every nonempty subset S of R that is bounded below has a greatest lower bound inf S.

Proof.
Let -S be the set -S consists of the negatives of the numbers in S. Since S is bounded below there is an m in R such that This implies that Thus -S is bounded above by -m and there exists sup(-S) by the completeness axiom. Let we need to prove (1) and (2)

The last two parts were left as exercise.

Proof.
Part (1)
Let This implies that Then

Part (2)
Assume This implies that which implies that -t is an upper bound of -S. Since This implies that

I think it makes sense but I have many experiences that the proof is wrong even though it makes sense to me!! So please comment on this!
munjo5746 is offline  
 
June 7th, 2013, 02:25 PM   #2
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Re: need help on the proof of the inf on a subset!!

Quote:
Originally Posted by munjo5746
Hi, I tried one exercise question from my text book!! Please correct me if I did something wrong!
The questions is about the greatest lower bound!! It is presented right after the Completeness Axiom!!
Some parts of the proof is written in the book but there were two parts left as exercise!!
Shouldn't that last sentence end In "!!!"?

Quote:
Corollary.
Every nonempty subset S of R that is bounded below has a greatest lower bound inf S.
This is a corollary to "Every nonempty subset S of R that is bounded above has a least upper bound sup S, right? (The "completeness axiom")

Quote:
Proof.
Let -S be the set -S consists of the negatives of the numbers in S. Since S is bounded below there is an m in R such that . This implies that Thus -S is bounded above by -m and there exists sup(-S) by the completeness axiom. Let we need to prove (1) and (2)

The last two parts were left as exercise.

Proof.
Part (1)
Let This implies that Then

Part (2)
Assume This implies that which implies that -t is an upper bound of -S. Since This implies that

I think it makes sense but I have many experiences that the proof is wrong even though it makes sense to me!! So please comment on this!
It looks good to me.
HallsofIvy is offline  
June 7th, 2013, 07:20 PM   #3
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Re: need help on the proof of the inf on a subset!!

Quote:
Originally Posted by HallsofIvy
Quote:
Originally Posted by munjo5746
Hi, I tried one exercise question from my text book!! Please correct me if I did something wrong!
The questions is about the greatest lower bound!! It is presented right after the Completeness Axiom!!
Some parts of the proof is written in the book but there were two parts left as exercise!!
Shouldn't that last sentence end In "!!!"?

Quote:
Corollary.
Every nonempty subset S of R that is bounded below has a greatest lower bound inf S.
This is a corollary to "Every nonempty subset S of R that is bounded above has a least upper bound sup S, right? (The "completeness axiom")

[quote:f7y205ci]Proof.
Let -S be the set -S consists of the negatives of the numbers in S. Since S is bounded below there is an m in R such that . This implies that Thus -S is bounded above by -m and there exists sup(-S) by the completeness axiom. Let we need to prove (1) and (2)

The last two parts were left as exercise.

Proof.
Part (1)
Let This implies that Then

Part (2)
Assume This implies that which implies that -t is an upper bound of -S. Since This implies that

I think it makes sense but I have many experiences that the proof is wrong even though it makes sense to me!! So please comment on this!
It looks good to me.[/quote:f7y205ci]
Thank you for your comment!!!
munjo5746 is offline  
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