My Math Forum need help on the proof of the inf on a subset!!

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 June 5th, 2013, 12:04 PM #1 Newbie   Joined: Apr 2013 Posts: 11 Thanks: 0 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 $\{-s:s \in S\};$-S consists of the negatives of the numbers in S. Since S is bounded below there is an m in R such that $m \leq s, \forall s \in S.$ This implies that $-m \geq -s, \forall s \in S.$ Thus -S is bounded above by -m and there exists sup(-S) by the completeness axiom. Let $s_0= sup(-S);$ we need to prove (1) $-s_0 \leq s, \forall s \in S,$ and (2) $if\ t \leq s, \forall s \in S,\ then\ t \leq -s_0.$ The last two parts were left as exercise. Proof. Part (1) Let $s_0= sup(-S).$ This implies that $s_0 \geq -s, \forall s \in S.$ Then $-s_0 \leq s, \forall s \in S.$ Part (2) Assume $t \leq s, \forall s \in S.$ This implies that $-t \geq -s, \forall s \in S,$ which implies that -t is an upper bound of -S. Since $s_0= sup(-S),$ $-t \geq s_0.$ This implies that $t \leq -s_0. Q.E.D$ 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!
June 7th, 2013, 01:25 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 $\{-s:s \in S\};$-S consists of the negatives of the numbers in S. Since S is bounded below there is an m in R such that $m \leq s, \forall s \in S$. This implies that $-m \geq -s, \forall s \in S.$ Thus -S is bounded above by -m and there exists sup(-S) by the completeness axiom. Let $s_0= sup(-S);$ we need to prove (1) $-s_0 \leq s, \forall s \in S,$ and (2) $if\ t \leq s, \forall s \in S,\ then\ t \leq -s_0.$ The last two parts were left as exercise. Proof. Part (1) Let $s_0= sup(-S).$ This implies that $s_0 \geq -s, \forall s \in S.$ Then $-s_0 \leq s, \forall s \in S.$ Part (2) Assume $t \leq s, \forall s \in S.$ This implies that $-t \geq -s, \forall s \in S,$ which implies that -t is an upper bound of -S. Since $s_0= sup(-S),$ $-t \geq s_0.$ This implies that $t \leq -s_0. Q.E.D$ 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.

June 7th, 2013, 06:20 PM   #3
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Joined: Apr 2013

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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 $\{-s:s \in S\};$-S consists of the negatives of the numbers in S. Since S is bounded below there is an m in R such that $m \leq s, \forall s \in S$. This implies that $-m \geq -s, \forall s \in S.$ Thus -S is bounded above by -m and there exists sup(-S) by the completeness axiom. Let $s_0= sup(-S);$ we need to prove (1) $-s_0 \leq s, \forall s \in S,$ and (2) $if\ t \leq s, \forall s \in S,\ then\ t \leq -s_0.$

The last two parts were left as exercise.

Proof.
Part (1)
Let $s_0= sup(-S).$ This implies that $s_0 \geq -s, \forall s \in S.$ Then $-s_0 \leq s, \forall s \in S.$

Part (2)
Assume $t \leq s, \forall s \in S.$ This implies that $-t \geq -s, \forall s \in S,$ which implies that -t is an upper bound of -S. Since $s_0= sup(-S),$ $-t \geq s_0.$ This implies that $t \leq -s_0. Q.E.D$

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]

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