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Proving inf A=-sup(-A)

  • #1
395
14

Homework Statement


Classic proof, I saw it before but I can't remember it.
A, a subset of R is bounded below. -A is -a where a is in A
I don't know why I'm having such a difficult time.


Homework Equations





The Attempt at a Solution


Because A is bounded below, there is some m such that m≤a for all a in A. Simply multiply by -1 to get -m≥-a for all -a in -A, proving that -A has an upper bound. By the completeness axiom, -A has a least upper bound. Lets call this u. u has two properties. u≥-a for all -a in -A. Also v≥u for all other upper bounds v. Multiply by -1. -u≤a for all a in A AND -v≤-u for all lower bounds -v. Therefore, -u is the inf of A. Therefore inf A=-sup(-A).

Is this okay? I feel uneasy.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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First, any number in -A is of the form -x for some x in A. Then, a is a lower bound for A: for any x in A, [itex]a\le x[/itex]. Therefore, [itex]-a\ge -x[/itex]. -a is an upperbound on -A.

Further, a is the greatest lower bound of A. If b is another lower bound on A, b< a. Let p be an upper bound on -A. [itex]p\ge x[/itex] for all x in A. Then, again since any number in -A is of the form -a for some a in A, [itex]p\ge -a[/itex]. Can you complete that?
 
  • #3
395
14
First, any number in -A is of the form -x for some x in A. Then, a is a lower bound for A: for any x in A, [itex]a\le x[/itex]. Therefore, [itex]-a\ge -x[/itex]. -a is an upperbound on -A.

Further, a is the greatest lower bound of A. If b is another lower bound on A, b< a. Let p be an upper bound on -A. [itex]p\ge x[/itex] for all x in A. Then, again since any number in -A is of the form -a for some a in A, [itex]p\ge -a[/itex]. Can you complete that?
We can't assume that A has a greatest lower bound, this is just one part of a proof of that.

Is there anything wrong with my proof? It seems very similar to yours.
 

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