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Show that a set of real numbers E is bounded if and only if there is a positive number r so that |x| <r for all x [tex]\epsilon[/tex] E.

My Attempt:

This is what I have done, and not sure if I am on the right path. Math proofs are pretty difficult for me so please any help would be greatly appreciated.

I knew I had to show both ways of the proof due to the "if and only if"

1. s => r

Set of real numbers E is bounded => a positive number r so that |x| < r for all x [tex]\epsilon[/tex]E.

[tex]\exists[/tex] bounded set E

Assume E: {0 < x < 1} (or { 0 [tex]\leq[/tex] x [tex]\leq[/tex] 1} wasn't sure which one to use)

By the Completeness Axiom, E has a least upper bound

1 is the least upperbound

1 > 0

r= 1

|x| < r

r is the least upper bound and E is bounded

2. r => s

a positive number r so that |x|< r for all x[tex]\epsilon[/tex] E. => set of real numbers E is bounded

[tex]\exists[/tex] r (r > 0)

Let there be a unbounded set of reals E

[tex]\exists[/tex] x (x [tex]\epsilon[\tex] E)

|x| < r (given) therefore by definition of upper bound, r is an upper bound

therefore E is not a unbounded set of reals

E is a bounded set of reals

I feel like there are a lot of holes and I just want to be able to improve my math proofing skills. Thanks for the help.

=)

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# Homework Help: Math Proof Help

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