Boundedness of Real Number Set E

[Ajot]

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 \epsilon 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 \epsilonE.

\exists bounded set E
Assume E: {0 < x < 1} (or { 0 \leq x \leq 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\epsilon E. => set of real numbers E is bounded

\exists r (r > 0)
Let there be a unbounded set of reals E
\exists x (x \epsilon[\tex] E)<br /> |x| &lt; r (given) therefore by definition of upper bound, r is an upper bound<br /> therefore E is not a unbounded set of reals<br /> E is a bounded set of reals<br /> <br /> <br /> 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.<br /> <br /> =)

 

[Petek]

To begin, what's your definition of a set of real numbers being bounded? One common definition is that a non-empty set of real numbers is said to be bounded if it is both bounded above and bounded below. If that's the definition you're using, then it follows that the set E is bounded if there exist real numbers m and M such that E \subseteq [m,M]. Now try to take it from there. Please post again if you need more help.

Petek

 

[HallsofIvy]

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 \epsilon 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 \epsilonE.

\exists bounded set E
Assume E: {0 < x < 1} (or { 0 \leq x \leq 1} wasn't sure which one to use)

You can't do this. You want to show that if E is any bounded set, then the conclusion is true. You cannot assume that E is any specific set.

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\epsilon E. => set of real numbers E is bounded

\exists r (r > 0)
Let there be a unbounded set of reals E
\exists x (x \epsilon[\tex] E)<br /> |x| &lt; r (given) therefore by definition of upper bound, r is an upper bound<br /> therefore E is not a unbounded set of reals<br /> E is a bounded set of reals<br /> <br /> <br /> 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.<br /> <br /> =)

 

[Ajot]

Okay, so let's focus on the first part:

I agree that the definition of a bounded set E is what Petek wrote. My concern is that in the problem it wants to prove a positive r such that |x|< r. To me that seems like an upper bound. Am I not getting something here?

Heres a second shot at the first part of my proof,

Let E be a bounded set by an upperbound where there exists Reals m, M such that E \subseteq [m,M]
Let x represent all the numbers in the set E so for all x \epsilon E
(here is where I don't know how to bring in the r)
M is the maximum of the set E therefore M+1 is not the maximum of the set and not an element of the set E
Let M+1 = r
If M < r then x < r
(Not sure if I can assume the below from what I have above)
|x| < r
Therfore, E is a bounded set by an upperbound r

(sorry if the symbols didn't work well, for some reason the ones I want don't show up correctly)

 

[Office_Shredder]

An upper bound is just a number M so that x<M if x is in E.

For example, the set of negative integers: 1 is an upper bound, but |x| can be arbitrarily large if x is a negative integer

For your proof attempt: M is not the maximum of the set, merely an upper bound. There might not even be a maximum of the set. This is a detail and can be fixed relatively easily.

More problematic is that you need to use the fact that x>m to get that |x| is bounded. Just because x is smaller than a number does not mean that |x| is

 

[HallsofIvy]

If A is bounded then, by your definition, there exist M such the x< M for all x in A and there exist m such that m< x for all x in A. Let r be the larger of M and |m|.

If there exist r such that |x|> r, then -r< x< r.