Associated Topics || Dr. Math Home || Search Dr. Math

### Bounded Set

```Date: 06/24/2003 at 13:57:27
From: Saint Monday
Subject: Proof about a bounded set

Let S be a set of real numbers. Prove that the following are
equivalent:

(a) S is bounded, i.e. there exists a number M > 0 such that abs(x)
<= M for every x in S.
(b) S has an upper bound and lower bound.

I need to find the infimum and supremum that implies that there exists
an upper and lower bound.

This is what I have so far:

abs(x) <= M is the same as -M <= x <= M
x <= M, upper bound
a <= x, lower bound

It was suggested to me to find a value for M that would satisfy this,
so would M = (abs(x) + 1) work?
```

```
Date: 06/24/2003 at 14:57:11
From: Doctor Mike
Subject: Re: Proof about a bounded set

Hi,

The infimum is the LARGEST possible lower bound, and the supremum is
the SMALLEST possible upper bound, so it is good to use those words
only when you mean exactly those specific concepts.

To show two statements are "equivalent" means to prove that if you
assume one of them is true, the other must be true also. Obviously,
such a proof comes in two parts. Namely,

Part 1 --- Assume (a) is true, and then prove that (b) is true.
This part you did pretty well on before. Assume for
notation purposes that the positive "M" is a bound for S.
Note, it may NOT be the SMALLEST bound for S. What we
are required to prove to show that (b) is true is that
S has an upper bound and has a lower bound. You correctly
observed that M must be an upper bound, and -M must be
a lower bound. Done with this part!

Part 2 --- Assume (b) is true, and then prove that (a) is true.
For this part, assume for notation purposes that "B" is
an upper bound for S, and that "C" is a lower bound for S.
That just means that for every x in S, C <= x and x <= B.

N = max( abs(B) , abs(C) )

That is, N is the larger of the absolute value of B and
the absolute value of C. We know that every x in S is
<= B so it must also be that x <= abs(B), which in turn
is <= N.  To summarize,

x <= B <= abs(B) <= N

Now let's look at what x is greater than or equal to.
x >= C is known.  We do not know whether C is positive
or negative or zero, but we DO know that -abs(C) must
be the same as C or smaller than C.  So, -abs(C) <= C.
Now take a very close look at this LONG thing :

x >= C >= -abs(C) >= -max(abs(B),abs(C)) = -N

So this x, which can be anything in S, is <= N and >= -N.
That's exactly what it means to say that N is a bound for S.

Go over this a few times to make sure you understand
each step. We are done with this part now.

When you are done with both directions (both parts), you are
completely done. I hope this clarifies what is going on in this
problem.  I hope also that you can carry on with this proof as a
"pattern" to do similar ones.

If you get to a problem where infimum and supremum really are
involved, the structure of an "equivalent" type of proof is still the
same: two parts. Just the details of the properties to be proved are
different. Good luck.

Write back again if you want to talk more

- Doctor Mike, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/24/2003 at 15:11:49
From: Saint Monday

You're a godsend, Dr. Mike.
Thank you so much for your time!
```
Associated Topics:
College Definitions
High School Definitions
High School Sets

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search