# Bounded set proof

1. Feb 28, 2008

### Seda

1. The problem statement, all variables and given/known data

If A and B are bounded sets, then A U B is a bounded set.
(Prove this statement)

2. Relevant equations

Definition of Union is a given.

A set A is bounded iff there exists some real value m such that lxl < m for all element x found in A.

3. The attempt at a solution

This makes sense to me. If set A is bounded by M and set B is bounded by N, then A U B will be bounded by which value is higher. I have to keep in mind that the definition of a bounded set has the "iff" term.

My attempt (this is quite odd looking to me, I don't know how to make it more straightfoward)

Let x exist in A. Then that means there is a value m where m>lxl by definition of a bounded set.

Let y exist in B. Then that means there is a value n where n>lyl by definition of a bounded set.

Thus, x and y exist in A U B by definition of union.

We know lxl and lyl are < whichever value of m or n is the larger of the two.

Thus, A U B is a bounded set.

The bolded step seems oddest, but critique on any part of the proof is welcome. Help!

2. Feb 28, 2008

### Mathdope

You should really start by assuming that x is in A U B. Then say why there is a constant K such that |x| < K.

3. Feb 28, 2008

### Seda

But I'm suppose to prove the implication the other way (if A and B are bounded, then A U B is bounded.), not the other way around. Regardless of whether or not the statement is actually biconditional.

4. Feb 28, 2008

### Mathdope

You are proving a statement about A U B, so you need to start with an arbitrary element in A U B. You can then invoke the known properties of A and B individually to prove the statement regarding the union.

5. Feb 28, 2008

### Seda

Reworked:

Let X exist in A U B
Thus X exists in A or B by definition of union.
Cases:
If x exists in A, then lxl < some real value m by the definition of bounded
If x exists in B, then lxl < some real value n by the definition of bounded.
If x exists in both A and B, then X is less than both m and n.
Subcases (in the case that x is in both A and B)
If m>n , x must be < n to fit the definition of bounded for both sets.
If n>m, x must be less then m to fit the definition of bounded for both sets.

Thus, for all possibities, the value x is less than some real value m or n. Thus, AUB is bounded since x exists in AUB.

THat seems to make more sense, even if it is a tad convuluted. Is that more fitting?

Last edited: Feb 28, 2008
6. Feb 28, 2008

### Mathdope

That's more like it, but the fact that x lies in the union simply means that |x| < max (m,n) which I would call K.

7. Feb 28, 2008

### Seda

Wouldn't it be min (m,n)?

Edit: Never mind, I see, the larger bound of A or B will become the bound of AUB by definition of union.

8. Feb 28, 2008

### ircdan

- you need to find a single value M s.t. for all x in A u B, we have |x| < M.
- i don't know why you are saying exists, a more common phrasing is,
suppose x is in A u B, or, assume x is in A u B, etc

your idea is right, but to help you out, i'll put parentheses around things that aren't part of the proof.

Proof. Suppose A and B are bounded. (now we write down what this means). Then there are M and N s.t. for all x in A, |x| < N, and for all x in B, |x| < M. Now suppose x is in A u B. (again, write down what this means). Then x is in A or x is in B.

(now think for a moment, looking at what you have already done, you know that if x is in A, then |x| < N, and you know that if x is in B, then |x| < M. We need some number K that is bigger than both N and M, so we want the larger of the two, well we can say it in words, let K be the larger of the two, or just K = max{M, N})

Set K = max{M, N}.

If x is in A, then |x| < N <= K.
If x is in B, then |x| < M <= K.

In any case, |x| < K, so A u B is bounded.

Edit: Note that it's important to specify a single value K because the definition requires it. There are times when you can't take a maximum and find a single value K(like if you had an infinite union of bounded sets, A U B U C U ..., it's not always possible to find a maximum, depends what the sets are)

Last edited: Feb 28, 2008
9. Feb 28, 2008

### Seda

Re-reworked

(The problem on my homework already states to suppose that A and B are bounded with that definition, so I wont need to restate that )

Let X exist in A U B. Also, let K= max (m,n)
Thus X exists in A or B by definition of union.
Cases:
If x is in A, then lxl < m thus lxl< K
If x is in B, then lxl < n thus lxl< K
If x is in both A and B, then lxl over both sets is always less than max (m, n) or lxl < K.

Thus, for all possibities, the value lxl is less than some real value K. Thus, AUB is bounded since x exists in AUB.

AM i getting closer?

Last edited: Feb 28, 2008
10. Feb 29, 2008

### HallsofIvy

Staff Emeritus
Since you are dealing with both x in A and x in B, you don't need to deal with "x in A and B" separately.