# Least upper bound analysis proof

1. Sep 13, 2009

### dancergirlie

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

Assume that A and B are nonempty sets, that A is bounded above, and that B is contained in A. Prove that B is bounded above and that the least upper bound of B is less than or equal to the least upper bound of A.

2. Relevant equations

Definition: Least Upper bound: Let s be the least upper bound then:
1)s is an upper bound (a set A contained in R is bounded above if there exists an element b in R such that a is less than or equal to b for all elements a in A.)
2)if b is any upper bound of A then s is less than or equal to b.

3. The attempt at a solution
a
Assume A and B are nonempty sets and that A is bounded above. Meaning, A is contained in R and that there exists an element c in R so that c is greater than or equal to a, for all a in A.
Now assume that B is contained in A, meaning every element in B is also an element of A. Since A has an upper bound (c) and every element of B is also an element of A, that means c is also an upper bound for for B. Therefore B is bounded above
**not sure if this is a good argument or not**

If B is contained in A that means, that there are no element in B that are greater than any of the elements in A. Therefore, the least upper bound of B must be less than or equal to the least upper bound of A.

**this seems rather short, I feel like i need to say more**

Any help/tips would be appreciated!!

2. Sep 13, 2009

### LCKurtz

Your first paragraph is correct but I would phrase it more mathematically. You are given:

$$B \subset A \subset R.$$

$$c$$ is an upper bound for $$A$$ means that for all $$a \in A, a \leq c$$.

Now suppose $$b \in B$$. Then $$b \in A$$ since $$B \subset A$$. Therefore b <= c since c is an upper bound for $$A$$. Hence c is an upper bound for $$B$$.

Notice that I haven't changed your argument at all, just expressed it more mathematically.

For the second part, your statement "If B is contained in A that means, that there are no element in B that are greater than any of the elements in A" is very imprecise. You might have better luck with an indirect argument. Start by assuming by way of contradiction that the least upper bound of $$B$$ is greater than that of $$A$$ and see what goes wrong.

3. Sep 13, 2009

### snipez90

You don't need an indirect argument for the second part. What you've essentially shown in the first part is that any upper bound for A is going to be an upper bound for B. In particular, the least upper bound of A is an upper bound of A and hence an upper bound for B. Now what can you say about the least upper bound for B?

4. Sep 14, 2009

### HallsofIvy

Staff Emeritus
Let $\alpha$ be the least upper bound for A. Then $\alpha$ is an upper bound for A. Since B is a subset of A, $\alpha$ is an upper bound for B. Therefore, the least upper bound for B is ......

That's even shorter!

5. Sep 14, 2009

### dancergirlie

thanks for the help everyone, it makes a lot more sense now :)