1. Oct 14, 2017

fishturtle1

1. The problem statement, all variables and given/known data
Consider the sets below. For each one, decide whether the set is bounded above. If it is, give the supremum in $\mathbb{R}$. Then decide whether or not the set is bounded below. If it is, give the infimum. Finally, decide whether or not the supremum is a maximum, and whether to not the infimum is a minimum.

c) the natural numbers $\mathbb{N}$
2. Relevant equations
In my class $\mathbb{N} = {0, 1, 2, ... }$

3. The attempt at a solution
Proof: $\mathbb{N}$ is not bounded above. We will show this with a contradiction. Suppose M is an upper bound on $\mathbb{N}$. Then ($\lceil M \rceil + 1) \space \epsilon \space \mathbb{N}$ and $\lceil M \rceil + 1 > M$, a contradiction. Therefore $\mathbb{N}$ does not have an upper bound and is not bounded above.

We will now show $\mathbb{N}$ is bounded below. Let $U = 0$ and let $x \space \epsilon \space \mathbb{N}$. Then $U = 0 \le x$. Therefore $0$ is a lower bound on $\mathbb{N}$.

In order to show inf$\mathbb{N} = 0$ we must show 0 is the greatest lower bound. We proceed by contradiction. Suppose $m$ is a lower bound on $\mathbb{N}$ such that $m > 0$. But $0 \space \epsilon \space \mathbb{N}$ so m is not a lower bound, a contradiction. We conclude that there does not exist a lower bound $m$ such that $m > 0$ and so inf$\mathbb{N} = 0$.

In order to show min$\mathbb{N} = 0$. we must show 0 is a lower bound on $\mathbb{N}$ and $0 \space \epsilon \space \mathbb{N}$. We've already shown 0 is a lower bound on $\mathbb{N}$ and $0 \space \epsilon \space \mathbb{N}$ is a true statement. Therefore min$\mathbb{N} = 0$. []

My question: Is this proof clear and easy to follow?

2. Oct 14, 2017

Staff: Mentor

Yes, this is correct.