# Upper Bound and Lower Bounds

• Sep 5th 2010, 01:15 PM
novice
Upper Bound and Lower Bounds
None of the sets $\displaystyle \mathbb{Z}, \mathbb{Q}$, and $\displaystyle \mathbb{N}$ is bounded above. The $\displaystyle \mathbb{N}$ is bounded below;$\displaystyle 1$ is a lower bound for $\displaystyle \mathbb{N}$ and so is any number less than $\displaystyle 1$. In fact, $\displaystyle 1$ is the larges or greatest lower bound.

Question: How could there be a number less than $\displaystyle 1$ when $\displaystyle 1$ is the smallest number in $\displaystyle \mathbb{N}$?
• Sep 5th 2010, 01:34 PM
Plato
Quote:

Originally Posted by novice
None of the sets $\displaystyle \mathbb{Z}, \mathbb{Q}$, and $\displaystyle \mathbb{N}$ is bounded above. The $\displaystyle \mathbb{N}$ is bounded below;$\displaystyle 1$ is a lower bound for $\displaystyle \mathbb{N}$ and so is any number less than $\displaystyle 1$. In fact, $\displaystyle 1$ is the larges or greatest lower bound.
Question: How could there be a number less than or equal to $\displaystyle 1$ when $\displaystyle 1$ is the smallest number in $\displaystyle \mathbb{N}$?

First, be careful. Your text seems to imply that $\displaystyle \mathbb{N}=\mathbb{Z}^+$. But in some texts $\displaystyle 0\in \mathbb{N}$.

Using the convention that if $\displaystyle k\in \mathbb{N}$ then $\displaystyle k\ge 1$.
In other words 1 is the first element in $\displaystyle \mathbb{N}$.
It depends on the superset as to how we answer your question.
In the real numbers, any number less than 1 is a lower bound for $\displaystyle \mathbb{N}$.
But in the set $\displaystyle \mathbb{N}$ there is only one lower bound, 1.

Does that help?
• Sep 5th 2010, 01:37 PM
emakarov
When one talks about a lower bound, it has to be in the context of some set P and its subset S. An x in P is a lower bound of S if x <= y for all y in S. To talk about a lower bound of $\displaystyle \mathbb{N}$ one has to designate a superset of $\displaystyle \mathbb{N}$.
• Sep 5th 2010, 02:05 PM
novice
Quote:

Originally Posted by Plato
Does that help?

Yes, tremendously.
• Sep 5th 2010, 05:48 PM
novice
Quote:

Originally Posted by Plato
First, be careful. Your text seems to imply that $\displaystyle \mathbb{N}=\mathbb{Z}^+$. But in some texts $\displaystyle 0\in \mathbb{N}$.

Using the convention that if $\displaystyle k\in \mathbb{N}$ then $\displaystyle k\ge 1$.
In other words 1 is the first element in $\displaystyle \mathbb{N}$.
It depends on the superset as to how we answer your question.
In the real numbers, any number less than 1 is a lower bound for $\displaystyle \mathbb{N}$.
But in the set $\displaystyle \mathbb{N}$ there is only one lower bound, 1.

Does that help?

Here is another example:
Quote:

The set $\displaystyle \{n^{(-1)^n$ $\displaystyle : n \in \mathbb{N}\}$ is not bounded above. Among its many lower bounds, 0 is the greatest lower bound.

This one does not have a minimum. The numbers alternately swing up to a larger number and swing back down to smaller fraction. The upper sides has no boundary, and the lower side becomes very small but never reached 0. So the set on the real number line is an interval $\displaystyle (0,\infty)$

Well, I think I figured it out on my own soon after I typed in the question.
At any rate, thank you for your time.