# Thread: Well ordering principle and the maximum principle

1. ## Well ordering principle and the maximum principle

Hello chaps

The question is as follows:
Use the well ordering principle to prove the maximum principle
My half-baked idea:
Let:

$S \neq \emptyset$ and $S \subset N$ (all natural numbers)

There is a hint which states we should let a new set $(B)$ be the upperbounds of $S$ in $N$ (natural numbers)

We know $B \neq \emptyset$, so by WOP it must have a minimum, which we shall call $m$.

We then assume:

$x < m \forall x \in S$

$x + 1 \leq m \Rightarrow x + 1 \leq B$
I'm not sure how we can go further with this. What else can we prove with this information? What am I missing?

2. What does "prove the maximum principle" mean?

3. Originally Posted by Plato
What does "prove the maximum principle" mean?
That was the exact question in my practical. My lecturer is German though. Perhaps his English is not entirely correct? I wouldn't know.

It's actually quite simple though. Prove that if you have a subset (of the natural numbers) which is non-empty and bounded, that the subset must have a maximum.

Then basically you assume it does not have a maximum and do a proof by contradiction that the earlier assumption was incorrect.

To continue from yesterday:

$x + 1 \leq m \forall x \in S$

then clearly:

$x \leq m - 1 \forall x \in S$

Which means $m - 1$ must be a maximum for the set $S$ and we have the necessary contradiction.

4. Here would be my approach.
Because $B \ne \emptyset \;\& \;B \subseteq \mathbb{N}$, then by W.O. $B$ has a least (or first) term. Call it $m$.
If $m\in S$ what does that mean?
If $m\notin S$ is $m-1\in S?$
In either case does $S$ have a maximum element?

5. Originally Posted by Plato
Here would be my approach.
Because $B \ne \emptyset \;\& \;B \subseteq \mathbb{N}$, then by W.O. $B$ has a least (or first) term. Call it $m$.
If $m\in S$ what does that mean?
If $m\notin S$ is $m-1\in S?$
In either case does $S$ have a maximum element?
If $m \in S$ then surely $m$ must be the maximum?

If $m \notin S$ then I don't think $m-1\in S$ (not impossible, but not definitely) unless we have some indication that there is some $x \in S$ where $x \leq m - 1$

6. Originally Posted by janvdl
If $m \notin S$ then I don't think $m-1\in S$ (not impossible, but not definitely) unless we have some indication that there is some $x \in S$ where $x \leq m - 1$
Well $m-1 which means $m-1\notin B$.
So $m-1$ is not an upper bound of $S$.
So some member of $k\in S$ has the property that $m-1.
But there is no integer between $m-1~\&~n$. Thus ??

7. Originally Posted by Plato
Well $m-1 which means $m-1\notin B$.
So $m-1$ is not an upper bound of $S$.
So some member of $k\in S$ has the property that $m-1.
But there is no integer between $m-1~\&~n$. Thus ??
Therefore $k = m$. There is no other way. And it must have a maximum!

That's an awesome approach!