# Thread: Maximal Element

1. ## Maximal Element

Fix N, some integer, and suppose L is a nonempty set of integers such that every a in L is <N. Show that L has a maximal element.

$L = \{n \in N|n\geq a \ \forall a\in L\}$

How do I show this?

2. Originally Posted by dwsmith
Fix N, some integer, and suppose L is a nonempty set of integers such that every a in L is <N. Show that L has a maximal element.
$L = \{n \in N|n\geq a \ \forall a\in L\}$
I think you need to edit the OP.
The set $L$ you have given does not meet the description: "Fix N, some integer, and suppose L is a nonempty set of integers such that every a in L is <N"

3. That is verbatim from the book.

4. I am assuming the question goes like this.
If $N\in\mathbb{Z}$ then define $\mathcal{L}=\{n\in\mathbb{Z}:n.
You can surely see that $N\notin\mathcal{L}\text{ and }(N-1)\in\mathcal{L}$. HOW & WHY?
What can you say about $N-1~?$

5. Originally Posted by Plato
I am assuming the question goes like this.
If $N\in\mathbb{Z}$ then define $\mathcal{L}=\{n\in\mathbb{Z}:n.
You can surely see that $N\notin\mathcal{L}\text{ and }(N-1)\in\mathcal{L}$. HOW & WHY?
What can you say about $N-1~?$
$n\leq N-1

N-1 can be greater than n or equal to it.

6. i am a bit confused. the set L = {n in Z: n < N} is the set of ALL integers less than N, whereas in the orginial question, L is presumed to be merely some (non-emtpy) set with the property that:

a in L--> a < N, which is NOT the same thing.

in other words, we are only talking about some subset of {n in Z: n < N}, and we don't know WHICH subset.

i would think it would be easier to assume our set L has no maximal element (that is, if a is in L, then there is b in L with a < b

note that a+1 ≤ b, necessarily), and from this assumption, show that some member in L must be larger than N,

which is a contradiction, so L must have a maximal element.