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.

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

How do I show this?

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- May 21st 2011, 08:48 AMdwsmithMaximal 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.

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

How do I show this? - May 21st 2011, 09:04 AMPlato
- May 21st 2011, 09:09 AMdwsmith
That is verbatim from the book.

- May 21st 2011, 10:08 AMPlato
I am assuming the question goes like this.

If $\displaystyle N\in\mathbb{Z}$ then define $\displaystyle \mathcal{L}=\{n\in\mathbb{Z}:n<N\}$.

You can surely see that $\displaystyle N\notin\mathcal{L}\text{ and }(N-1)\in\mathcal{L}$. HOW & WHY?

What can you say about $\displaystyle N-1~?$ - May 21st 2011, 10:11 AMdwsmith
- May 22nd 2011, 09:07 AMDeveno
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.