Prove that if $S$ has a greatest element $b$, then $b = lubS$.
These are the definitions I used:
Def. Given a partially ordered set ($P, \le$), then an element $b$ of a subset $S$ of $P$ is the greatest element of $S$ if $\forall s \in S, s \le b$.
Def. $lubS$ is the unique number having the following properties:
1) $\forall x \in S, x \le lubS$
2) If$\;\exists b' \in S \, s.t. \forall x \in S, x \le b', then \, lubS \le b' $
Assume $s \in S$.
Let $b \in S \,be \,s.t. \forall s \in S, s \le b$.
Then b is an upper bound for S. So we have 1).
Suppose by contradiction that $\exists b' \in S \,s.t. \forall x \in S, x \le b' \land b > b'$.
Then $b \notin S$. This contradicts the fact that $b \in S$.
Then $\exists b' \in S \,s.t. \forall x \in S, x \le b' \implies lubS \le b'$. So we have 2).
Then $\forall s \in S, \exists b \in S, s \le b \implies b=lubS.\square$
Is my proof complete and correct? What other ways are there to prove this? I have seen other lub proofs that are similar but vary slightly in the second property. For example, they use 2) as "if $M' < M, then\, \exists x \in E \,s.t. x > M'$." My textbook uses the theorem
If $M$ is the least upper bound of the set $S$ and $\epsilon$ is a positive number, then there is at least one number $s$ in $S$ such that $M - \epsilon < s \le M$.
and then proves things like the $lubS = 3$ where $S=\{0,1,2,3\}$ by taking $\epsilon = 0.0001$ and $3 - 0.0001 < s \le 3$. I am not convinced fully of this because I have seen other proofs that do this for arbitrary positive $\epsilon$.
Can someone explain to me the differences/similarities among these definitions? What is the most appropriate/general approach to proving least upper and greatest lower bounds? Thanks.
EDIT
The correct definition for $lubS$ is
Def. $lubS$ is the unique number having the following properties:
1) $\forall x \in S, x \le lubS$
2) If$\;\exists b' \in \mathbb{R} \, s.t. \forall x \in S, x \le b', then \, lubS \le b' $
For your first question, be careful with the definition of a LUB of a set $S$ - this is not necessarily an element of $S$ itself. Think of the partial order $([0,1],\leq)$ with $S=[0,1)$.
However, in the case that $S$ has a maximal element $b$ it is true that $b=LUB(S)$. Your proof is essentially correct, but you should remove the assumption $b'\in S$ since this is not necessary.
Your second definition of the LUB is less general than the first, this time for the particular case that your set $S$ is some subset of the real line. But the concept of a LUB is much broader. For more information on all this look on wikipedia.