Various definitions of Supremum stated formally

Question

In literature, I found 4 different definitions for Supremum. I tried to express these formally.

1. Question: Are these definitions correct?
2. Question: How can we prove their equivalence formally?
3. Question: Are there even more definitions than these 4?
4. Question: To definition (3), do I have to write both, that $\epsilon$ is an element of $\Bbb{F}$ and $>0$, and - if yes - how?
5. Question: To definition (4), what if Supremum doesn't exist?

Let $\Bbb{F}$ be an ordered field. Let $A \subset \Bbb{F}$ be nonempty. Let $s \in \Bbb{F}$.

(1) $s=\sup A$ $:\iff (\forall a \in A(a \le s)) \land ( \forall b \in \Bbb{F}((\forall a \in A(a \le b)) \implies (s \le b) ))$

(2) $s=\sup A$ $:\iff (\forall a \in A(a \le s)) \land (\forall b \in \Bbb{F} ( (b < s) \implies \exists a \in A (a > b)))$

(3) $s=\sup A$ $:\iff (\forall a \in A(a \le s)) \land (\forall \epsilon > 0(\exists a \in A(a > s-\epsilon)) )$

(4) $s=\sup A$ $:\iff s=min\{c \in \Bbb{F} | (\forall a \in A(a \le c)\}$

Answer 1

Question 1: Yes these definitions are correct

Question 2: Seeing that definitions 1 and 2 is equivalent is easy. Notice that both start by saying “$s$ is an upper bound for $A$”. Then the definitions split as follows:

1. For any other upper bound $b$, we have that $s \leq b$
2. If $b$ is smaller than $s$, then $b$ is not an upper bound for $A$

Notice that (2) is the contrapositive of (1). Thus they are equivalent.

Definition 3 also starts by asserting that $s$ is an upper bound for $A$, but then goes on to say that for every $\epsilon > 0$, we may find an element $a \in A$ such that $s - \epsilon < a$. This is easily seen to be equivalent to condition (2) above.

Finally definition 4 is seen to be equivalent to definition 1.

Question 3: Sure, we could come up with all kinds of different equivalent conditions.

Question 4: I’m not sure what you mean.

Question 5: Suppose that $A$ is bounded above, but fails to have a supremum (for example, the set $\{x \in \mathbb{Q} : x^2 < 2 \}$). Then let’s look at the set $\{x \in \mathbb{F} : \text{ for all } a \in A, a \leq x\}$. Since $A$ has no supremum, this set will fail to have a smallest element.