I'm trying to prove the following result, which has been presented as an example in my book:
Let $F$ be an ordered field. Then, $F$ is unbounded.
Proof Attempt:
Suppose that $F$ is bounded. Then:
$\exists K \in F: \forall x \in F: x \leq K$
$\exists k \in F: \forall x \in F: k \leq x$
So, the claim is that:
$F = [k,K] = \{x \in F: k \leq x \leq K\}$
Since $K \in F$, $K + 1 \in F$ and $K+1 \notin [k,K]$. That's a contradiction. Since $k \in F$, $k-1 \in F$ and $k-1 \notin F$. That's a contradiction. Hence, $F$ cannot be bounded.
Is the argument above correct? Would there have been a better way to formulate it or no?
Edit:
Let $F$ be an ordered field. Let $A$ be a nonempty subset of $F$. Then:
A is bounded above if $\exists K \in F: \forall x \in A: x \leq K$
A is bounded below if $\exists k \in F: \forall x \in A: k \leq x$
A is bounded if it is bounded below and bounded above.
A is unbounded if it is not bounded.
Those are the definitions given in my book so that's what I'm working with.