# Proof that ordered fields are unbounded

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:

1. A is bounded above if $$\exists K \in F: \forall x \in A: x \leq K$$

2. A is bounded below if $$\exists k \in F: \forall x \in A: k \leq x$$

3. A is bounded if it is bounded below and bounded above.

4. A is unbounded if it is not bounded.

Those are the definitions given in my book so that's what I'm working with.