Proof that ordered fields are unbounded

Question

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.

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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.

Answer 1

One contradiction is enough. So e.g. assume $\exists K \in F: \forall x \in F:x \le K$. Then $0 \le 1$ (an axiom of ordered fields), so $K=K+0 \le K+1$. And this we combine with $K+1 \le K$ which holds by assumption and so $K+1=K$ by the P.O. axioms which yields a final contradiction $0=1$. So $K$ cannot exist.

Answer 2

Your proof assumes too much (first, how do you know $1>0$, second, why do you know that $k+1>k$, etc)

I think an easier statement to prove is that any ordered field has the integers embedded in them (I'll only show that the naturals are embedded, but you can show the rest by considering $-1$).

We know that $0\neq1$, so $0<1$ or $0>1$. WLOG $0<1$ (if $0>1$, just generate the negative integers instead)

I claim that $k\in F$ for $k\in\mathbb N$. We already know $0,1\in F$. Assuming $0,...,k\in F$, note that $(k-1)+1<k+1$, so $k+1$ is distinct from $0,...,k$, and greater than all of them.

By induction, we showed there exists an embedding of the ordered naturals. You can show there's an ordered embedding of the integers, which would complete the proof.

Answer 3

Your proof is good but I think this may be more direct.

We have (presumably) already proven that $1>0$[1].

so for any $K \in F$ we have $K+1 > K$.

Thus no $K\in F$ can be an upper bound for $K$ as there will always exist a $K+1 > K$.

Sometimes a proof can be a simple statement.

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[1] $x > 0 \iff -x < 0$ Pf: $x>0\implies 0=x-x> -x\implies x> -x+x =0$.

$0x = 0$ Pf: $0=0x +(-0x)= (0+0)x+(-0x)=0x + 0x + (-0x)=0x + 0 = 0x$.

$(-a)b=a(-b)=-ab$ Pf: $ab + (-a)b=(a+(-a))b=0b=0;ab+a(-b)=a(b+(-b))=a0=0$ and $ab+(-ab)=0$ so $(-a)b=a(-b)=-ab$.

$-(-x)=x$ Pf: $-x+x = 0 = -x +(-(-x))$.

Axiom: If $a< b$ and $c > 0$ then $ac < bc$. So if $x > 0$ then $x^2=x*x>x*0=0$. If $x = 0$ then $x^2 = 0*0=0$ And if $x<0$ then $-x > 0$ and $0 < (-x)^2 = (-x)(-x)=-(x(-x))= -(-(x*x))=x*x=x^2$.