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?


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.

  • $\begingroup$ It may be. I'm not sure what your notation of F=[k,K]$ means or what your definition of "bounded" is. If "bounded" means that there is a $K \in F$ for which $x \le K$ for all $x\in K$ then that would mean $K+ 1 \le K$ which would be $1 \le 0$ which is a contradiction. $\endgroup$ – fleablood Feb 29 '20 at 18:18
  • $\begingroup$ Uh let me update the post so my notation is way clearer. $\endgroup$ – Abhi Feb 29 '20 at 18:21

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.

  • $\begingroup$ That's a very nice answer. Would my proof, as it stands, work though? I mean, I guess I could've shortened it but probably could work, right? $\endgroup$ – Abhi Feb 29 '20 at 18:04
  • $\begingroup$ @AbhijeetVats It could work, but I'd make the axioms and other facts you're using more explicit.. $\endgroup$ – Henno Brandsma Feb 29 '20 at 18:05
  • $\begingroup$ Ah alright, that's fair enough. $\endgroup$ – Abhi Feb 29 '20 at 18:05
  • $\begingroup$ $0\le 1$ is not an axiom of ordered fields. $0\ne 1$ is either a definition of fields or it is easily proven $0=1$ will yield a trivial field with one element. $x^2 \ge 0$ is a proposition of fields that must be proven. Thus if you assume an ordered field has more than one element then $1=1^2\ge 0$ and $1\ne 0$. (Hypothetically, A trivial field with only one element, $1=0$ is ordered.) $\endgroup$ – fleablood Feb 29 '20 at 18:29
  • $\begingroup$ @fleablood I don't allow $0=1$ for fields. Maybe it's a Dutch thing :)? $x^2 \in P$ is an axiom on the Wikipedia page. $\endgroup$ – Henno Brandsma Feb 29 '20 at 18:31

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.

  • $\begingroup$ In fact $0 < 1$ in an ordered field is true; even an axiom.. $\endgroup$ – Henno Brandsma Feb 29 '20 at 17:51
  • $\begingroup$ @HennoBrandsma Is it? Oops. $\endgroup$ – Don Thousand Feb 29 '20 at 17:58
  • $\begingroup$ @HennoBrandsma According to wikipedia, it is not an axiom. Although there is no interesting structure that would be barred if it was an axiom. $\endgroup$ – Don Thousand Feb 29 '20 at 18:00
  • $\begingroup$ For $1 \in F$, $1^2=1 \in P$ and $P$ is the set of positive elements.. $\endgroup$ – Henno Brandsma Feb 29 '20 at 18:02
  • $\begingroup$ You can prove that $1 > 0$. The field axioms (at least as far as I've encountered them) just state that $1 \neq 0$. In any case, thanks for the additional answer. I'll definitely upvote it. $\endgroup$ – Abhi Feb 29 '20 at 18:03

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.


[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 $0 < (-x)^2 = (-x)(-x)=-(x(-x))= -(-(x*x))=x*x=x^2$.

  • $\begingroup$ Oh nice. That's pretty cool. It seems like I can do the shorter proofs for the simpler problems but my proofs become absurdly convoluted for the harder ones. I'm just starting out in analysis and linear algebra, so I'm not sure how i can fix that. $\endgroup$ – Abhi Feb 29 '20 at 18:55
  • 1
    $\begingroup$ That's pretty common as when one is started one doesn't know what can and can't be assumed. In which case, it's better to prove too much than to assume what you are not allowed to assume. But sometimes the subtle (such as something isn't true for arbitrary $x$ is true for all $x$. So if arbitrary $x$ is not an upper bound because specific $x+1>x$, then nothing is an upper bound) can be hard to feel "right"...("yeah, that $x$ isn't an upper bound but what about something else....") $\endgroup$ – fleablood Feb 29 '20 at 19:06

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