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Let $a$ and $b$ be elements in an ordered field, prove that if $a \ge c$ for every $c$ such that $c \lt b$, then $a\ge b$.

My proof idea below:

Let $S = \{x | x<b\}$. Then $a$ is an upper bound for $S$. If I can show that $b$ is the least upper bound for $S$, then it follows from the definition of least upper bound that $a\ge b$.

However, I have a hard time proving the claim that $b$ is the least upper bound for $S$. Am I on the right direction? Can anyone help? Thank you.

Huiwen Zheng
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  • There has been some back and forth in the problem statement, but now I am absolutely sure the problem statement is correct. – Huiwen Zheng Oct 18 '16 at 19:37
  • You don't know that the field has the least upper bound property. – fleablood Oct 18 '16 at 19:57
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    @fleablood That is irrelevant. In an ordered field, sets of the form ${ x : x < b}$ always have a least upper bound (namely $b$). A proof of that fact follows from marty cohen's answer. – Daniel Fischer Oct 18 '16 at 20:14

5 Answers5

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Suppose $a < b$. Let $d = b-a$, so that $d > 0$.

Let $c = b-d/2$. Then $c < b$, but $c = b-d/2 = b-d+d/2 =a+d/2 > a $ which contradicts the assumption that $a \ge c$ for every $c < b$.

Therefore $a \ge b$.

marty cohen
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  • My worry about this approach is that the problem statement only mentions that a, b are elements in an ordered field. I don't know if I can assume the operation "divide an element by 2" is valid for all ordered fields. – Huiwen Zheng Oct 18 '16 at 01:23
  • @marty cohen That is such a nice proof. Really it relies on the fact that it is for all c, that c < b. You cannot have an a$\lt$ b and remain greater than or equal to all c, because you can always find a larger c, correct? – RJM Oct 18 '16 at 01:23
  • @Huiwen Zheng wouldn't $\mathbb{Q}$ be an ordered field with out the least upper bound property? The field axioms would support the division by 2, though. – RJM Oct 18 '16 at 01:33
  • @RobertJMcGinness you are definitely correct, thank you! – Huiwen Zheng Oct 18 '16 at 01:42
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    @TheGreatDuck This answer is correct. The question has from the beginning contained "for every $c$". In an ordered field $F$ with $a,b \in F$, we have $\bigl((\forall c \in F)(c < b \implies c \leqslant a)\bigr) \implies (b \leqslant a)$. – Daniel Fischer Oct 18 '16 at 11:58
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    @TheGreatDuck I have no idea what you're trying to say. You have the wrong inequality between $c$ and $a$, it always was $a \geqslant c$ or $a > c$ in the question. And the implication was always "if $a \geqslant c$ (or later $a > c$) for every $c$ such that $c < b$". – Daniel Fischer Oct 18 '16 at 14:49
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This post is intended to point out a flaw in an earlier version of the question by providing a counter-example to an impossible proof.

Let $a = b^-$, where $b^- = c$ and c is the surreal number infinitely close to b such that no number is between c and b and c < b. Then a < b. Conjecture contradicted. Please think more carefully next time.

user64742
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  • it makes sense. But in fact this is one of the problems on my midterm and I am pretty sure I didn't copy and paste it wrongly. I will wait and see some more suggestions. But thank you. – Huiwen Zheng Oct 18 '16 at 01:02
  • I definitely agree you are stating a truth, I will go back and check with my instructor. – Huiwen Zheng Oct 18 '16 at 01:51
  • @TheGreatDuck I see the question was edited. However, consider the following: ($\exists a \ge c, \forall c \lt b$) $\implies a \ge b$. Choose a, such that a = c for some c < b. Let $c_1 = \frac{c+b}{2}$. $c_1 \gt c$ and $a \lt c_1$, thus there exists a number in the ordered field less than b, that does not equal a. Therefore it is absurd to assume $\exists a = c, \forall c \lt b$. The assumption is false, so the implication is vacuously true. It relies on the "for all." So your statement is not true for all c, which implies $a \ge b$. – RJM Oct 18 '16 at 02:42
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    This is plain wrong. If $b,c$ are elements of an ordered field with $c < b$, then there are infinitely many $x$ in the field with $c < x < b$. (Ordered fields have characteristic $0$.) – Daniel Fischer Oct 18 '16 at 11:56
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As noted above, $a$ must be greater than $c$ for this to possibly be provable. Possibly check with your instructor? Typos do happen...

Joel Eliason
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  • Yes, I checked with my instructor and I've edited the question. – Huiwen Zheng Oct 18 '16 at 01:55
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    @HuiwenZheng It is sufficient to have the nonstrict inequality, since we're talking about an ordered field. If we were talking about an arbitrary ordered set, we'd need the strict inequality (or some condition guaranteeing that $b$ has no immediate predecessor, then the nonstrict inequality suffices). – Daniel Fischer Oct 18 '16 at 12:00
  • @DanielFischer yes, I double checked with my instructor, the problem statement is correct – Huiwen Zheng Oct 18 '16 at 19:33
  • @DanielFischer ok, that makes more sense. – Joel Eliason Oct 18 '16 at 21:06
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Let $S=\{c:c<b\} .$ By hypothesis, $\forall c\in S\;(c<a).$ So if $a\in S$ then $a<a,$ which cannot be, because "$<$" is irreflexive. The whole field is equal to $S\cup \{b\} \cup \{d:d>b\} $ because "$<$" satisfies trichotomy. Since $a\not \in S $ we have $a\in \{b\}\cup \{d:d>b\}.$ QED.

Note that this applies to any linearly ordered set.

DanielWainfleet
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think about the definition of a ordered field. Since a and b are elements of the field, (b-a) is in F. thus (b-a)'s multiplicative inverse is in F. call this inverse x. we get x(b-a) = e. xb-xa=e. note that e is not the smallest elements of the ordered field, why? pick y>e, then y inverse must be smaller than e (this may not be true in a general field, but should be true in an ordered field, by the properties of the ordered field.) then from here, you can construct an element between a and b and the conclusion will follow. This is what i would do. but not 100% sure tho.

Xinzhe Zuo
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