3

Suppose we claim that if there is a set $E := \{a \in \mathbb{R} : a < \epsilon, \forall \epsilon \in \mathbb{Q}^{+} \}$, then it must be true that $a \leq 0$.

I aim to prove this using only the ordered field axioms and the completeness axiom (as per the "request" by my professor).

I am going to prove it by contradiction. So suppose $\exists a>0$ so that $0 < a < \epsilon, \forall \epsilon \in \mathbb{Q}^{+}$. Now since $a < \epsilon, \forall \epsilon \in \mathbb{Q}^{+}$, we have $a < \frac{\epsilon}{2}$ since $\frac{\epsilon}{2} \in \mathbb{Q}^{+}$. So $2a < \epsilon$. It follows that we can always find $a' \in E$ such that $a<a'$ for any $a \in E$.

Now here is my problem. I know that the last sentence in the previous paragraph somehow contradicts the completeness axiom. But I cannot see why.

Do you have any suggestions?

  • The claim in your first sentence does not make any sense. What $a$ is it that you are claiming to be $\leq0$? – Mariano Suárez-Álvarez Sep 10 '15 at 04:11
  • As a side note, the last sentence in your proof attempt is ambiguous. It could easily be interpreted as the (true) statement that $$\forall a\in E\cap(0,\infty),\exists a'\in E:a<a'$$ or the (false) statement that $$\exists a'\in E:\forall a\in E\cap(0,\infty),a<a'.$$ Also, in your first sentence, I think you mean to say that "if $a\in E,$ then it must be true that $a\le 0.$" – Cameron Buie Sep 10 '15 at 04:11
  • @MarianoSuárez-Alvarez I assumed he meant if $a \in E$ then $a \le 0$. – Rudy the Reindeer Sep 10 '15 at 04:16

2 Answers2

1

The completeness axiom states that any set that's bounded above has a least upper bound.

You use this to produce the contradiction as follows:

Let $a$ be the least upper bound of $E$. You claim $a \le 0$. By contradiction assume that it is not. Then there exists a $q \in \mathbb Q$ such that $0<q<a$ since between any two real numbers there exists a rational number (this is a consequence of the completeness axiom).

Consider any $x \in E$. Then we have $x < q$ since $q > 0$. But then $q$ is an upper bound of $E$ and furthermore smaller than $a$. This contradicts $a$ being the least upper bound. Hence $a$ must be less than or equal to $0$.

Rudy the Reindeer
  • 46,359
0

The contradiction you are looking for is as follows : Let x= $\text{lub}(E)$. If $E$ has a positive member then $x>0$.Now if $x\ge q$ for any $q \in Q^+$ then $x/2 \ge q/2 >y$ for all $y \in Q^+$, so $x/2$ is an upper bound for $E$ and is less than $\text{lub}( E)$, which is absurd, so we must have $x<q$ for all $q \in Q^+$. Therefore $x \in E$. But as you have shown,$ 0 <x \in E \implies 2 x \in E$. But then $x$ is not $\text{lub}( E)$ because $x$ is not an upper bound for $E$ because $E$ has a member $(2x)$ which is greater than $x$. (Equivalently we could say $ 0<x=\text{lub}( E)$ and $2x \in E$ implies $0< 2 x \le \text{lub}( E)= x$ which is absurd.

DanielWainfleet
  • 57,985
  • We can extend the reals to an ordered field F that has a positive element p which is less than every positive rational. And 1/p is larger than every rational. But F cannot have the completeness property, and the sequence 1/2,1/3.1/4,... has no definable limit in F. – DanielWainfleet Sep 10 '15 at 05:00