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Suppose we define the real numbers as an ordered field satisfying the least upper bound property. Is there only one set $\mathbf E$ that satisfies the axioms listed below?

  • For all $x\in\mathbb R$, exactly one of the following holds: $x=0$, or $x\in \mathbf E$, or $-x\in\mathbf E$.
  • If $a\in\mathbf E$ and $b\in\mathbf E$, then $a + b\in\mathbf E$.
  • If $a\in\mathbf E$ and $b\in\mathbf E$, then $a \cdot b\in\mathbf E$.

I'm interested in defining the real numbers axiomatically. It is common to state the properties of the positive real numbers as above, but I have never seen a proof that the above axioms uniquely characterise the positive real numbers. (Perhaps this is because they don't. If this is the case, then I would be interested in learning about an additional property that would uniquely characterise the set of positive reals.)

Joe
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    Does any of these answer your question? https://math.stackexchange.com/q/2237833/42969, https://math.stackexchange.com/q/3248955/42969. – Martin R Sep 15 '21 at 07:37
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    @MartinR: Thanks for the links. Unfortunately, none of them answer my question because I am interested in whether there is only one subset of $\mathbb R$ satisfying the trichotomy law, and closure under multiplication and addition. The only reason I mentioned how I was defining the real numbers was so that any potential answerer would know what they are allowed to assume. – Joe Sep 15 '21 at 13:00
  • There are many subfields of $\mathbb {R} $ which satisfy the order properties. Two famous ones are $\mathbb{Q} $ and the field of real algebraic numbers. – Paramanand Singh Sep 15 '21 at 13:51
  • Also given any real algebraic number $a$ one can order $\mathbb {Q} (a) $. – Paramanand Singh Sep 15 '21 at 13:53

1 Answers1

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For $x>0$ define $\sqrt{x}$ to be the least upper bound of the set $\{y|y^2\leq x\}$.

Suppose we have a set $E$ satisfying your axioms.

If $x>0$ then $$x=(\sqrt{x})(\sqrt{x})=(-\sqrt{x})(-\sqrt{x}).$$
Either $ \sqrt{x}\in E$ or $-\sqrt{x}\in E$ so either way $x\in E$.

Conversely if $x<0$ then $-x\in E$ so $x\notin E$.

Thus $E$ is just the positive reals.

tkf
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