Is $[0,\infty)$ the maximum candidate in $\mathbb{R}$ for the positive numbers?

Question

Let us regard $0$ as a positive number, at least for this question, and define that a subset $S$ of a ring $R$ is a candidate for the positive numbers, or simply a "candidate," iff

1. $S$ is a semiring: that is, $0 \in S$ and $S$ is closed under sums and products.

2. $x+y = 0$ implies $x = 0$ and $y = 0$, for all $x,y \in S$.

For example, taking $\mathbb{Z}$ as our ring $R$, we have that for all $n \in \mathbb{N},$ the following are candidates.

1. $\{m \in \mathbb{Z} \mid m \geq n\} \cup \{0\}$

2. $n \mathbb{N}.$

Furthermore, $\mathbb{N}$ is the maximum candidate.

Proof. Suppose towards a contradiction that some candidate $S \subseteq \mathbb{Z}$ has an element not in $\mathbb{N},$ call it $k$. Begin by observing the following.

1. Since $k \in S$, therefore $k^2 \in S$ (closure under products). Note also that $k^2$ is non-zero.

2. On the other hand, we have that since $k \in S$, therefore $-k^2 \in S,$ because $-k^2$ can be expressed as $|k|k,$ which is just a finite sum $k + \cdots + k$ (closure under sums).

Putting these together, we conclude that $k^2 = 0$ (since $x+y = 0$ implies $x=0$ for all $x,y \in S$). But this contradicts the observation that $k^2$ is non-zero.

Question. With a bit of prelude, the above argument basically works in $\mathbb{Q}$ as well. This reveals that $\mathbb{Q} \cap [0,\infty)$ is the maximum candidate in $\mathbb{Q}.$ Is $[0,\infty)$ the maximum candidate in $\mathbb{R}$? Does it even have one?

Answer 1

Your definition of a candidate is not dependent on $R$ at all.

If $t$ is an abstract transcendental element over $\mathbb{Q}$ then the polynomial semiring $\mathbb{N}[t]$ is a candidate. So if we choose a negative transcendental real value for $t$, say $-\pi$, then $\mathbb{N}[-\pi]$ is also a candidate.

And clearly $\mathbb{R}^+$ is a maximal candidate in $\mathbb{R}$ and does not contain $-\pi$. So there is no maximum candidate.

NB: It is not the case that we can extend $\mathbb{N}$ with any negative irrational element. Suppose, for example, $S$ contains $-\sqrt[3]{2}$. Then as $S$ is closed under multiplication, it also contains $-\sqrt[3]{2} \times -\sqrt[3]{2} \times -\sqrt[3]{2} =-2$, and then axiom 2 is violated.

(I would guess that negative square roots, for example, would be permissible, but haven't worked this out fully, so I don't know what the strongest available result here is. All transcendental numbers are OK, though.)