Physlib.ProbabilisticTheory.OrderUnit.Archimedean

The real numbers $\mathbb{R}$ constitute an Archimedean order-unit space. In this structure, the distinguished order unit is $1 \in \mathbb{R}$, and the ordering is the standard less-than-or-equal-to relation $\le$. It satisfies the following axioms: 1. The unit is non-negative: $1 \ge 0$. 2. For every $x \in \mathbb{R}$, there exists a natural number $n$ such that $x \le n \cdot 1$. 3. The Archimedean property holds: if $x \in \mathbb{R}$ satisfies $x \le \epsilon \cdot 1$ for all real $\epsilon > 0$, then $x \le 0$.

For an element $A$ in an order-unit space $E$, the set of order-unit bounds is the set of all non-negative real numbers $r \in \mathbb{R}$ such that $-r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}$, where $\mathbf{1}$ denotes the distinguished order unit of $E$.

For an element $A$ in an Archimedean order-unit space $E$ with a distinguished order unit $\mathbf{1}$, the order-unit norm $\|A\|_1$ is defined as the infimum of the set of order-unit bounds of $A$: $$\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$$

Let $E$ be an Archimedean order-unit space with distinguished order unit $\mathbf{1}$. For any element $A \in E$, the set of order-unit bounds $\{r \in \mathbb{R} \mid r \ge 0, -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}\}$ is bounded below in $\mathbb{R}$.

In an Archimedean order-unit space $E$ with a distinguished order unit $\mathbf{1}$, let $A$ be an element of $E$. If $r \in \mathbb{R}$ is an order-unit bound of $A$ (that is, $r \ge 0$ and $-r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}$), then the order-unit norm $\|A\|_1$ satisfies $\|A\|_1 \le r$.

Let $E$ be an order-unit space with a distinguished order unit $\mathbf{1}$. For any element $A \in E$, the set of order-unit bounds, defined as $\{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}\}$, is non-empty.

Let $E$ be an Archimedean order-unit space. For any element $A \in E$, its order-unit norm $\|A\|_1$ is non-negative, i.e., $0 \le \|A\|_1$. This follows from the fact that $\|A\|_1$ is defined as the infimum of a set of non-negative real numbers.

In an Archimedean order-unit space $E$, the order-unit norm of the zero element $0 \in E$ is $0$, i.e., $\|0\|_1 = 0$.

In an Archimedean order-unit space $E$ with order unit $\mathbf{1}$, for any element $A \in E$, the set of order-unit bounds of $-A$ is equal to the set of order-unit bounds of $A$. Here, the set of order-unit bounds of an element $X \in E$ is defined as the set of non-negative real numbers $r$ such that $-r \cdot \mathbf{1} \le X \le r \cdot \mathbf{1}$.

Let $E$ be an Archimedean order-unit space with a distinguished order unit $\mathbf{1}$. For any element $A \in E$, the order-unit norm of $-A$ is equal to the order-unit norm of $A$: $$\|-A\|_1 = \|A\|_1$$ where the order-unit norm $\|A\|_1$ is defined as the infimum of the set of real numbers $r \ge 0$ such that $-r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}$.

Let $E$ be an order-unit space with a distinguished order unit $\mathbf{1}$. For any elements $A, B \in E$ and real numbers $r, s \in \mathbb{R}$, if $r$ is an order-unit bound for $A$ and $s$ is an order-unit bound for $B$, then their sum $r + s$ is an order-unit bound for the sum $A + B$. (Note: A real number $r$ is an order-unit bound for an element $X$ if $r \ge 0$ and $-r \cdot \mathbf{1} \le X \le r \cdot \mathbf{1}$).

Let $E$ be an Archimedean order-unit space with a distinguished unit $\mathbf{1}$. For any element $A \in E$ and any real number $\varepsilon > 0$, there exists an order-unit bound $r$ for $A$ such that $r < \|A\|_1 + \varepsilon$. (Note: A real number $r$ is an order-unit bound for $A$ if $r \ge 0$ and $-r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}$. The order-unit norm $\|A\|_1$ is defined as the infimum of the set of all such bounds.)

In an Archimedean order-unit space $E$, the order-unit norm $\|\cdot\|_1$ satisfies the triangle inequality: for any elements $A, B \in E$, $$\|A + B\|_1 \le \|A\|_1 + \|B\|_1$$ where the order-unit norm $\|X\|_1$ is defined as $\inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le X \le r \cdot \mathbf{1}\}$.

Let $E$ be an Archimedean order-unit space with a distinguished unit $1$. For any real numbers $r$ and $s$, if $r \leq s$, then $r \cdot 1 \leq s \cdot 1$ holds with respect to the order on $E$.

In an Archimedean order-unit space $E$ with a distinguished unit $\mathbf{1}$, every element $A \in E$ is bounded above by its order-unit norm $\|\cdot\|_1$ scaled by the unit. Specifically, $A \le \|A\|_1 \cdot \mathbf{1}$, where $\|A\|_1$ is the order-unit norm defined as the infimum of all $r \ge 0$ such that $-r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}$.

Let $E$ be an Archimedean order-unit space with a distinguished order unit $\mathbf{1}$. For any element $A \in E$, the inequality $-\|A\|_1 \cdot \mathbf{1} \leq A$ holds, where $\|A\|_1$ denotes the order-unit norm of $A$.

For any element $A$ in an Archimedean order-unit space $E$ with a distinguished order unit $\mathbf{1}$, the order-unit norm $\|A\|_1$ is itself an order-unit bound of $A$. That is, $\|A\|_1$ satisfies the condition $-\|A\|_1 \cdot \mathbf{1} \le A \le \|A\|_1 \cdot \mathbf{1}$, where the norm is defined as $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$.

Let $E$ be an Archimedean order-unit space with a distinguished order unit $\mathbf{1}$. For any element $A \in E$ and any real number $\varepsilon$, if the order-unit norm $\|A\|_1$ is strictly less than $\varepsilon$, then $A \le \varepsilon \cdot \mathbf{1}$.

Let $E$ be an Archimedean order-unit space with order unit $\mathbf{1}$. For any element $A \in E$ and any real number $\epsilon \in \mathbb{R}$, if the order-unit norm $\|A\|_1$ of $A$ is strictly less than $\epsilon$, then $-\epsilon \cdot \mathbf{1} \le A$.

Let $E$ be an Archimedean order-unit space with order unit $\mathbf{1}$. For any element $A \in E$ and any real number $r$, the order-unit norm satisfies $\|A\|_1 \le r$ if and only if $0 \le r$ and $-r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1}$.

In an Archimedean order-unit space $E$, for any element $A \in E$, the order-unit norm $\|A\|_1$ is equal to $0$ if and only if $A = 0$. Here, the order-unit norm is defined as $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$, where $\mathbf{1}$ is the distinguished order unit.

For any element $A$ in an Archimedean order-unit space $E$ and any non-negative real number $r \ge 0$, the order-unit norm of the scalar multiplication $r \cdot A$ is bounded above by $r$ times the order-unit norm of $A$: $$\|r \cdot A\|_1 \le r \cdot \|A\|_1$$ where $\|\cdot\|_1$ denotes the order-unit norm.

Let $E$ be an Archimedean order-unit space and $\|\cdot\|_1$ be the order-unit norm defined on $E$. For any element $A \in E$ and any positive real number $r > 0$, the norm of the scalar multiple $r \cdot A$ is equal to $r$ times the norm of $A$, i.e., $\|r \cdot A\|_1 = r \|A\|_1$.

Let $E$ be an Archimedean order-unit space with order-unit norm $\|\cdot\|_1$. For any real number $r \in \mathbb{R}$ and any element $A \in E$, the order-unit norm of the scalar product $r \cdot A$ is given by $\|r \cdot A\|_1 = |r| \cdot \|A\|_1$.

Let $E$ be an Archimedean order-unit space. For any element $A \in E$ with order-unit norm $\|A\|_1 \neq 0$, there exists an element $B \in E$ such that $\|B\|_1 \le 1$ and $\|A\|_1 \cdot B = A$. Here, the order-unit norm is defined as $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$, where $\mathbf{1}$ is the distinguished order unit of the space.

For an Archimedean order-unit space $E$ with order unit $\mathbf{1}$, the order-unit norm $\|\cdot\|_1$, defined by $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$, constitutes an additive group norm on $E$. This means it satisfies the properties $\|0\|_1 = 0$, $\|A + B\|_1 \le \|A\|_1 + \|B\|_1$, $\|-A\|_1 = \|A\|_1$, and $\|A\|_1 = 0 \iff A = 0$.

Given an Archimedean order-unit space $E$ with a distinguished order unit $\mathbf{1}$, the order-unit norm $\|\cdot\|_1$, defined for any $A \in E$ as $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$, equips $E$ with the structure of a normed additive commutative group.

For an Archimedean order-unit space $E$ with a distinguished order unit $\mathbf{1}$, the order-unit norm $\|\cdot\|_1$, defined for any $A \in E$ as $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$, equips $E$ with the structure of a normed space over $\mathbb{R}$.

In an Archimedean order-unit space $E$, the positive cone $\{x \in E \mid x \ge 0\}$ is a closed set with respect to the topology induced by the order-unit norm $\|\cdot\|_1$, where $\|A\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le A \le r \cdot \mathbf{1} \}$.

Let $E$ be an Archimedean order-unit space. In the topology induced by the order-unit norm $\|\cdot\|_1$, every upper interval $[a, \infty) = \{x \in E \mid a \le x\}$ is a closed set for any $a \in E$. Here, the order-unit norm is defined as $\|x\|_1 = \inf \{r \in \mathbb{R} \mid r \ge 0 \text{ and } -r \cdot \mathbf{1} \le x \le r \cdot \mathbf{1} \}$, where $\mathbf{1}$ is the distinguished order unit of $E$.