Proof of Arbitrary Positive Slack, and Approximation of Suprema and Infima, in an Ordered Field

The order of $F$ is a total order (Ordered Fields §ordered-field). We use Rules of Arithmetic and Order in an Ordered Field and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares for $F$.

Slack above. Suppose $x\le y+\varepsilon$ for every positive $\varepsilon$, and suppose that $x\le y$ fails. Then $y<x$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Adding $-y$ gives $0<x-y$ (Rules of Arithmetic and Order in an Ordered Field §order-sum). Let $\varepsilon=(x-y)/2$; by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving, $0<\varepsilon<x-y$. Adding $y$ gives $y+\varepsilon<(x-y)+y=x$ (Rules of Arithmetic and Order in an Ordered Field §order-sum; as $(x-y)+y=x$). Together with $x\le y+\varepsilon$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $x<x$, contradicting Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive. Hence $x\le y$.

Slack below. Suppose $y-\varepsilon\le x$ for every positive $\varepsilon$. Adding $\varepsilon$ gives $y=(y-\varepsilon)+\varepsilon\le x+\varepsilon$ (Rules of Arithmetic and Order in an Ordered Field §order-sum; as $(y-\varepsilon)+\varepsilon=y$). So $y\le x$ by Slack above, with the roles of $x$ and $y$ exchanged.

Vanishing. Suppose $0\le x$ and $x\le\varepsilon=0+\varepsilon$ for every positive $\varepsilon$. Then $x\le0$ by Slack above with $y=0$, so $x=0$ by antisymmetry (Partial and Total Orders on a Set and the Associated Strict Relation §partial).

Strict above. Let $S$ have a supremum and let $b<\sup S$. Suppose there were no $s\in S$ with $b<s$. Then for every $s\in S$, $b<s$ fails, so $s\le b$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Thus $b$ is an upper bound of $S$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds). Since $\sup S$ is the least element of the set of upper bounds (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum), $\sup S\le b$. From $b<\sup S$ and $\sup S\le b$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $b<b$, contradicting Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive.

Epsilon above. Let $0<\varepsilon$. Adding $-\varepsilon$ to $0<\varepsilon$ gives $-\varepsilon<0$ (Rules of Arithmetic and Order in an Ordered Field §order-sum), so $\sup S-\varepsilon<\sup S+0=\sup S$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Apply Strict above with $b=\sup S-\varepsilon$.

Strict below. Let $S$ have an infimum and let $\inf S<b$. Suppose there were no $s\in S$ with $s<b$. Then for every $s\in S$, $s<b$ fails, so $b\le s$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Thus $b$ is a lower bound of $S$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds). Since $\inf S$ is the greatest element of the set of lower bounds (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum), $b\le\inf S$. Then Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $b<b$, a contradiction.

Epsilon below. Let $0<\varepsilon$. Then $\inf S=\inf S+0<\inf S+\varepsilon$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Apply Strict below with $b=\inf S+\varepsilon$.