Proof of Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers

Each result cited is universally quantified over the data in its own statement.

By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, $\mathbb{R}$ is an ordered field, so its order is a total order (Ordered Fields §ordered-field). We use Rules of Arithmetic and Order in an Ordered Field for $\mathbb{R}$.

Slack above. Suppose $x\le y+\varepsilon$ for every real $\varepsilon>0$, 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 real $\varepsilon>0$. Adding $\varepsilon\le\varepsilon$ 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 real $\varepsilon>0$. 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 $S$ be bounded above. As $S$ is a nonempty subset of $\mathbb{R}$ that is bounded above, it has a supremum by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum, so $\sup S$ is defined. Let $b\in\mathbb{R}$ with $b<\sup S$. 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 $S$ be bounded above, so that $\sup S$ is defined as in Strict above, and let $\varepsilon$ be real with $0<\varepsilon$. 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 $S$ be bounded below. As $S$ is a nonempty subset of $\mathbb{R}$ that is bounded below, it has an infimum by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §infimum, so $\inf S$ is defined. Let $b\in\mathbb{R}$ with $\inf S<b$. 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), $b\le\inf S$. From $b\le\inf S$ and $\inf S<b$, 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 below. Let $S$ be bounded below, so that $\inf S$ is defined as in Strict below, and let $\varepsilon$ be real with $0<\varepsilon$. 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 $b=\inf S+\varepsilon$.