The absolute value or modulus \(\lvert x\rvert\) of a real number \(x \in \mathbb{R}\) is \(x\) when \(x\) is positive, \(-x\) when \(x\) is negative and \(0\) when \(x\) is zero.

\[\lvert x\rvert \, := \begin{cases} -x & x<0 \\ 0 & x=0 \\ x & x>0 \end{cases}\]

Let \(x,y \in \mathbb{R}\) be real numbers. The distance \(d(x,y)\) between \(x\) and \(y\) is the absolute value of their difference.

\[d(x,y) := \lvert x-y\rvert\]

Let \(x,y,\varepsilon \in \mathbb{R}\) be real numbers and \(\varepsilon>0 .\) \(x\) and \(y\) are \(\boldsymbol{\varepsilon}\)-close if and only if \(d(x,y) \leq \varepsilon .\)

Let \(S \subseteq \mathbb{R}\) and \(x \in \mathbb{R}.\) \(x\) is an upper bound of \(S\) if \(s \leq x \, \forall s \in S\) and \(x\) is a lower bound of \(S\) if \(x \leq s \, \forall s \in S.\)

Let \(S \subseteq \mathbb{R}\) and \(\alpha \in \mathbb{R}.\) \(\alpha\) is the supremum or least upper bound of \(S\) if \(\alpha\) is an upper bound of \(S\) and for all upper bounds \(z\) of \(S,\) \(\alpha \leq z.\) \(\alpha\) is the infimum or greatest lower bound of \(S\) if \(\alpha\) is a lower bound of \(S\) and for all lower bounds \(z\) of \(S,\) \(z \leq \alpha.\)

Let \(S \subseteq \mathbb{R}\) and \(\alpha \in \mathbb{R}.\) \(\alpha\) is the maximum of \(S\) if \(\alpha \in S\) and \(s \leq \alpha \, \forall s \in S.\) \(\alpha\) is the minimum of \(S\) if \(\alpha \in S\) and \(\alpha \leq s \, \forall s \in S.\) If \(S=\varnothing\) or \(S\) is not bounded above then the maximum does not exist, and if \(S=\varnothing\) or \(S\) is not bounded below then the minimum does not exist. A non-empty bounded set need not have a maximum or minimum either; for example, the open interval \((0,1)\) has neither.