[proofplan] We prove the two implications separately. If $f$ is bounded, then the definition gives a uniform absolute-value bound $M\ge 0$, and the elementary order property of the absolute value converts this into the two-sided bound $-M\le f(x)\le M$. Conversely, if $f$ is trapped between two [real numbers](/page/Real%20Numbers) $a$ and $b$, then the single real number $M=\max\{|a|,|b|\}$ bounds $|f(x)|$ for every $x\in X$, so $f$ is bounded. [/proofplan]

[step:Convert an absolute-value bound into two-sided order bounds]Assume that $f:X\to\mathbb R$ is bounded. By the definition of boundedness for a real-valued function, there exists a real number $M\ge 0$ such that, for every $x\in X$, \begin{align*} |f(x)|\le M. \end{align*} For any fixed $x\in X$, the real-order characterisation of absolute value gives \begin{align*} -M\le f(x)\le M. \end{align*} Define $a:=-M\in\mathbb R$ and $b:=M\in\mathbb R$. Then, for every $x\in X$, \begin{align*} a\le f(x)\le b. \end{align*} Thus the required two-sided real bounds exist.[/step]

[guided]Assume that $f:X\to\mathbb R$ is bounded. The definition of boundedness for a real-valued function means that there is one real number $M\ge 0$ which controls the absolute value of $f(x)$ uniformly over all points of $X$. Explicitly, there exists $M\in\mathbb R$ with $M\ge 0$ such that, for every $x\in X$, \begin{align*} |f(x)|\le M. \end{align*} Now fix an arbitrary point $x\in X$. Since $f(x)\in\mathbb R$, the elementary order property of the absolute value says that $|f(x)|\le M$ is equivalent to the two inequalities \begin{align*} -M\le f(x)\le M. \end{align*} Because the point $x$ was arbitrary, the same two-sided estimate holds for every $x\in X$. Define the lower and upper bounds by \begin{align*} a:=-M \end{align*} and \begin{align*} b:=M. \end{align*} Then $a,b\in\mathbb R$, and for every $x\in X$, \begin{align*} a\le f(x)\le b. \end{align*} This proves the forward implication.[/guided]

[step:Convert two-sided order bounds into an absolute-value bound]Assume that there exist $a,b\in\mathbb R$ such that, for every $x\in X$, \begin{align*} a\le f(x)\le b. \end{align*} Define \begin{align*} M:=\max\{|a|,|b|\}. \end{align*} Then $M\in\mathbb R$ and $M\ge 0$. Let $x\in X$. Since $a\le f(x)\le b$, we have $f(x)\le b\le |b|\le M$. Also $a\le f(x)$ implies $-f(x)\le -a\le |a|\le M$. Therefore both $f(x)\le M$ and $-f(x)\le M$ hold, so \begin{align*} |f(x)|\le M. \end{align*} Because $x\in X$ was arbitrary, $|f(x)|\le M$ for every $x\in X$. Hence $f$ is bounded.[/step]

[guided]Assume that there are real numbers $a,b\in\mathbb R$ such that, for every $x\in X$, \begin{align*} a\le f(x)\le b. \end{align*} To prove boundedness, we need a single non-negative real number that bounds $|f(x)|$ uniformly for all $x\in X$. The lower bound $a$ controls how negative $f(x)$ can be, while the upper bound $b$ controls how positive $f(x)$ can be. A single number that controls both sides is \begin{align*} M:=\max\{|a|,|b|\}. \end{align*} Then $M\in\mathbb R$, $M\ge 0$, $|a|\le M$, and $|b|\le M$. Now fix an arbitrary point $x\in X$. From the upper bound $f(x)\le b$, and from the inequality $b\le |b|$, we get \begin{align*} f(x)\le b\le |b|\le M. \end{align*} From the lower bound $a\le f(x)$, multiplying by $-1$ reverses the inequality and gives $-f(x)\le -a$. Since $-a\le |a|$, we obtain \begin{align*} -f(x)\le -a\le |a|\le M. \end{align*} Thus both $f(x)\le M$ and $-f(x)\le M$ hold. Equivalently, the real number $f(x)$ satisfies \begin{align*} |f(x)|\le M. \end{align*} Because $x\in X$ was arbitrary, this estimate holds for every $x\in X$. Therefore $f$ is bounded by the definition of boundedness for real-valued functions.[/guided]

[step:Account for the empty domain] If $X=\varnothing$, the universal statements over $X$ used above are vacuous. In particular, the forward implication still follows from any boundedness witness, and the reverse implication still follows from the displayed construction whenever two-sided bounds are given. Also, the existential two-sided condition is satisfiable directly by choosing $a:=0$ and $b:=0$. Thus the equivalence holds for all sets $X$, including the empty set. [/step]