A bounded set is a set whose elements stay within a finite distance of a chosen reference point. This is one of the first places where the language of [metric spaces](/page/Metric%20Space) turns geometric intuition into a reusable definition: an interval may have endpoints or fail to have endpoints, a subset may be open or closed, but boundedness asks a separate question about how far the set can spread. In real analysis, boundedness is the hypothesis that prevents sequences, functions, and domains from escaping to infinity while other limiting processes are studied.

Boundedness is not a topological notion by itself. It depends on a metric or a norm, so it belongs naturally beside [normed vector spaces](/page/Normed%20Vector%20Space), compact sets, and the study of [sequences](/page/Sequence). In Euclidean space it is tied to compactness through the [Heine-Borel theorem](/theorems/309); in general metric spaces it separates from compactness and [total boundedness](/page/Total%20Boundedness), making it a useful test case for understanding which results are genuinely Euclidean.

## Definition

The basic problem is to say that a set has finite extent without referring to coordinates or to an ambient line. A metric provides the only structure needed: it lets us measure how far a point of the set is from a reference point. For a point $x_0\in X$ and a radius $R>0$, the notation $B(x_0,R)$ means the open metric ball

\begin{align*} B(x_0,R)=\{x\in X:d(x,x_0)<R\}. \end{align*}

[definition: Bounded Set in a Metric Space] Let $(X,d)$ be a metric space and let $A \subset X$. The set $A$ is bounded in $(X,d)$ if $A=\varnothing$, or if there exist $x_0 \in X$ and $R > 0$ such that \begin{align*} A \subset B(x_0,R). \end{align*} [/definition]

Most applications in elementary analysis take place in $\mathbb{R}^n$, where distances are computed from the Euclidean norm. A separate Euclidean formulation is needed to make the estimate usable in coordinates, in inequalities, and in comparisons with balls and boxes. The next definition also fixes the convention that bounded means bounded for the usual [Euclidean metric](/page/Euclidean%20Metric) unless another metric is named.

[definition: Bounded Set in Euclidean Space] Let $A \subset \mathbb{R}^n$. The set $A$ is bounded if there exist $x_0 \in \mathbb{R}^n$ and $R>0$ such that \begin{align*} |x-x_0| < R \qquad \text{for every } x \in A. \end{align*} [/definition]

The reference point is often taken to be the origin. That choice is convenient for computations, but it is not part of the concept. To compare different choices of centre and to state size estimates without naming any centre at all, it is useful to measure the largest distance between two points of the set.

[definition: Diameter] Let $(X,d)$ be a metric space. Write $\mathcal{P}(X)$ for the power set of $X$, the collection of all subsets of $X$. The diameter functional is the map \begin{align*} \operatorname{diam}:\mathcal{P}(X)\to[0,\infty] \end{align*} defined by \begin{align*} \operatorname{diam}(A)=\sup\{d(x,y):x,y\in A\}, \end{align*} with the convention \begin{align*} \operatorname{diam}(\varnothing):=0. \end{align*} [/definition]

With this convention, a subset has finite diameter exactly when this functional takes a finite value on it. The empty set is assigned diameter $0$ so that finite-size statements do not need to exclude it separately.

The definitions above can be checked immediately on the sets that first motivate the idea. Bounded intervals fit inside large balls, while rays do not.

[example: Interval and Ray in the Real Line] In $\mathbb{R}$ with its usual metric, the interval $(0,1)$ is bounded. If $x\in(0,1)$, then $0<x<1$, so \begin{align*} |x-0|=x<1<2. \end{align*} Thus every $x\in(0,1)$ lies in $B(0,2)$, and hence \begin{align*} (0,1)\subset B(0,2). \end{align*} By contrast, the ray $[0,\infty)$ is unbounded. To see this, let $x_0\in\mathbb{R}$ and $R>0$ be arbitrary, and set \begin{align*} y=|x_0|+R+1. \end{align*} Then $y\ge 0$, so $y\in[0,\infty)$. If $x_0\ge 0$, then $|x_0|=x_0$, and \begin{align*} |y-x_0|=|x_0+R+1-x_0|=R+1>R. \end{align*} If $x_0<0$, then $|x_0|=-x_0$, and \begin{align*} |y-x_0|=|-x_0+R+1-x_0|=R+1-2x_0>R. \end{align*} In either case, $y\notin B(x_0,R)$. Therefore no ball in $\mathbb{R}$ contains all of $[0,\infty)$, so the ray is unbounded. This example separates finite extent from the presence of endpoints. [/example]

Limits in analysis are usually expressed through sequences, so size control for a sequence must mean more than each individual term being a point of the space. The relevant question is whether all terms live in a single bounded region. This definition lets boundedness enter compactness criteria and convergence arguments for sequences.

[definition: Bounded Sequence] Let $(X,d)$ be a metric space and let $(x_k)_{k=1}^{\infty}$ be a sequence in $X$. The sequence $(x_k)_{k=1}^{\infty}$ is bounded if the set \begin{align*} \{x_k:k\in \mathbb{N}\} \end{align*} is a bounded subset of $X$. [/definition]

For real-valued functions, the object whose size matters is usually the image of the domain. A function may be defined on a complicated set, but boundedness asks whether all its output values fit into a finite interval.

[definition: Bounded Function] Let $X$ be a set and let $f:X\to \mathbb{R}$ be a function. The function $f$ is bounded on $X$ if the set $f(X)\subset \mathbb{R}$ is bounded. [/definition]

For functions into $\mathbb{R}^m$ or a normed [vector space](/page/Vector%20Space), the same definition is used with the metric induced by the norm on the codomain. A quick example shows why the definition looks at the image rather than at the domain.

[example: Bounded and Unbounded Images on the Same Domain] Let $X=(0,1)\subset\mathbb{R}$. Define $f:X\to\mathbb{R}$ by $f(x)=x$, and define $g:X\to\mathbb{R}$ by $g(x)=1/x$. We first compute the image of $f$. If $x\in X$, then $0<x<1$ and $f(x)=x$, so $f(x)\in(0,1)$. Conversely, if $y\in(0,1)$, then $y\in X$ and $f(y)=y$, so $y\in f(X)$. Hence \begin{align*} f(X)=(0,1). \end{align*} For every $y\in f(X)$, we have $0<y<1$, and therefore \begin{align*} |y-0|=y<1<2. \end{align*} Thus $f(X)\subset B(0,2)$, so $f$ is bounded on $X$. Now consider $g$. To show that $g(X)$ is not bounded, let $x_0\in\mathbb{R}$ and $R>0$ be arbitrary. Set \begin{align*} t=|x_0|+R+1. \end{align*} Then $t>1$, so $x=1/t$ satisfies $0<x<1$ and hence $x\in X$. For this $x$, \begin{align*} g(x)=\frac{1}{1/t}=t=|x_0|+R+1. \end{align*} Since $x_0\le |x_0|$, we have \begin{align*} g(x)-x_0=|x_0|+R+1-x_0\ge R+1>R. \end{align*} In particular $g(x)-x_0>0$, so \begin{align*} |g(x)-x_0|=g(x)-x_0>R. \end{align*} Therefore $g(x)\notin B(x_0,R)$. Since no choice of $x_0$ and $R$ captures all of $g(X)$, the function $g$ is not bounded on $X$. Thus boundedness of a real-valued function is a condition on its output values, not merely on the size of its domain. [/example]

## Equivalent Characterisations