Punctured Neighbourhood¶

Definition¶

Let $X\subseteq\mathbb{R}$ and let $x_0 \in \mathbb{R}\text{.}$

The point $x_0$ is a limit point of $X$ if, for every $\delta>0\text{,}$ there is an $x\in X$ with $0<|x-x_0|<\delta\text{.}$ If $x_0\in X$ is not a limit point of $X\text{,}$ it is an isolated point of $X\text{.}$

For $r>0\text{,}$ the set $(x_0-r,\,x_0+r)=\{x\in\mathbb{R}\colon |x-x_0|<r\}$ is a neighbourhood of $x_0\text{.}$ Removing $x_0$ itself gives the punctured neighbourhood of $x_0\text{,}$ which is any set of the form

\{x \in \mathbb{R} \colon 0 < |x - x_0| < r\},

for some $r > 0\text{.}$

Interpretation¶

A punctured neighbourhood describes the points arbitrarily close to a given point $x_0$ while excluding $x_0$ itself. It is the set of inputs that a limit at a point looks at, so local behaviour near $x_0$ can be studied without evaluating the function at $x_0$ .

A number line with the interval from x0 minus delta to x0 plus delta and the point x0 itself removed.

In the figure the radius is $\delta$ : the punctured neighbourhood is $(x_0-\delta,x_0)\cup(x_0,x_0+\delta)$ $δ)$ , an open interval about $x_0$ with its centre removed. The absolute value expresses distance from the centre; $0<|x-x_0|$ excludes the centre and $|x-x_0|<\delta$ keeps the point inside the interval.

In these terms, $x_0$ is a limit point of $X$ when every punctured neighbourhood of $x_0$ contains a point of $X$ . This is exactly what " $x \to$ $x\to x_0$ within $X$ " needs, and $x_0$ itself need not belong to $X$ : $0$ is a limit point of $(0,1)$ , so $\lim_{x\to0}f(x)$ makes sense for a function defined on $(0,1)$ . By contrast, every integer $n$ is an isolated point of $\mathbb{Z}$ , since $(n-\tfrac{1}{2},n)\cup(n,n+\tfrac{1}{2})$ contains no integer; for a function defined only on $\mathbb{Z}$ the limit at a point is therefore not defined at all, and only the limit as $n\to\infty$ of a sequence remains.

Limit points return in Math III under the name accumulation points, defined in the same way for more general sets.

Key Results¶

The definition of a limit at a point requires $x_0$ to be a limit point of the domain, which is exactly what makes " $x$ approaches $x_0$ " meaningful.
The limit laws for functions are stated for functions defined on a punctured neighbourhood of $x_0\text{.}$
Excluding the centre point avoids any reliance on the value $f(x_0)\text{.}$
A function is never differentiable at an isolated point of its domain.

Examples¶

$(-1,0)\cup(0,1)$ is a punctured neighbourhood of the origin.
Every point of $[0,1]$ is a limit point of $[0,1]\text{,}$ and so is every point of $(0,1)\text{.}$
$3$ is an isolated point of $[0,1]\cup\{3\}\text{.}$

Prerequisites¶

One-Sided Limit

Sources¶

Math I: Calculus · Lecture 5

Aliases: punctured neighborhood, deleted neighbourhood, deleted neighborhood Keywords: limit point, isolated point, neighbourhood Symbols: delta, element of, infinity, integers, real numbers, subset, union

Last updated: 20 October 2026 03:02