The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$

Throughout, $\mathbb{R}$ is the complete ordered field (Complete ordered field (least-upper-bound property), Ordered field) with its order (Order on the reals) and its absolute value (Absolute value in an ordered field).

Let $x\in \mathbb{R}$ and let $\epsilon \in \mathbb{R}$ with $\epsilon >0$. The $\epsilon$-neighbourhood of $x$ is

$N_{\epsilon }(x):=\{y\in \mathbb{R}:\mathrm{\mid }y-x\mathrm{\mid }<\epsilon \},$

and the punctured $\epsilon$-neighbourhood of $x$ is

$N_{\epsilon }^{\ast }(x):=N_{\epsilon }(x)\setminus \{x\}=\{y\in \mathbb{R}:0<\mathrm{\mid }y-x\mathrm{\mid }<\epsilon \}\mathrm{.}$

The two descriptions of $N_{\epsilon }^{\ast }(x)$ agree because $\mathrm{\mid }y-x\mathrm{\mid }=0$ holds exactly when $y=x$ (Basic properties of the absolute value).

A neighbourhood is an open interval. For every $x$ and every $\epsilon >0$,

$N_{\epsilon }(x)=(x-\epsilon ,x+\epsilon ),$

the interval of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length. Indeed Basic properties of the absolute value gives, for $\epsilon >0$, the equivalence $\mathrm{\mid }y-x\mathrm{\mid }<\epsilon ⟺-\epsilon <y-x<\epsilon$, and adding $x$ throughout turns the right-hand side into $x-\epsilon <y<x+\epsilon$ (Ordered field).

The centre lies in its own neighbourhoods. $x\in N_{\epsilon }(x)$, since $\mathrm{\mid }x-x\mathrm{\mid }=\mathrm{\mid }0\mathrm{\mid }=0<\epsilon$ (Basic properties of the absolute value).

Punctured neighbourhoods are never empty. The element $y:=x+\epsilon \mathrm{/}2$ satisfies $\mathrm{\mid }y-x\mathrm{\mid }=\epsilon \mathrm{/}2$, which is $>0$ and $<\epsilon$, so $y\in N_{\epsilon }^{\ast }(x)$ (Basic properties of the absolute value, Ordered field).

Monotonicity in the radius. If $0<\delta \le \epsilon$ then $N_{\delta }(x)\subseteq N_{\epsilon }(x)$, because $\mathrm{\mid }y-x\mathrm{\mid }<\delta \le \epsilon$ (Ordered field).

Nesting at an interior point. If $y\in N_{\epsilon }(x)$ and $0<\delta \le \epsilon -\mathrm{\mid }y-x\mathrm{\mid }$, then

$N_{\delta }(y)\subseteq N_{\epsilon }(x)\mathrm{.}$

Indeed for $z\in N_{\delta }(y)$ the triangle inequality (The triangle inequality) gives $\mathrm{\mid }z-x\mathrm{\mid }=\mathrm{\mid }(z-y)+(y-x)\mathrm{\mid }\le \mathrm{\mid }z-y\mathrm{\mid }+\mathrm{\mid }y-x\mathrm{\mid }<\delta +\mathrm{\mid }y-x\mathrm{\mid }\le \epsilon$. Note that $\epsilon -\mathrm{\mid }y-x\mathrm{\mid }>0$ precisely because $y\in N_{\epsilon }(x)$, so such a $\delta$ always exists.

• The radius is a real number, not a rational. Nothing on this page tests a condition against rational radii only. That convention belongs to Limits and Cauchy sequences of reals, where the quantifier is over rational $\epsilon$ and the passage between the rational and the real form is the sanctioned remark of Sequences of reals: bounded, eventually, frequently, tails, subsequences. Here $\epsilon$ ranges over the positive reals throughout, and every statement above is proved for an arbitrary positive real.

• Why the punctured version is separated out. A limit point of a set is a point every punctured neighbourhood of which meets the set (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), and deleting the centre is exactly what stops a point of the set from qualifying automatically. The unpunctured condition defines the weaker notion of an adherent point, and the difference between the two is precisely an isolated point.

• Nesting is the workhorse. Almost every openness verification on this page has the shape "given $y$ in the set, shrink the radius by the distance already travelled", which is the nesting property above. It is recorded here once so that no later proof has to redo the triangle inequality in passing.