Every nonempty closed subset $A$ of $\mathbb{R}$ is the zero set of $x\mapsto d(x,A)$ and the intersection of the open sets $\{x:d(x,A)<1\mathrm{/}(n+1)\}$, worked for $[0,1]$ and for $\{0\}$

Let $\mathbb{R}$ carry its usual metric $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ and its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and write $1\mathrm{/}(n+1)$ for the inverse of the canonical natural $\iota (n+1)$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field). Let $A\subseteq \mathbb{R}$ be nonempty and closed, and put $d(x,A):=\inf \{\mathrm{\mid }x-a\mathrm{\mid }:a\in A\}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Then, as the general metric theorem (In a metric space every closed set is a zero set and a $G_{\delta }$, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal) specialises:

1. $x\mapsto d(x,A)$ is continuous and $A=Z(d(\cdot ,A))$ (Zero sets and cozero sets of continuous real-valued functions), so $A$ is a zero set.
2. $A={\bigcap }_{n\in \mathbb{N}}\{x\in \mathbb{R}:d(x,A)<1\mathrm{/}(n+1)\}$, an intersection of open sets, so $A$ is a $G_{\delta }$ of the topological space $\mathbb{R}$ ($G_{\delta }$ and $F_{\sigma }$ subsets of a topological space, agreeing with the real-line notion) and hence a $G_{\delta }$ subset of $\mathbb{R}$ in the sense of $F_{\sigma }$ and $G_{\delta }$ subsets of $\mathbb{R}$, the two notions being the same one.

Two worked instances:

• $A=[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Here $d(x,[0,1])=\{\begin{array}{ll}-x& x<0\\ 0& 0\le x\le 1\\ x-1& x>1\end{array}$ so $Z(d(\cdot ,[0,1]))=[0,1]$ and, for every real $\epsilon >0$, $\{x:d(x,[0,1])<\epsilon \}=(-\epsilon ,1+\epsilon )$. Taking $\epsilon =1\mathrm{/}(n+1)$ gives $[0,1]={\bigcap }_{n\in \mathbb{N}}(-1\mathrm{/}(n+1),1+1\mathrm{/}(n+1))\mathrm{.}$
• $A=\{0\}$. Here $d(x,\{0\})=\mathrm{\mid }x\mathrm{\mid }$, so $\{0\}={\bigcap }_{n\in \mathbb{N}}(-1\mathrm{/}(n+1),1\mathrm{/}(n+1)),$ the standard presentation of a point of $\mathbb{R}$ as a $G_{\delta }$.

The converse fails. A $G_{\delta }$ subset of $\mathbb{R}$ need not be closed: $(0,1)$ is open, hence a $G_{\delta }$ by the constant sequence, and it is not closed.

• The index starts at $n=0$, where the radius is $1\mathrm{/}(0+1)=1$, so the first set in each intersection is $(-1,2)$ for $[0,1]$ and $(-1,1)$ for $\{0\}$. Writing $1\mathrm{/}n$ instead would divide by zero (The canonical natural $\iota (n)=n\cdot 1_F$ of a field).

• The two presentations are not independent. A zero set is always a $G_{\delta }$ (Zero sets and cozero sets of continuous real-valued functions), so claim 2 follows from claim 1; the explicit intersection is written out because it is the presentation an argument actually uses, and because it makes the radii visible.

• Why this is the metric case of perfect normality. That every closed set is a $G_{\delta }$ is one of the two conjuncts of perfect normality (In a metric space every closed set is a zero set and a $G_{\delta }$, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal); the other, that $\mathbb{R}$ is normal, comes from the same theorem's completely normal clause. So $\mathbb{R}$ is $T_6$, and everything below it in the chain.