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 $\{x : d(x,A) < 1/(n+1)\}$, worked for $[0,1]$ and for $\{0\}$

Let $\mathbb{R}$ carry its usual metric $d_{\mathbb{R}}(s,t) = |s-t|$ and its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ 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 $1/(n+1)$ 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 $d(x,A) := \inf\{\, |x - a| : a \in A \,\}$ (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. $A = \bigcap_{n \in \mathbb{N}} \{\, x \in \mathbb{R} : d(x,A) < 1/(n+1) \,\}$, 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 $d(x,[0,1]) = \begin{cases} -x & x < 0 \\ 0 & 0 \le x \le 1 \\ x - 1 & x > 1 \end{cases}$ so $Z(d(\cdot,[0,1])) = [0,1]$ and, for every real $\varepsilon > 0$, $\{\, x : d(x,[0,1]) < \varepsilon \,\} = (-\varepsilon,\ 1 + \varepsilon)$. Taking $\varepsilon = 1/(n+1)$ gives $[0,1] \;=\; \bigcap_{n \in \mathbb{N}} \big(-1/(n+1),\ 1 + 1/(n+1)\big).$
• $A = \{0\}$. Here $d(x,\{0\}) = |x|$, so $\{0\} \;=\; \bigcap_{n \in \mathbb{N}} \big(-1/(n+1),\ 1/(n+1)\big),$ 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 $1/(0+1) = 1$, so the first set in each intersection is $(-1, 2)$ for $[0,1]$ and $(-1,1)$ for $\{0\}$. Writing $1/n$ 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.