A Urysohn function for $(-\mathrm{\infty },0]$ and $[1,\mathrm{\infty })$ in $\mathbb{R}$, written down and checked against the definition

In $\mathbb{R}$ with its usual topology, normal by In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal, the sets $A:=(-\mathrm{\infty },0]$ and $B:=[1,\mathrm{\infty })$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) are disjoint and closed. Define $g:\mathbb{R}\to [0,1]$ by

$g(x):=\max \{0,\min \{1,x\}\}\mathrm{.}$

This $g$ is a witness for Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0,1]$, and conversely such a space is normal applied to $A$ and $B$: continuous, with $A\subseteq g^{-1}(\{0\})$ and $B\subseteq g^{-1}(\{1\})$.

• For $0<x<1$, neither clamp is active and $g(x)=x$, so $g$ interpolates linearly across the gap between $A$ and $B$. Nothing in Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0,1]$, and conversely such a space is normal requires linearity: replacing $x$ by $x^2$ on $[0,1]$ while keeping the same constant values outside that interval gives a different Urysohn function for this pair.

• No dyadic recursion is visible here. This $g$ is written down directly, not produced by the construction of Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0,1]$, and conversely such a space is normal; the companion example works through several levels of that construction by hand for the same pair $A,B$.