Urysohn's Lemma and the Tietze Extension Theorem: Examples and Counterexamples

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 $A,B$.

Take $A:=(-\mathrm{\infty },0]$ and $B:=[1,\mathrm{\infty })$ in $\mathbb{R}$, disjoint closed sets of the normal space $\mathbb{R}$ (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). For $r$ a dyadic rational of $[0,1]$ with $r<1$ (The dyadic rationals of $[0,1]$, their finite levels $D_n$, and their density in $[0,1]$) put

$U_r:=(-\mathrm{\infty },\frac{1+r}{2}),U_1:=\mathbb{R}\mathrm{.}$

These are open, and satisfy $\overline{U_r}\subseteq U_s$ for every $r<s$ in $D$ and $U_1=\mathbb{R}$, so $(U_r{)}_{r\in D}$ is a legitimate instance of the family hypothesised in If $(U_r{)}_{r\in D}$ are open with $\overline{U_r}\subseteq U_s$ whenever $r<s$ and $U_1=X$, then $x\mapsto \inf \{r\in D:x\in U_r\}$ is a continuous map $X\to [0,1]$, and no choice principle is used and could arise from the construction inside 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 $A,B$. In particular:

$U_0=(-\mathrm{\infty },\frac{1}{2}),U_{1\mathrm{/}4}=(-\mathrm{\infty },\frac{5}{8}),U_{1\mathrm{/}2}=(-\mathrm{\infty },\frac{3}{4}),U_{3\mathrm{/}4}=(-\mathrm{\infty },\frac{7}{8}),U_1=\mathbb{R}\mathrm{.}$

• The five requested sets are read off the general formula. $U_0,U_{1\mathrm{/}4},U_{1\mathrm{/}2},U_{3\mathrm{/}4}$ are nested, each strictly inside the next by step 1.1, and $U_1=\mathbb{R}$ is the point at which the family widens all at once, in line with the discussion 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's own Remarks of why the recursion tracks $X\setminus B$ rather than $X$ until the very last step.

• This is one legitimate family among many. 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 never claims uniqueness, and the family here is not literally the output of the dependent-choice recursion in that item's proof — it is a hand-picked family satisfying the same two hypotheses, chosen because its members have closed forms.

Let $(X,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) with its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) and hence normal (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal), and let $A,B\subseteq X$ be disjoint, nonempty and closed. Define $h:X\to [0,1]$ by

$h(x):=\frac{d(x,A)}{d(x,A)+d(x,B)}\mathrm{.}$

Then $h$ 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 $B$, and it is written down by a single formula: no choice principle, dependent or otherwise, is spent in producing it, in contrast with the general construction inside that theorem.

• Nothing is selected. Every value $h(x)$ is computed from $d(x,A)$ and $d(x,B)$, both determined by $x$, $A$, $B$ and $d$ alone; no step above fixes a witness from a nonempty set of alternatives. This is the same observation In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal and 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 make about their own constructions, and it is why the metric case of every separation theorem on this page needs no choice hypothesis at all.

• The nonemptiness of $A$ and $B$ is not decoration. If $A=\mathrm{\varnothing }$ then $d(x,A)$ is undefined for every $x$ ($\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y)$, so the distance to a fixed nonempty set is $1$-Lipschitz presupposes a nonempty set), and the constant function $1$ serves instead; the formula above is written for the case that matters, where both sets carry a point to measure distance from.

$\mathbb{R}$ with its usual topology is metrizable (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), hence perfectly normal by 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: every closed $C\subseteq \mathbb{R}$ is a zero set (Zero sets and cozero sets of continuous real-valued functions) and a $G_{\delta }$ ($G_{\delta }$ and $F_{\sigma }$ subsets of a topological space, agreeing with the real-line notion). Taking $C:=\{0\}$ makes both witnesses explicit: $C=Z(f)$ for $f(x):=\mathrm{\mid }x\mathrm{\mid }$, and $C={\bigcap }_{n\in \mathbb{N}}(-\frac{1}{n+1},\frac{1}{n+1})$.

This is exactly what Under dependent choice a space is perfectly normal if and only if it is normal and every closed set is a zero set predicts of a perfectly normal space, illustrated by the metric case that theorem's own proof does not need to run through, since perfect normality of $\mathbb{R}$ is already established directly from the metric.

• No general closed subset of $\mathbb{R}$ is exceptional here. The argument above uses nothing about $\{0\}$ beyond it being closed and nonempty in a metric space; the same two formulas, with $d(x,C)$ in place of $\mathrm{\mid }x\mathrm{\mid }$, exhibit any closed $C\subseteq \mathbb{R}$ as a zero set and a $G_{\delta }$, choice-free.

• This does not exercise the harder half of Under dependent choice a space is perfectly normal if and only if it is normal and every closed set is a zero set. That theorem's forward direction builds a zero set from a $G_{\delta }$ presentation via a countable family of Urysohn functions; here the zero set is read off directly from the metric, with no such construction and no dependent choice.

Let $A:=[0,1]\subseteq \mathbb{R}$, closed, and $f:A\to \mathbb{R}$, $f(x):=x^2$, continuous. $\mathbb{R}$ is normal (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal), so Under dependent choice, a continuous real-valued map on a closed subspace of a normal space extends to the whole space, and a map into an open interval extends into that same open interval guarantees a continuous $F:\mathbb{R}\to \mathbb{R}$ with $F{\mathrm{\mid }}_A=f$, with no formula supplied. One is written down here directly:

$F(x):=(\max \{0,\min \{1,x\}\}{)}^2\mathrm{.}$

• The trick generalises. For any closed bounded interval $[a,b]$ and any continuous $f:[a,b]\to \mathbb{R}$, precomposing $f$ with the clamp $x\mapsto \max \{a,\min \{b,x\}\}:\mathbb{R}\to [a,b]$ gives a continuous extension of $f$ to all of $\mathbb{R}$, by the same two-step argument as above. Nothing about $x^2$ is used beyond its own continuity on $[0,1]$.

• This is not the extension the recursive proof of Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a,b]$ extends continuously to the whole space, and this property characterises normality would produce. That proof builds an extension as a uniformly convergent series of Urysohn functions, and it makes no claim of matching the clamp-composition extension above; both are legitimate continuous extensions, and neither is canonical.

The continuous function $f:(0,1]\to \mathbb{R}$, $f(x):=1\mathrm{/}x$, extends to a continuous function $F:\mathbb{R}\to \mathbb{R}$.

This is the single witness behind FALSE: Every continuous real-valued function on a subspace of a normal space extends continuously to the whole space, presented on its own as the counterexample it is: it shows that dropping the hypothesis "$A$ closed" from Under dependent choice, a continuous real-valued map on a closed subspace of a normal space extends to the whole space, and a map into an open interval extends into that same open interval is not a minor loosening but breaks that extension statement outright, on the very space $\mathbb{R}$ where it is otherwise available.

Which statement this witness refutes, and which it does not. The corollary is the $\mathbb{R}$-valued form, and $f$ meets every one of its hypotheses except closedness of $A$, so it isolates that hypothesis exactly. It does not refute Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a,b]$ extends continuously to the whole space, and this property characterises normality itself with the closedness hypothesis removed: that theorem is stated for maps into a bounded interval $[a,b]$, and $f$ is unbounded, so $f$ fails its codomain hypothesis as well. A witness violating two hypotheses cannot isolate one.

• The only hypothesis of the $\mathbb{R}$-valued extension statement that fails here is closedness of $A$. $\mathbb{R}$ is normal and $f$ is continuous on $A$; the closure of $A$ in $\mathbb{R}$ is $[0,1]$, and it is exactly the missing point $0$ where $f$ has nowhere finite to go. Against Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a,b]$ extends continuously to the whole space, and this property characterises normality itself the witness fails a second hypothesis, since that theorem takes values in a bounded interval and $f$ does not, which is why the statement refuted above is framed against Under dependent choice, a continuous real-valued map on a closed subspace of a normal space extends to the whole space, and a map into an open interval extends into that same open interval.

• The obstruction is boundedness, not the existence of a limit. Step 1.2 shows $f$ is unbounded on every neighbourhood of the missing point directly from the reciprocal's growth, with no appeal to ${\lim }_{x\to 0^{+}}f(x)$ failing to exist as a real number.

Sierpinski space $S=\{a,b\}$, $a\mathrm{\ne }b$, with ${\mathcal{T}}_{\mathrm{S}\mathrm{i}\mathrm{e}\mathrm{r}}=\{\mathrm{\varnothing },\{b\},S\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly) but not completely regular (Completely regular spaces and Tychonoff ($T_{3\frac{1}{2}}$) spaces), by FALSE: Every normal space is completely regular. The refutation there in fact shows $S$ is not even regular — the only open set containing $a$ is $S$ itself — and complete regularity then fails because it implies regularity (Every completely regular space is regular, and every Tychonoff space is $T_3$). And $S$ is not $T_1$: the singleton $\{b\}$ is not closed in $S$.

This is exactly why Under dependent choice a normal $T_1$ space is completely regular, so $T_4\Rightarrow T_{3\frac{1}{2}}$, and together with the implications already proved this is the whole classical chain proves its arrow for a normal $T_1$ space and not for a bare normal space: $S$ is a normal space for which the corollary's conclusion — complete regularity — genuinely fails, and the one hypothesis it lacks is $T_1$.

• This is not a new computation. Every fact used above is proved in FALSE: Every normal space is completely regular; this item only reads that refutation as the positive example it also is, and connects it to the hypothesis of Under dependent choice a normal $T_1$ space is completely regular, so $T_4\Rightarrow T_{3\frac{1}{2}}$, and together with the implications already proved this is the whole classical chain by name.

The subspace $A:=K\cup \{0\}\subseteq (\mathbb{R},{\mathcal{T}}_K)$ of the $K$-topology (The $K$-topology on $\mathbb{R}$, generated by the open intervals together with their complements of $K=\{1\mathrm{/}(n+1):n\in \mathbb{N}\}$, is $T_1$ and Hausdorff but not regular) is closed, and the continuous $k:A\to [0,1]$ with $k\equiv 0$ on $K$ and $k(0):=1$ extends to a continuous $F:\mathbb{R}\to [0,1]$.

This is false, and the failure is not a failure of closedness: $A$ is closed. What fails is normality of the ambient space — $(\mathbb{R},{\mathcal{T}}_K)$ is $T_1$ but not normal — in contrast with the companion witness on this page that instead varies the closedness hypothesis of Tietze's theorem while keeping the ambient space normal.

• $(\mathbb{R},{\mathcal{T}}_K)$ is not normal, and this is why Tietze's theorem does not apply here. It is $T_1$ (step 1.1's citation of [L1]) but not regular (used directly in step 4.1); by A normal $T_1$ space is regular, hence $T_3$, hence Urysohn, Hausdorff, $T_1$ and $T_0$, a normal $T_1$ space is regular, so a $T_1$ space that is not regular cannot be normal. Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a,b]$ extends continuously to the whole space, and this property characterises normality requires normality of the ambient space, and that hypothesis is exactly what fails.

• Closedness of $A$ is not in question. $A$ is closed by step 1.1, and $k$ is continuous on $A$ by step 3.1; every hypothesis of Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a,b]$ extends continuously to the whole space, and this property characterises normality except normality of $\mathbb{R}$ with the ${\mathcal{T}}_K$ topology holds here.