TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)‡ rests on unproved material

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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

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $(X, T)$ be a topological space.

If $X$ is normal (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly) and $A, B \subseteq X$ are disjoint closed sets, there is a continuous $f : X \to [0, 1]$ (Continuity of a map of topological spaces at a point and globally, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with $A \subseteq f^{- 1} ({0})$ and $B \subseteq f^{- 1} ({1})$ .
Conversely, if every pair of disjoint closed subsets of $X$ admits a continuous function into $[0, 1]$ separating them in the sense of clause 1, then $X$ is normal. This direction uses no choice principle.

Where the choice principle of clause 1 is spent, and why not less. The construction below builds, for each $n \in N$ , an assignment of an open set to every dyadic rational of level $n$ , extending the level- $(n - 1)$ assignment; at each single level the finitely many new open sets are chosen at once by Every natural-number-indexed list of nonempty sets has a choice function on its family of values, a theorem of ZF, but stringing together infinitely many such levels, each depending on the one before, is exactly the situation dependent choice is for. The published Urysohn's lemma is not a theorem of ZF, nor of ZF plus countable choice ‡ records, with its sources, that $Z F$ and even $Z F$ together with the Axiom of Countable Choice do not suffice, and that dependent choice does; nothing here claims dependent choice is necessary for clause 1, only that the construction given is carried out in $Z F + D C$ .

Facts & Assumptions

Given: A topological space $(X, T)$ and dependent choice.

[A1]

$D C$ : for every nonempty set $P$ , every relation $R \subseteq P \times P$ entire on $P$ (every $p \in P$ has some $q \in P$ with $p R q$ ), and every $a \in P$ , there is a sequence $(p_{k})_{k \in N}$ with $p_{0} = a$ and $p_{k} R p_{k + 1}$ for every $k$ (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[L1]

Shrinking: if $X$ is normal, $C \subseteq X$ is closed and $O \subseteq X$ is open with $C \subseteq O$ , then there is open $W$ with $C \subseteq W \subseteq \overline{W} \subseteq O$ (A space is normal if and only if every closed $A$ inside an open $U$ admits an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq U$ ).

[L2]

Finite choice: a function $F$ with domain a natural number $n$ , all of whose values are nonempty sets, admits a choice function for the family $F [n]$ of its values (Every natural-number-indexed list of nonempty sets has a choice function on its family of values, Choice function), a theorem of ZF.

[L3]

The dyadic rationals $D = ⋃_{n} D_{n}$ of $[0, 1]$ are an increasing union of finite levels; for $n \in N$ , $D_{n + 1} = D_{n} \cup {t_{j} : 0 \leq j < 2^{n}}$ , where $t_{j}$ is strictly between the $D_{n}$ -consecutive pair $r_{j} : = j / 2^{n}$ and $s_{j} : = (j + 1) / 2^{n}$ , the $2^{n}$ points $t_{j}$ are pairwise distinct and disjoint from $D_{n}$ , and every two elements of $D$ lie together in some common $D_{n}$ (The dyadic rationals of $[0, 1]$ , their finite levels $D_{n}$ , and their density in $[0, 1]$ ).

[L4]

Chaining: if $V_{0}, \dots, V_{k}$ ( $k \geq 0$ ) are subsets of $X$ with $\overline{V_{i}} \subseteq V_{i + 1}$ for every $i < k$ , then $\overline{V_{0}} \subseteq V_{k}$ , since $V_{i} \subseteq \overline{V_{i}} \subseteq V_{i + 1}$ for each $i$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space) makes $\overline{V_{0}} \subseteq V_{1} \subseteq V_{2} \subseteq \dots \subseteq V_{k}$ a chain of inclusions.

[L5]

The generic construction: if $(U_{r})_{r \in D}$ is a family of open subsets of $X$ with $\overline{U_{r}} \subseteq U_{s}$ whenever $r < s$ in $D$ and $U_{1} = X$ , then $g (x) : = \inf ({r \in D : x \in U_{r}} \cup {1})$ is a continuous map $X \to [0, 1]$ (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).

[L6]

The order rays $(- \infty, \frac{1}{2})$ and $(\frac{1}{2}, \infty)$ are open in the usual topology of $R$ (The absolute value makes $R$ 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, clause 3), so their traces $[0, \frac{1}{2})$ and $(\frac{1}{2}, 1]$ are open in the subspace topology of $[0, 1]$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). They are disjoint and contain $0$ and $1$ , respectively.

[L7]

Preimages of open sets under a continuous map are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b), Continuity of a map of topological spaces at a point and globally).

[L8]

$A \subseteq \overline{A}$ for every $A \subseteq X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

Proof

technique · constructive

1.1

Assume $X$ is normal and $A, B \subseteq X$ are disjoint closed sets (the hypothesis of clause 1).

assume-hyp

1.2

Assume instead that every pair of disjoint closed subsets of $X$ admits a continuous function into $[0, 1]$ separating them as in clause 1 (the hypothesis of clause 2).

assume-hyp

2.1

Under step 1.1: $A \subseteq X ∖ B$ , since $A \cap B = \emptyset$ , and $X ∖ B$ is open since $B$ is closed; by [L1] applied to the closed set $A$ and the open set $X ∖ B$ , fix open $Φ_{0} (0)$ with $A \subseteq Φ_{0} (0) \subseteq \overline{Φ_{0} (0)} \subseteq X ∖ B$ , and put $Φ_{0} (1) : = X ∖ B$ , defining $Φ_{0} : D_{0} \to T$ on $D_{0} = {0, 1}$ .

step 1.1L1chooseconstruct

2.2

Under step 1.2: let $C, E \subseteq X$ be disjoint closed sets; fix a continuous $h : X \to [0, 1]$ with $C \subseteq h^{- 1} ({0})$ and $E \subseteq h^{- 1} ({1})$ .

step 1.2choose

3.1

Under step 1.1: $A \subseteq Φ_{0} (0)$ ; $\overline{Φ_{0} (0)} \subseteq Φ_{0} (1)$ ; and $Φ_{0} (1) = X ∖ B$ .

step 2.1

3.2

Under step 1.2, continuing: by [L6], $[0, \frac{1}{2})$ and $(\frac{1}{2}, 1]$ are open in $[0, 1]$ , disjoint, with $0 \in [0, \frac{1}{2})$ and $1 \in (\frac{1}{2}, 1]$ ; put $O_{1} : = h^{- 1} ([0, \frac{1}{2}))$ and $O_{2} : = h^{- 1} ((\frac{1}{2}, 1])$ , open in $X$ by [L7].

step 2.2L6L7

4.1

Under step 1.1: for $n \in N$ , call $Φ : D_{n} \to T$ admissible at level $n$ when (i) $\overline{Φ (r)} \subseteq Φ (s)$ for every $r < s$ in $D_{n}$ ; (ii) $A \subseteq Φ (0)$ ; (iii) $Φ (1) = X ∖ B$ . Put $P : = {(n, Φ) : n \in N, Φ admissible at level n}$ , and for $(n, Φ), (n^{'}, Φ^{'}) \in P$ say $(n, Φ) R (n^{'}, Φ^{'})$ when $n^{'} = n + 1$ and $Φ^{'} ∣_{D_{n}} = Φ$ . By step 3.1, $(0, Φ_{0}) \in P$ .

step 3.1construct

4.2

Under step 1.2: $C \subseteq O_{1}$ , since $h \equiv 0 \in [0, \frac{1}{2})$ on $C$ ; $E \subseteq O_{2}$ , since $h \equiv 1 \in (\frac{1}{2}, 1]$ on $E$ ; and $O_{1} \cap O_{2} = h^{- 1} ([0, \frac{1}{2}) \cap (\frac{1}{2}, 1]) = h^{- 1} (\emptyset) = \emptyset$ .

step 2.2step 3.2L6

5.1

Under step 1.1: let $(n, Φ) \in P$ . For each $j$ with $0 \leq j < 2^{n}$ , with $r_{j}, s_{j}, t_{j}$ as in [L3]: since $r_{j} < s_{j}$ in $D_{n}$ , admissibility (i) gives $\overline{Φ (r_{j})} \subseteq Φ (s_{j})$ , so by [L1] the set of open $W$ with $\overline{Φ (r_{j})} \subseteq W \subseteq \overline{W} \subseteq Φ (s_{j})$ is nonempty.

step 4.1L1L3

5.2

Under step 1.2: since $C, E$ were an arbitrary disjoint closed pair, step 4.2 exhibits disjoint open supersets for every such pair, so $X$ is normal by Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly; this is clause 2, and no step of it used [A1].

step 4.2

6.1

Under step 1.1, continuing under step 5.1: by [L2] applied to the function assigning, to each $j < 2^{n}$ , the nonempty set of open $W$ with $\overline{Φ (r_{j})} \subseteq W \subseteq \overline{W} \subseteq Φ (s_{j})$ , fix a simultaneous choice, giving open $W_{j}$ with $\overline{Φ (r_{j})} \subseteq W_{j} \subseteq \overline{W_{j}} \subseteq Φ (s_{j})$ for every $0 \leq j < 2^{n}$ .

step 5.1L2choose

7.1

Under step 1.1: define $Φ^{'} : D_{n + 1} \to T$ by $Φ^{'} ∣_{D_{n}} : = Φ$ and $Φ^{'} (t_{j}) : = W_{j}$ for $0 \leq j < 2^{n}$ ; this is well defined since $D_{n + 1} = D_{n} \cup {t_{j} : 0 \leq j < 2^{n}}$ with the $t_{j}$ pairwise distinct and disjoint from $D_{n}$ by [L3]. Then $(n, Φ) R (n + 1, Φ^{'})$ .

step 6.1L3construct

8.1

Under step 1.1, with $Φ, Φ^{'}$ as in step 7.1: for the $D_{n + 1}$ -consecutive pair $(r_{j}, t_{j})$ : $\overline{Φ^{'} (r_{j})} = \overline{Φ (r_{j})} \subseteq W_{j} = Φ^{'} (t_{j})$ by step 6.1; for the pair $(t_{j}, s_{j})$ : $\overline{Φ^{'} (t_{j})} = \overline{W_{j}} \subseteq Φ (s_{j}) = Φ^{'} (s_{j})$ by step 6.1.

step 7.1step 6.1

9.1

Under step 1.1: for $x < y$ in $D_{n + 1}$ , the finitely many elements of $D_{n + 1} \cap [x, y]$ , listed increasingly as $x = u_{0} < u_{1} < \dots < u_{k} = y$ , are $D_{n + 1}$ -consecutive at each step $u_{i} < u_{i + 1}$ , and each such pair is one of the pairs of step 8.1 (every $D_{n + 1}$ -consecutive pair has at least one member among the new points $t_{j}$ , since a new point was inserted into every $D_{n}$ -consecutive gap); so $\overline{Φ^{'} (u_{i})} \subseteq Φ^{'} (u_{i + 1})$ at each step, and [L4] gives $\overline{Φ^{'} (x)} = \overline{Φ^{'} (u_{0})} \subseteq Φ^{'} (u_{k}) = Φ^{'} (y)$ .

step 8.1L3L4

10.1

Under step 1.1: $A \subseteq Φ^{'} (0) = Φ (0)$ , since $0 \in D_{n}$ is unaffected by the extension; $Φ^{'} (1) = Φ (1) = X ∖ B$ , since $1 \in D_{n}$ is likewise unaffected; with step 9.1 this is admissibility of $Φ^{'}$ at level $n + 1$ , so $(n + 1, Φ^{'}) \in P$ .

step 7.1step 9.1L3

11.1

Under step 1.1: by steps 5.1, 6.1, 7.1 and 10.1, every $(n, Φ) \in P$ has some $(n + 1, Φ^{'}) \in P$ with $(n, Φ) R (n + 1, Φ^{'})$ ; so $R$ is entire on $P$ .

step 7.1step 10.1

12.1

Under step 1.1: $P$ is nonempty by step 4.1 and $R$ is entire on $P$ by step 11.1; by [A1] applied with $a : = (0, Φ_{0})$ , there is a sequence $((m_{k}, Ψ_{k}))_{k \in N}$ with $(m_{0}, Ψ_{0}) = (0, Φ_{0})$ and $(m_{k}, Ψ_{k}) R (m_{k + 1}, Ψ_{k + 1})$ for every $k$ .

step 4.1step 11.1A1construct

13.1

Under step 1.1: since $(n, Φ) R (n^{'}, Φ^{'})$ forces $n^{'} = n + 1$ , and $m_{0} = 0$ , induction on $k$ gives $m_{k} = k$ for every $k \in N$ ; so each $Ψ_{k} : D_{k} \to T$ is admissible at level $k$ , and $Ψ_{k + 1} ∣_{D_{k}} = Ψ_{k}$ for every $k$ .

step 12.1

14.1

Under step 1.1: for $r \in D$ , fix $n$ with $r \in D_{n}$ [L3] and define $V_{r} : = Ψ_{n} (r)$ ; by step 13.1, for $n \leq n^{'}$ with $r \in D_{n}$ , $Ψ_{n^{'}} (r) = Ψ_{n} (r)$ (chaining $Ψ_{n^{'}} ∣_{D_{n}} = Ψ_{n}$ through the intermediate levels), so $V_{r}$ does not depend on the level $n$ chosen.

step 13.1L3construct

15.1

Under step 1.1: for $r < s$ in $D$ , fix $n$ with $r, s \in D_{n}$ [L3]; then $\overline{V_{r}} = \overline{Ψ_{n} (r)} \subseteq Ψ_{n} (s) = V_{s}$ by admissibility (i) of $Ψ_{n}$ . Also $A \subseteq V_{0}$ and $V_{1} = X ∖ B$ , by admissibility (ii) and (iii) of $Ψ_{n}$ for any $n$ .

step 14.1step 13.1L3

16.1

Under step 1.1: define $U_{r} : = V_{r}$ for $r \in D$ with $r < 1$ , and $U_{1} : = X$ . For $r < s$ in $D$ : if $s < 1$ , $\overline{U_{r}} = \overline{V_{r}} \subseteq V_{s} = U_{s}$ by step 15.1; if $s = 1$ , $\overline{U_{r}} = \overline{V_{r}} \subseteq V_{1} = X ∖ B \subseteq X = U_{1}$ by step 15.1. So $\overline{U_{r}} \subseteq U_{s}$ whenever $r < s$ in $D$ , and $U_{1} = X$ .

step 15.1construct

17.1

Under step 1.1: by [L5] applied to $(U_{r})_{r \in D}$ of step 16.1, $f (x) : = \inf ({r \in D : x \in U_{r}} \cup {1})$ is a continuous map $X \to [0, 1]$ .

step 16.1L5

17.2

Under step 1.1: for $b \in B$ and $r \in D$ with $r < 1$ : fix $n$ with $r \in D_{n}$ [L3]; since $1 \in D_{n}$ also, admissibility (i) of $Ψ_{n}$ applied to $r < 1$ gives $\overline{Ψ_{n} (r)} \subseteq Ψ_{n} (1) = X ∖ B$ , that is $\overline{V_{r}} \subseteq X ∖ B$ ; since $V_{r} \subseteq \overline{V_{r}}$ by [L8] and $U_{r} = V_{r}$ by step 16.1, $U_{r} \cap B = \emptyset$ , so $b \notin U_{r}$ .

step 14.1step 13.1step 16.1L3L8

18.1

Under step 1.1: for $a \in A$ : $a \in V_{0}$ by step 15.1, and $U_{0} = V_{0}$ by step 16.1 (as $0 < 1$ ), so $a \in U_{0}$ and $0 \in {r \in D : a \in U_{r}}$ ; hence $f (a) \leq 0$ , and $f (a) \geq 0$ since $f$ maps into $[0, 1]$ by step 17.1, so $f (a) = 0$ .

step 17.1step 16.1step 15.1

18.2

Under step 1.1: for $b \in B$ : by step 17.2, $b \notin U_{r}$ for every $r \in D$ with $r < 1$ , and $b \in U_{1} = X$ by step 16.1; so ${r \in D : b \in U_{r}} \cup {1} = {1}$ , giving $f (b) = \inf {1} = 1$ .

step 17.2step 16.1

19.1

Steps 17.1, 18.1 and 18.2 show that, under the hypothesis of step 1.1, $f$ is a continuous map $X \to [0, 1]$ with $A \subseteq f^{- 1} ({0})$ and $B \subseteq f^{- 1} ({1})$ , which is clause 1.

step 17.1step 18.1step 18.2

20.1

Steps 19.1 and 5.2 establish clauses 1 and 2 respectively.

step 19.1step 5.2discharge-construct∎

Remarks

The lemma is stated for a normal space, not a $T_{4}$ space. $T_{1}$ is used nowhere above; it is needed only to turn a point into a closed set, which is the extra step the next corollary spends. The published Urysohn's lemma is not a theorem of ZF, nor of ZF plus countable choice ‡ states the classical $T_{4}$ form; the form proved here is the more general one, and the two are not in tension — the $T_{4}$ form follows by adding the $T_{1}$ hypothesis, which is not used in this proof at all.
Only clause 1 costs a choice principle, and it is spent at exactly one place: the single application of dependent choice in step 12.1, which strings together the countably many admissible levels built one finite step at a time in steps 5.1–10.1. Every other existential instantiation above (steps 2.1, 2.2 and 6.1) draws from a single nonempty set or, in step 6.1, from a finite family of them via Every natural-number-indexed list of nonempty sets has a choice function on its family of values, and neither costs anything beyond ZF.
Why the construction tracks $X ∖ B$ rather than $X$ at $r = 1$ . Recording $Φ_{n} (1) : = X ∖ B$ throughout the recursion, rather than $X$ , is what makes admissibility clause (i) alone carry the whole $B$ -avoidance property: since $1 \in D_{n}$ for every $n$ , clause (i) applied to any $r < 1$ already gives $\overline{Φ_{n} (r)} \subseteq Φ_{n} (1) = X ∖ B$ , with no separate bookkeeping. Only at the very end, in step 16.1, is the top value widened from $X ∖ B$ to $X$ , which is exactly what 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 requires.

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Sources

Urysohn's lemma (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §33 (standard reference, not scraped)
S. Willard, General Topology, §15 (standard reference, not scraped)
J. P. May, An Outline Summary of Basic Point Set Topology, §6 (standard reference, not scraped)
Axiom of dependent choice (Wikipedia) (standard reference, not scraped)

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][0,1], and conversely such a space is normal