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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. At each dyadic level finitely many new open sets are selected by Every natural-number-indexed list of nonempty sets has a choice function on its family of values, a theorem of ZF. Dependent choice assembles a sequence of admissible levels, each extending its predecessor. This proves sufficiency of DC for this construction; it asserts no necessity or independence result.

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

1.1assume-hyp

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

1.2assume-hyp

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).

2.1step 1.1L1chooseconstruct

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 $D_{0} = {0, 1}$ .

2.2step 1.2choose

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 $E \subseteq h^{- 1} ({1})$ .

3.1step 2.1

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

3.2step 2.2L6L7

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].

4.1step 3.1construct

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, $(0, Φ_{0}) \in P$ .

4.2step 2.2step 3.2L6

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 $O_{1} \cap O_{2} = h^{- 1} ([0, \frac{1}{2}) \cap (\frac{1}{2}, 1]) = h^{- 1} (\emptyset) = \emptyset$ .

5.1step 4.1L1L3

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.

5.2step 4.2

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].

6.1step 5.1L2choose

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 $0 \leq j < 2^{n}$ .

7.1step 6.1L3construct

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 $(n, Φ) R (n + 1, Φ^{'})$ .

8.1step 7.1step 6.1

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.

9.1step 8.1L3L4

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 $\overline{Φ^{'} (x)} = \overline{Φ^{'} (u_{0})} \subseteq Φ^{'} (u_{k}) = Φ^{'} (y)$ .

10.1step 7.1step 9.1L3

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 $(n + 1, Φ^{'}) \in P$ .

11.1step 7.1step 10.1

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 $P$ .

12.1step 4.1step 11.1A1construct

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 $k$ .

13.1step 12.1

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 $k$ .

14.1step 13.1L3construct

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.

15.1step 14.1step 13.1L3

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 $n$ .

16.1step 15.1construct

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 $U_{1} = X$ .

17.1step 16.1L5

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 $X \to [0, 1]$ .

17.2step 14.1step 13.1step 16.1L3L8

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 $b \notin U_{r}$ .

18.1step 17.1step 16.1step 15.1

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 $f (a) = 0$ .

18.2step 17.2step 16.1

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 $f (b) = \inf {1} = 1$ .

19.1step 17.1step 18.1step 18.2

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.

20.1step 19.1step 5.2discharge-construct∎

Steps 19.1 and 5.2 establish clauses 1 and 2 respectively.

Remarks

The lemma is stated for a normal space. $T_{1}$ is used nowhere above; it is needed only to turn a point into a closed set, the extra step used by the separation-chain corollary.
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.

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

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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