Math. Log. Quart. 52, No. 5, 464 – 469 (2006) / DOI 10.1002/malq.200610006

A constructive treatment of Urysohn’s Lemmain an apartness space

Douglas Bridges∗ and Hannes Diener∗∗

Department of Mathematics and Statistics, University of Canterbury,Private Bag 4800, Christchurch, New Zealand

Received 31 March 2006, revised 2 August 2006, accepted 8 August 2006Published online 1 October 2006

Key words Constructive mathematics, Urysohn’s Lemma, apartness spaces.MSC (2000) 03F60, 03F65

This paper is dedicated to Prof. Dr. Gunter Asser, whose work in founding this journal and maintaining it overmany difficult years has been a major contribution to the activities of the mathematical logic community.

At first (maybe even at second) sight it appears highly unlikely that Urysohn’s Lemma has any significantconstructive content. However, working in the context of an apartness space and using functions whose valuesare a generalisation of the reals, rather than real numbers, enables us to produce a significant constructiveversion of that lemma.

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The classical form of Urysohn’s Lemma states that

Lemma 1 A topological space X is normal if and only if for any two disjoint closed subsets S, T of X thereexists a continuous function f : X −→ [0, 1] such that f(S) = {0} and f(T ) = {1}.

In view of the following Brouwerian example this naıve form of Urysohn’s Lemma is not going to be provableconstructively – that is, with intuitionistic logic.1) Given an arbitrary syntactically correct statement P , constructthe two disjoint closed subsets

S = {x ∈ R : x = 0 ∨ (x = 1 ∧ P )}, T = {x ∈ R : x = 2 ∨ (x = 1 ∧ ¬P )}

of the real line. If S and T can be separated, as in Lemma 1, by a continuous function f , then looking at f(1)would enable us to decide whether 1 /∈ S or 1 /∈ T , and therefore whether ¬P ∨ ¬¬P . Thus a constructive proofof Lemma 1 would lead to one of the weak law of excluded middle, ¬P ∨ ¬¬P .

In this note we derive a form of Urysohn’s Lemma constructively. Apart from our use of intuitionistic logic,one distinctive feature of our result is that it is set in the context of an apartness space, rather than the morefamiliar one of a topological space. (Although every apartness space has a natural topology, it is the apartness,rather than the topology, that is the primary focus in our study of the space.2))

We consider an inhabited set X with a relation �= of inequality between points and a relation ��X(usuallywritten simply ��) of pre-apartness between subsets. These relations give rise to two of the following threecomplements of a subset S of X :

1. the logical complement ¬S = {x ∈ X : x /∈ S},

2. the complement ∼S = {x ∈ X : (∀y ∈ S)(x �= y)},

3. the apartness complement −S = {x ∈ X : {x} �� S}.

∗ Corresponding author: e-mail: D.Bridges@math.canterbury.ac.nz∗∗ e-mail: H.Diener@math.canterbury.ac.nz1) For more information about constructive mathematics see [1, 2, 5].2) A systematic treatment of apartness spaces is in preparation [6].

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Math. Log. Quart. 52, No. 5 (2006) 465

We require the pre-apartness to satisfy these four axioms:B1: X �� ∅.

B2: A �� B ⇒ A ⊂ ∼B.

B3: (A �� (B ∪ C) ⇔ (A �� B ∧ A �� C)) ∧ ((A ∪ B) �� C ⇔ (A �� C ∧ B �� C)).B4: −A ⊂ ∼B ⇒ −A ⊂ −B.

We then call the pair (X, ��), or when no confusion is likely, simply the set X itself, a pre-apartness space.Defining

x �� A ⇔ {x} �� A,

we obtain a so-called “point-set pre-apartness” associated with the given set-set one. We note two simple, usefulconsequences of the axioms:

−S ⊂ ∼S ⊂ ¬S

and

(A �� B ∧ S ⊂ A) ⇒ S �� B.

The apartness complements form a base of open sets of a topology on X , the apartness topology, consistingof the nearly open sets. This is the topology that we have in mind when we write about topological concepts,such as “interior” and “closure”, in the context of an apartness space. For later reference we note two facts:

1. If A ⊂ X is nearly open, then ¬A = ∼A (see [6, Chapter 2, Proposition 34]).

2. For x ∈ X and A ⊂ X ,

x �� A ⇔ (∃U ⊂ X)(x ∈ −U ⊂ ∼A)

(see [6, Chapter 2, Proposition 27]).By a (set-set) apartness on X we mean a relation �� between subsets of X that satisfies B1 – B3 and

B5: x ∈ −A ⇒ (∃S ⊂ X)(x ∈ −S ∧ X = −A ∪ S).It then also satisfies B4 and so is a pre-apartness.

The canonical example of an apartness space is a uniform space (X,U), in which we define the inequality andapartness by

x �= y ⇔ (∃U ∈ U)((x, y) /∈ U)

and

A �� B ⇔ (∃U ∈ U)(A × B ⊂ ∼U),

respectively. In this case the apartness is symmetric:

S �� T ⇒ T �� S.

The strongest of all the separation properties normally considered for an apartness space X is the Efremoviccondition:EF: A �� B ⇒ (∃E ⊂ X)(A �� ¬E ∧ E �� B),

which holds in a large variety of apartness spaces, including uniform spaces. For the proof of our main result wenote that if EF holds, then

(1) A �� B ⇔ A �� ∼∼B.

Before dealing with Urysohn’s Lemma, we introduce an extension of the notion of “real number”, introducedby Troelstra. A set S ⊂ Q is a weak left cut ifwlc1: there exist s ∈ S and t /∈ S;

wlc2: s ∈ S whenever s < t and ¬¬(t ∈ S);wlc3: for each s ∈ S there exists s′ ∈ S with s′ > s.

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466 D. Bridges and H. Diener: A constructive treatment of Urysohn’s Lemma in an apartness space

The set Re of weak left cuts is called the weakly extended real line. Every real number x can be regardedalso as an element xe of Re in the obvious way. Any uniformly continuous map f : Q −→ R has a canonicalextension f∗ : Re −→ Re. Hence the usual operations such as +, · can be canonically extended to Re. With a bitmore work one can show that < extends from Q to Re. For more details see [7].

A mapping f of a point-set pre-apartness space X into Re is said to be continuous if

(∀x ∈ X)(∀A ⊂ R)(f(x) �� A ⇒ x �� f−1(Ae)).

If X is a set-set pre-apartness space and

(∀A, B ⊂ R)(A �� B ⇒ f−1(Ae) �� f−1(Be)),

then f is said to be strongly continuous.These notions of continuity are analogues of the corresponding notions for a mapping of X into R (cf. [4]).We can now state Urysohn’s Lemma for apartness spaces.

Theorem 2 Let X be a symmetric pre-apartness space satisfying the Efremovic condition, and let S, T be sub-sets of X such that S �� T . Then there is a strongly continuous mapping Φ : X −→ [0, 1]e such that Φ(x) = 0e

for all x ∈ S, and Φ(y) = 1e for all y ∈ T .

P r o o f. The first part of the proof follows the idea in [9, p. 19]. Applying EF twice, construct subsets E1, E1/2

of X such that

S �� ¬E1 ∧ E1 �� T

and

S �� ¬E1/2 ∧ E1/2 �� ¬E1.

Then S ⊂ −¬E1/2 and

−¬E1/2 ⊂ ∼¬E1/2 ⊂ ∼∼E1/2,

so −¬E1/2 �� ¬E1, by (1); whence −¬E1/2 ⊂ −¬E1. On the other hand, since −¬E1 ⊂ ∼¬E1 ⊂ ∼∼E1,we see from (1) that −¬E1 �� T . Thus, altogether, we have

S �� ¬E1/2 ∧ (−¬E1/2 ⊂ −¬E1 �� T ).

Now apply EF to obtain E3/4 ⊂ X such that −¬E1 �� ¬E3/4 and E3/4 �� T ; then (1) yields

−¬E1 ⊂ −¬E3/4 ⊂ −¬E3/4 �� T.

Note that as −¬E1/2 ⊂ −¬E1 �� ¬E3/4, we have −¬E1/2 �� ¬E3/4. On the other hand, since S �� ¬E1/2,we can likewise produce E1/4 ⊂ X such that S �� ¬E1/4 and

S ⊂ −¬E1/4 �� ¬E1/2.

Carrying on in this way, for each dyadic rational t ∈ (0, 1) we produce a subset Et of X such that S �� ¬Et,Et �� T , and −¬Es �� ¬Et for each dyadic rational s ∈ (0, 1) with s < t. For each real number t ∈ [0, 1] wenow define

Ut =⋃{−¬Er : r ∈ (0, 1) is a dyadic rational and r � t}.

Note that if 0 < s � t < 1, then Us ⊂ Ut, and that if t ∈ [0, 1] is a dyadic rational, then Ut = −¬Et. Furthermorewe can show that the sets Us are stable in the following sense:

p < q ⇒ ¬¬Up ⊂ Uq.

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Math. Log. Quart. 52, No. 5 (2006) 467

To see this consider p < q. Choose dyadic r1, r2 such that p < r1 < r2 < q. As −¬Er1 �� ¬Er2 and we assumedthat the space satisfies EF we can find a set E such that −¬Er1 �� E and ¬E �� ¬Er2 . So

¬¬Up ⊂ ¬¬Ur1 = ¬¬ − ¬Er1 ⊂ ¬¬¬E = ¬E,

and therefore ¬¬Up �� ¬Er2 . Hence ¬¬Up ⊂ −¬Er2 ⊂ Uq.So far, we have been able, with the help of EF, to mimic the standard classical proof of Urysohn’s Lemma

(see [10, pp. 89 – 91]). However, things get problematic from now on. For a start, following the classical path ofdefining Us to be ∅ when s < 0 and to be X when s > 1, we obtain Us only on the set (−∞, 0] ∪ [0,∞), not onthe whole real line. (It is well known that one cannot decide, for any given real number x, that x � 0 or x � 0.)Even worse, the classical definition of the Urysohn function requires the existence of

inf{t ∈ [0, 1] : x ∈ Ut}

which is of no constructive use since we cannot prove, for a general x ∈ X , that f(x) exists. However, we stillhave the sets Ut to play with, and we can define a mapping f of X into the extended reals by

Φ(x) = {t ∈ Q ∩ [−∞, 1] : t � 0 ∨ (∃s > t)(x ∈ Us)}.

We need to be sure that this set-valued mapping actually takes extended reals as its values. Conditions wlc1

and wlc3 are fulfilled, by the definition of Φ. For wlc2 assume that there are rational numbers s, t such that s < tand ¬¬(t ∈ Φ(x)). If s � 0, there is nothing to prove, so we may assume that 0 < s. Choose dyadic ratio-nals r1, r2 such that s < r1 < r2 < t. Since Up ⊂ Uq whenever p < q, if x /∈ Ur1 , then there does not exist q > tsuch that x ∈ Uq, so t /∈ Φ(x); this contradicts our assumption that ¬¬(t ∈ Φ(x)). Hence ¬¬(x ∈ Ur1), whichimplies that x ∈ Ur2 and therefore that s ∈ Φ(x).

It remains for us to show that Φ is strongly continuous. Accordingly, let A, B be subsets of R such that A �� B.First consider the case where A and B are contained in disjoint closed subintervals of [0, 1]. We may as-sume that there exist a, b such that 0 < a < b < 1, A ⊂ [0, a], and B ⊂ [b, 1]. Pick dyadic rationals r1, r2

such that a < r1 < r2 < b. Then Φ−1(Ae) ⊂ Ur1 = −¬Er1 and Φ−1(Be) ⊂ −Ur2 = −− ¬Er2 ⊂ ∼∼¬Er2 .Since −¬Er1 �� ¬Er2 and therefore, by (1), −¬Er1 �� ∼∼¬Er2 , it follows that Φ−1(Ae) �� Φ−1(Be).

Now consider the case of general A, B ⊂ R. By definition of the apartness on R, there exists a positiveinteger N such that |s − t| > 4/N for all s ∈ A and t ∈ B. Setting

IA = {j ∈ N : A ∩ ( jN , j+2

N ) �= ∅}, IB = {k ∈ N : B ∩ ( kN , k+2

N ) �= ∅},

consider any j ∈ IA and any k ∈ IB . There exist s ∈ A ∩ ( jN , j+2

N ) and t ∈ B ∩ ( kN , k+2

N ). Then

4N < |s − t| � |s − j+1

N | + |k−jN | + |t − k+1

N | < 1N + |k−j

N | + 1N ,

so

|k−jN | > 2

N

and therefore |k − j| > 2. Hence the intervals ( jN , j+2

N ) and ( kN , k+2

N ) are separated by a distance at least 1/N .It follows from the special case we considered above that

Φ−1(( jN , j+2

N )e) �� Φ−1(( kN , k+2

N )e).

Applying the finite unions axiom for apartness, we now obtain

Φ−1(Ae) ⊂ ⋃j∈IA

Φ−1( jN , j+2

N )e ��⋃

k∈IBΦ−1( k

N , k+2N )e ⊃ Φ−1(Be).

Thus Φ is strongly continuous.

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468 D. Bridges and H. Diener: A constructive treatment of Urysohn’s Lemma in an apartness space

We now prove a converse of Urysohn’s Lemma.

Proposition 3 Let X be an inhabited point-set apartness space, and define a set-set relation on X by set-ting S �� T if and only if there is a continuous mapping Φ : X −→ [0, 1]e such that Φ(x) = 0e for all x ∈ Sand Φ(y) = 1e for all y ∈ T . Let the inequality on X be defined by

x �= y ⇔ {x} �� {y}.

Then �� is a symmetric pre-apartness relation that satisfies the Efremovic condition.

P r o o f. It is clear that �� is symmetric. To show that it satisfies B1, we define Φ(x) = 0e for each x ∈ X;we also have, trivially, Φ(y) = 1e for all y ∈ ∅. Next, if there exists a continuous mapping Φ : X −→ [0, 1]e suchthat Φ(x) = 0e for all x ∈ S and Φ(y) = 1e for all y ∈ T , then x �= y; hence S ⊂ ∼T , and B2 holds.

It is also more or less immediate that if R �� (S ∪ T ), then R �� S and R �� T . Conversely, let Φ, Ψ becontinuous mappings of X into [0, 1]e such that Φ(x) = 0e, Ψ(x) = 0e, Φ(y) = 1e and Ψ(z) = 1e, for all x ∈ R,y ∈ S and z ∈ T . Define

Θ(x) = max{Ψ(x), Φ(x)} = Ψ(x) ∪ Φ(x).

Then Θ is a continuous mapping of X into [0, 1]e, Θ(x) = 0e for all x ∈ R and Θ(y) = 1e for all y ∈ S ∪ T ;so R �� (S ∪ T ). Since �� is symmetric, this completes the verification of B3.

To deal with B4, we need to work a little harder. Let x �� A and −A ⊂ ∼B. Then there exists a continuousmapping Φ : X −→ [0, 1]e such that Φ(x) = 0e and Φ(a) = 1e for all a ∈ A. Consider any y ∈ B and assumethat Φ(y) � (2

3 )e. Then define a continuous mapping Ψ : X −→ [0, 1]e by

Ψ(z) = 3e(Φ(x) ∩ (23 )e) − 2e.

Since Ψ(y) = 0e and Ψ(a) = 1e for all a ∈ A, we conclude that y ∈ −A; so y ∈ ∼B, which contradicts ourchoice of y. It follows from this contradiction that ¬Φ(x) � (2

3 )e. Thus Φ(b) � (13 )e for any b ∈ B.3) Now

define a continuous function Θ : X −→ [0, 1]e by

Θ(x) = 3e((13 )e ∪ Ψ(x)).

Since Θ(x) = 0e and Θ(b) = 1e for all b ∈ B, we have x �� B. This completes the proof of B4.Finally, to verify EF, let S �� T , and construct a continuous mapping Φ : X −→ [0, 1]e such that Φ(s) = 0e

and Φ(t) = 1e for all s ∈ S and t ∈ T . Define

E = {x ∈ X : Φ(x) � (12 )e}.

Let x ∈ ¬E. The assumption that Φ(x) � (12 )e leads to a contradiction so we conclude that Φ(x) > (1

3 )e. Usingthe same technique as before, define a continuous mapping Ψ : X −→ [0, 1]e by

Θ(x) = 3e((13 )e ∩ Ψ(x))

Since Ψ(s) = 0e for all s ∈ S and Ψ(t) = 1e for all t ∈ ¬E, we conclude that S �� ¬E. A similar argumentshows that E �� T . Thus �� satisfies EF.

Acknowledgements The first author thanks Luminita Vıta for many stimulating conversations about this paper and aboutapartness in general. He also thanks the FoRST New Zealand for supporting her as a New Zealand Science & TechnologyPostdoctoral Research Fellow from 2002 to 2005, thereby facilitating our development of the theory of apartness spaces.The second author thanks the University of Canterbury for supporting him by a Doctoral Scholarship. Both authors thank thereferee for pointing out a gap in the argument in our original version of the paper.

3) Notice that this is not a consequence of (∀x ∈ Re)(( 1

3)e ≤ x ∨ x ≤ ( 2

3)e) which does not hold for the extended reals, but of s < t

and ¬(r > s) implying r < t for r, s, t ∈ Re.

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Math. Log. Quart. 52, No. 5 (2006) 469

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