We work within the framework of Russian constructivism in the style of Richman’s [46, 7] and synthetic computability [3]. The following principles are valid in Russian constructivism.
1. Countable choice AC(N), in fact even dependent choice.
2. Markov Principle: if not all terms of a binary sequence are one, then there exists a term which is zero.
3. Enumerability Axiom: there are countably many countable subsets ofN.
At first we restrict attention to T = Σ = Σ01, the standard choice. Markov Principle says Σ ⊆ Ω¬¬, so the conditions of Theorem 2.53 are satisfied. In particular, closed subsets are precisely those which have open complements, andR, as well as metric spaces in general, are Hausdorff.
Countable choice (more precisely, its instance AC(N, 2N։ Σ01)) and Σ = Σ01 imply that the set of countable subsets ofN is just O(N) (Proposition 1.4). By the Enumerability Axiom there is an enumeration W :N → O(N). Furthermore, the Enumerability Axiom is equivalent to Richman’s axiom CFP which states that there is an enumeration φ0, φ1, . . . of those partial mapsN ⇀ N that have countable graphs. A partial map f :N ⇀ N has an enumerable graph if and only if “f (n) is defined” is semidecidable for all n ∈N.
The classical nature of Markov Principle and the non-classical nature of Enumerability Axiom combine into a strange mix of consequences. We show that WSO is valid, but N• is not metrized or compact.
Proposition 5.1 The principle WSO holds.
Proof. A detailed proof can be found in [3, 4.26], but we also present a proof here. By Lawvere’s positive radius is contained in V .
Proof. Let Tot ⊆ N be the set of those n ∈ N for which φn is a total function, and let
Tot and r are surjective, so is ψ.
Furthermore, define the map s :N• →N• by s(α)(k) := α k(k+1)2
. We have s(∞) = ∞ while for n ∈N s(n) computes the n-th term of the sequence (0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4 . . .) which repeats every natural number one more time than its value.
Let
We have
Sm:= s−1(m) =nm(m+1)
2 ,m(m+1)2 + 1, . . . ,(m+1)(m+2)2 − 1o
=m(m+1)2 +N<m+1. For every i ∈ Smthere is ni∈N such that ψni= i. Each of the numbers niis an element of U when we can find a witness k for the condition ψk= ψni in the definition of U . Note that different nis cannot receive the same witness k, and as there are m + 1 numbers ni, but only m possible witnesses k, not all of nks are in U , therefore not all elements of Sm are in V . Obviously ∞ ∈ V . For any r ∈R>0 the ball B (∞, r) is not contained in V because that would also mean that Sm is contained in V for large enough m.1
This proposition tells us that N• is not (weakly) metrized (and by Theorem 4.22, because WSO holds, that it is not compact). The immediate consequence is that the Baire spaceB and the Cantor spaceC are not (weakly) metrized2 or compact either since if they were, so would be their retractN•.
With a little more effort we can extend this example considerably. Recall that the domain of a convergent sequence in a complete metric space can be extended toN•. In particular, if a complete metric space X has an accumulation point, an application of dependent choice yields an injective map fromN•into X. Should the image of this map have the subspace intrinsic topology, X cannot be metrized since otherwise N• would be metrized as well by Corollary 4.39. We show this idea works.
Lemma 5.3 Let K ⊆ X. Then K is subcompact in X if and only if its double complement KC C = X \ (X \ K) is subcompact in X.
Proof. For any open set U ∈ O(X) we have
KC C ⊆ U =⇒ K ⊆ U =⇒ KC C ⊆ UC C =⇒ KC C⊆ U
(the last implication holding by Markov principle), so K is subcompact if and only if KC C is.
Lemma 5.4 Recall the successor map succ :N• → N• and the predecessor map pred : N• → N•. For an arbitrary u ∈N• the following holds.
1.
N•[pred(u),succ(u)]= N•<pred(u)∪N•>succ(u)
C
=
= N•<pred(u)∪N•(pred(u),u)∪N•(u,succ(u))∪N•>succ(u)
C
=
pred(u), u, succ(u) C C
2. The setN•[pred(u),succ(u)] is subcompact inN•.
1The complement of the set V from the proof is neither finite nor infinite. In recursion theory such sets are called immune. In fact, for any V satisfying the claim of the proposition in any model of synthetic topology in which WSOholds and Σ ⊆ Ω¬¬, the complement of V must be immune. Indeed, it cannot be bounded (much less finite) since otherwise we would have a ball with positive radius around ∞ in V . And if an injective sequenceN → VC existed we could replace it with a strictly increasing sequence which would induce a map e :N• →N•. The set e−1(V ) = {∞} would then be open inN•, in violation of WSO.
2The fact thatB is not (weakly) metrized is essentially a result of Friedberg’s who constructed an effective but not partial recursive operator [17].
Proof.
1. The first equality holds by definition of ≤. It is easy to see N•(pred(u),u) =N•(u,succ(u)) = ∅
which takes care of the second equality. For the final one, observe that for t ∈N• t > a ∧ t < b ⇐⇒ ¬¬(t > a) ∧ ¬¬(t < b) = ¬(t ≤ a ∨ t ≥ b) by Markov principle, so
¬
t < pred(u) ∨ t > pred(u) ∧ t < u
∨ t > u ∧ t < succ(u)
∨ t > succ(u)
⇐⇒ t ≥ pred(u) ∧ ¬¬ t ≤ pred(u) ∨ t ≥ u
∧ ¬¬ t ≤ u ∨ t ≥ succ(u)
∧ t ≤ succ(u)
⇐⇒ ¬¬
t ≥ pred(u) ∧ t ≤ pred(u) ∨ t ≥ u
∧ t ≤ u ∨ t ≥ succ(u)
∧ t ≤ succ(u)
⇐⇒ ¬¬
t = pred(u) ∨ t ≥ u
∧ t ≤ u ∨ t = succ(u)
⇐⇒ ¬¬ t = pred(u) ∨ t = u ∨ t = succ(u) .
2. The set
pred(u), u, succ(u)
is finite, hence compact, and since
N•[pred(u),succ(u)]=
pred(u), u, succ(u) C C
by the previous item, the setN•[pred(u),succ(u)] is subcompact inN• by Lemma 5.3.
Theorem 5.5 Let X = (X, dX) be a metric space with an accumulation point p ∈ X.
1. If X is complete, there exist maps a :N•→R≥0, b : {−1} ∪N•→R≥0 and y :N• → X with the following properties (with t ∈N• when indexing a, y, and t ∈ {−1} ∪N• when indexing b; also pred(0) := −1 on {−1} ∪N•):
• t < ∞ ⇐⇒ at> 0 ⇐⇒ bt> 0 ⇐⇒ dX(p, yt) > 0,
• the maps a, b are (strictly) decreasing: for t, u ∈ N•
t < u =⇒ at> au, t ≤ u =⇒ at≥ au, and for t, u ∈ {−1} ∪N•,
t < u =⇒ bt> bu, t ≤ u =⇒ bt≥ bu,
• dX(p, yt) ≤ 2−succ(t),
• at≥ bpred(t)≥ dX(p, yt) ≥ asucc(t)≥ bt; if (and only if ) t ∈N, then the inequalities are strict, i.e. at> bt−1> dX(p, yt) > at+1> bt.
2. If the maps a, b, y as above exist, then there is a (strongly) injective nonexpansive map e :N•→ X such that e(∞) = p, and the image of e is a strongly closed subset of X.
Proof.
1. We inductively define sequences a :N → R>0, b :Z≥−1→R>0and y :N → X as follows. Let a0:= 1. Now take n ∈N, and assume that we know an. Define bn−1:= a2n, and let yn ∈ X be some element for which 0 < dX(p, yn) < bn−1 (it exists because p is an accumulation point). Also, let an+1 := dX(p,y2 n). Dependent choice gives us the actual sequences which have the following properties for all suitable n:
bn+1<bn
4 , an+1< bn−1
2 =an
4 , bn= an+1
2 < an+1, an ≤ 2−2n, bn≤ 2−2n−3, dX(p, yn) < bn−1 ≤ 2−2n−1≤ 2−n−1,
an> dX(p, yn) > bn, [
n∈N
R(an,bn)=R(0,1).
These sequences can be extended to a, b :N•→R≥0, y :N•→ X (which we by a slight abuse of notation denote by the same letters) by Proposition 4.8. Notice that a∞ = b∞ = 0 and y∞= p, and that a, b, y satisfy all required properties.
2. Define e :N•→ X by e(t) := yt. For any i, j ∈N, i 6= j, we have
dX(yi, yj) ≤ dX(p, yi) + dX(p, yj) ≤ 2−i−1+ 2−j−1 ≤ 2− inf{i,j}= dC(i, j).
By Theorem 3.50 the map e is Lipschitz with coefficient 1, i.e. nonexpansive.
To prove that e is strongly injective, take t, u ∈ N•, and assume t # u, i.e. t < u ∨ t > u.
Without loss of generality t < u (in particular t ∈N and t + 1 ≤ u). Then dX(yt, yu) ≥ dX(p, yt) − dX(p, yu) > at+1− au≥ 0.
It remains to show that the image of e is strongly closed. We need to see that it is closed, and that it has the subspace intrinsic topology. Closedness is easy: N•is an inhabited CTB, so its image is as well, and therefore strongly located in X. Since R is Hausdorff, e(N•) is closed.
As for the subspace topology, take any U ∈ O(e(N•)). For x ∈ X we define Ax:= {t ∈N•| at≥ dX(p, x) ≥ bt} .
We claim that for all x ∈ BX(p, 1) the set Ax is inhabited, though not by much. More precisely, we claim that there exists u ∈N•such that u ∈ Ax⊆N•[pred(u),succ(u)]. Define α to be a sequenceN → 2 such that αn = 1 =⇒ dX(p, x) < an+1and αn= 0 =⇒ dX(p, x) > bn. Such α exists by countable choice because bn < an+1. Let u := r(α) where r : 2N→N• is the standard retraction. In particular this means that if u ∈N, then αu is the first zero in the sequence α.
• u ∈ Ax
First we prove au ≥ dX(p, x). If u = 0, we are done since x ∈ BX(p, 1). Assume now u ≥ 1, and au< dX(p, x). This means there exists n ∈N such that 2−n< dX(p, x). If n < u, then αn = 1, so dX(p, x) < an+1 < dX(p, yn) ≤ 2−n−1< 2−n, a contradiction.
If n ≥ u, then u ∈N, and αu = 0 while αu−1= 1 (recall u ≥ 1), so dX(p, x) < auwhich is a contradiction also.
Second, assume dX(p, x) < bu. Then bu> 0, so u ∈N. Consequently αu= 0, meaning dX(p, x) > bu, a contradiction.
• Ax⊆N•[pred(u),succ(u)]
Take any t ∈ Ax; then at≥ dX(p, x) ≥ btholds.
Assume t < pred(u). Then t ∈N, and t + 1 < u, so αt+1 = 1 which means dX(p, x) <
at+2< bt, a contradiction.
Assume t > succ(u). Then u ∈ N, and u + 2 ≤ t. We have αu = 0 which means dX(p, x) > bu> au+2≥ at, a contradiction.
• Axis subcompact inN•
The set Axis closed inN•[pred(u),succ(u)] (because ≥ is a closed relation since < is open) which is in turn subcompact inN• by Lemma 5.4.
Consequently e(Ax) is subcompact in e(N•), so the set
V := {x ∈ BX(p, 1) | e(Ax) ⊆ U }
is open in BX(p, 1), therefore also in X (since Σ is a dominance). We prove the theorem once we conclude V ∩ e(N•) = U .
• Ayu = {u} for all u ∈N•
Since au ≥ dX(p, yu) ≥ bu, we have u ∈ Ayu. Conversely, take any t ∈ Ayu. If t < u, then t ∈N, and we obtain a contradiction dX(p, yu) > at+1≥ au≥ dX(p, yu). If t > u, we similarly obtain a contradiction dX(p, yu) > au+1≥ at≥ dX(p, yu). Thus t = u.
• V ∩ e(N•) = U
The claim follows from e(Ayu) ⊆ U ⇐⇒ e({u}) ⊆ U ⇐⇒ yu∈ U .
Lemma 5.6 Let A ⊆ X, AC = ∅, and let X = (X, dX) be a metric space, with A = (A, dA) its metric subspace.
1. If A is metrized, then so is X.
2. If A is compact, then so is X.
Proof.
1. Take any U ∈ O(X). Then U ∩ A is open in A, so under the assumption that A is metrized, we can write it as an overtly indexed union
U ∩ A =[
i∈I
BA(ai, ri)
where ai∈ A, ri∈R.
We claim that U =S
i∈IBX(ai, ri). Since the sets both on the left and the right are open, and therefore ¬¬-stable, we can just as well prove that their complements match. But for any V ∈ O(X) and x ∈ X we have
¬(x ∈ V ∧ x ∈ A) ⇐⇒ ¬(¬¬(x ∈ V ) ∧ ¬¬(x ∈ A)) ⇐⇒ ¬¬¬(x ∈ V ) ⇐⇒ ¬(x ∈ V ), so (V ∩A)C= VC. Since U ∩A =S
i∈IBA(ai, ri) and S
i∈IBX(ai, ri)
∩A =S
i∈IBA(ai, ri), the result follows.
2. If A is compact, it is subcompact in X, so by Lemma 5.3 AC C = X is subcompact in X, i.e. X is compact.
Theorem 5.7 Let X = (X, d) be a metric space with a metrically dense subovert subset (e.g. a metrically separable space).
1. If X has an accumulation point, then it is neither metrized nor compact.
2. Say that X is locally semilocated when for every x ∈ X there exists r ∈ R>0 such that the distance d {y ∈ X | 0 < d(x, y) < r} , x
is an extended real number. The following are equivalent.
• X is metrized and locally semilocated.
• The metric d is equivalent to the discrete metric.
In particular, if either (and therefore both) of these conditions holds, then X has decidable equality.
Proof.
1. Let p ∈ X be an accumulation point in X, and bX the completion of X (we identify X with a metric subspace of bX). By Theorem 5.5 there exist suitable sequences a, b, y which induce an injective nonexpansive map e : N•→ bX with a strongly closed image. Let X′ := X ∪ e(N•), and X′the metric subspace in bX with the underlying set X′. By Theorem 5.5(2) the image of e is strongly closed also in X′. If X is metrized/compact, then so is X′ by Lemma 5.6 which implies that e(N•) is metrized/compact by Theorem 4.38. Since e is bijective onto its image, compactness of e(N•) implies compactness ofN•, and since e is furthermore nonexpansive, metrization of e(N•) implies metrization ofN•by Lemma 4.33. Contradiction.
2. Notice that d is equivalent to the discrete metric if and only if for every x ∈ X there exists ǫ ∈R>0such that B (x, ǫ) = {x}, or equivalently, {y ∈ X | 0 < d(x, y) < ǫ} = ∅.
• (⇒)
Take any x ∈ X, and let r ∈ R>0 witness local semilocatedness around x. Let A :=
{y ∈ X | 0 < d(x, y) < r}, and define ǫ := inf{d(A, x), r}; then ǫ is a nonnegative real number. Since d(A, x) is the strict infimum, if d(A, x) = 0, then x is an accumulation point, and X could not be metrized. Thus (by Markov principle) d(A, x) > 0, so ǫ > 0.
If there is y ∈ X, 0 < d(x, y) < ǫ, then y ∈ A, therefore d(x, y) ≥ ǫ, a contradiction, so this ǫ works.
• (⇐)
Suppose for every x ∈ X there is ǫ ∈ R>0 such that B (x, ǫ) = {x}. Then r := ǫ witnesses local semilocatedness of x (namely, A := {y ∈ X | 0 < d(x, y) < r} is empty, so dX(A, x) = ∞). Moreover, we see that the only metrically dense subset of X is X itself, so X is overt, and for any y ∈ X,
⊤ ⇐⇒ 0 < d(x, y) ∨ d(x, y) < ǫ =⇒ x 6= y ∨ x = y,
so X has decidable equality. Thus X is metrized by the discrete metric (by Proposi-tion 4.3), therefore also by d since it is equivalent to it.
The conclusion is that in Russian constructivism most interesting metric spaces are not metrized.
Granted, we considered only the case Σ = Σ01 (the smallest Σ for whichN is overt), but if a space is not metrized with respect to Σ01, then it is not metrized with respect to any larger Σ. This suggests that Type I computability is not an optimal choice for computation with metric spaces.