Recall that Specker’s theorem, in its original formulation, asserted the existence of anondecreasing
Specker sequence. To reflect this, we now acknowledge a further two weak anti-Specker properties which are nondecreasing counterparts ofASandAS¬. Theincreasing anti-Specker propertyfor a
(compact) metric spaceXwith total order6states:
AS↑X IfX∪{ξ}is a one-point extension ofXwithx < ξfor eachx∈X, and(zn)n>1
is anondecreasingsequence inX∪{ξ}that is eventually bounded away from
each point ofX, thenzn=ξeventually.
We will consider the following more general formulation, which follows readily fromASltd:
AS↑ AS↑Xholds for every totally ordered compact metric spaceX.
It may be thatAS↑and (say)AS↑[0,1]are equivalent, as was the case for the anti-Specker properties
examined earlier in this chapter. However, determining the fact of this matter remains an open problem: the need to preserve the ordering of the relevant sequences complicates the situation somewhat.
AS↑X¬ If(zn)n>1is anondecreasingsequence inX, then it is impossible for(zn)to be eventually bounded away from each point ofX.
AS↑¬ AS↑X¬holds for every totally ordered compact metric spaceX.
These properties are very natural to formulate and, indeed, constitute more faithful antitheses of Specker’s theorem than our prior anti-Specker properties; however, they are much harder to apply than, say,ASltd. Nevertheless, we are able to useAS↑in a compactness-like role to establish another
Heine-Borel property; namely, the (very weak)countable Heine-Borel property for disjoint intervals
of [BB08b, p. 585]. This states, for a subsetK⊂R:
HBdiK If(In)n>1is a sequence of pairwise-disjoint, inhabited, bounded open intervals such thatK⊆S∞
i=1Ii, then there existsk∈N
+such thatK⊆Sk
i=1Ii.
We will prove that this property holds for compact subsetsK⊂R; to do so, we make use of the
following lemma.
Lemma 3.23: BISH ` Let(In)n>1be a sequence of pairwise-disjoint, inhabited, bounded open intervals, andKa compact subset ofS∞
i=1Ii. Then for each`∈N+, eitherK\I`=∅, orK\I`is inhabited and compact.
Proof. Fix`and writeI`≡(a,b). By Bishop’s lemma (Proposition 3.1.1of [BV06, p. 64]), there existsx∈Ksuch that ifa6=x, thenρ(a,K)>0. Findksuch thatx∈Ik. Were it the case that a ∈ Ik, we would be able to find a point (slightly greater thana) belonging to bothI` andIk, contradicting the hypothesis that these intervals are disjoint; hence,a /∈Ik. This separatesaand x, so we see thatρ(a,K)> 0; a similar argument shows thatρ(b,K)> 0 also. Fix any positive r <min{ρ(a,K),ρ(b,K)}.
K a
b I`
r r
Figure 3.2:The intervalI`and its endpoints, which are separated from the compo- nents ofK.
Given anyx∈K, we can decide which of the following two cases holds:
v a−r < x < b+r, whencex∈I`;
v x < aorx > b, whencex /∈I`(and sox∈K\I`).
(That is,I`is in a certain sensedetachablefromK.) In particular, if infK∈I`and supK∈I`, we see thatK⊂I`and thusK\I`=∅. Otherwise, one of infKand supKbelongs toK\I`, which is
therefore inhabited. We henceforth assume thatK\I`is inhabited, and show furthermore that it is compact.
K\I`is complete.Let(xn)n>1be a Cauchy sequence inK\I`, and letxbe the limit of(xn)in
K. Were it the case thatx∈I`, we would have|xn−x|>2rfor alln— a contradiction. Hence
x∈K\I`.
K\I`is totally bounded.Given >0, letδ=min{, 2r}, and take a finiteδ-approximationYto
K. For each pointy∈Y, determine whethery∈I`ory∈K\I`, and discardyin the former case.
The resulting set is a finite-approximation toK\I`.
We now prove our intended result.
Proposition 3.24: BISH+AS↑ ` HBdiKfor each inhabited, compactK⊂R.
Proof. Fix a compactK⊂R, chooseξ∈Rsuch thatξ−supK >1, and suppose that(In)n>1 is a sequence of intervals satisfying the hypotheses ofHBdiK. Construct a nondecreasing sequence
(zn)n>1∈K∪{ξ}, along with a sequence(Kn)n>1of compact subsets ofKand a sequence(`n)n>1 of indices, as follows:
TakeK1=K, and letz1=infK1. (Notice that, sinceK1is a totally bounded subset ofR,Proposition 2.2.5of [BV06, p. 39] guarantees the existence of infK1; then, by completeness, infK1∈K1.) Now
for eachn>1, givenKn ⊆Kandzn =infKn∈Kn, find`nsuch thatzn∈I`n, and consider
Kn+1≡Kn\I1\I2\· · ·\I`n.
v IfKn+1is empty, setzm = ξfor allm> n+1. (It doesn’t matter how(Kn)and(`n) behave after this point — sayKm=∅and`m=`n.)
v IfKn+1is inhabited and compact, setzn+1=infKn+1∈Kn+1and repeat this construction.
SinceKn+1⊂Kn, the sequence(zn)thus obtained is nondecreasing. Note also that, for eachnwithzn+1∈Kn+1, we can show the following:
(∗) `m>mfor allm6n+1.
For, sincezn+1∈/ I1,I2, . . . ,I`n butzn+1∈I`n+1, we have`n+1> `n. Furthermore,zm+1∈Km+1
for allmwith 1 6 m 6 n; so, repeating the same argument a total ofntimes, we see that `n+1> `n >· · ·> `1>1. This proves (∗), and also that
Kn+1=K\I1\I2\· · ·\I`n
for eachn∈N+.
Now fix any pointx∈K: we show that(zn)is eventually bounded away fromx. Find an index
k∈N+such thatx∈Ik. SinceI
kis open, there existsr >0 such thatB(x,r)⊂Ik. Now consider
zm+1for anym>k:
v Ifzm+1=ξ, we have|zm+1−x|>1.
v Ifzm+1 ∈ Km+1, then`m > m > kby (∗). In this casezm+1 ∈/ I1,I2, . . . ,I`m, and in
particular,zm+1∈/ Ik. Hence|zm+1−x|> r.
In either case,|zm+1−x| > min{1,r}; since this holds for allm>k, we conclude that(zn)is indeed eventually bounded away fromx. ApplyingAS↑, we find some minimal indexN∈N+such
thatzN+1=ξ. Then KN+1=K\ `N [ i=1 Ii=∅,
and we conclude thatK⊆S`N
i=1Ii.
The problem of identifying how much weaker thanASltdandAS¬these increasing anti-Specker
from an increasing anti-Specker property to its less restricted analogue? This corresponds to a problem we will encounter inChapter 4; namely, that of finding how much weaker the principle
Speck[0,1]is than its counterpartSpeck↑[0,1].
The results of this chapter are summarised inFigure 4.6; however, this figure also incorporates
severalweak fan theoremsand corresponding results which we do not introduce untilSection 4.1;
In this chapter, we identify three equivalence classes of principles that are, largely, opposing counterparts of ones we have encountered already. In doing so, we will consider principles that do not hold classically or intuitionistically, but many of which are true withinRUSS. Accordingly, the
linchpins of this investigation will be two variations upon Specker’s theorem for the unit interval. Recall the first of these fromSection 2.1:
Speck[0,1] There exists a sequence in[0, 1]that is eventually bounded away from each
point of[0, 1].
We will begin by exploring the relationship betweenSpeck[0,1]and antitheses of various versions
of Brouwer’s fan theorem. We say that a barBfor2∗isnonuniformif, for eachn∈N, there exists
a path of lengthnin2∗that missesB. With this in mind, we will be interested in principles of the
form:
anti-FT? There exists a nonuniform ?-bar for2∗.
Diener studied these principles in [Die08, pp. 65–70], proving (among other things) the equivalence ofanti-FTc,anti-FTFullandSpeck[0,1]. However, the proof we present here is novel and illuminates
the relationships between these principles from a new perspective. In particular, we demonstrate how the Specker property arises by appealing directly to intuitions about the behaviour of paths in the complete binary fan, rather than by using Bishop’s lemma (and in doing so utilise a different embedding of paths into the unit interval1). The hope is that this supplements Diener’s proof in
giving a glimpse at the underlying structural relationships between nonuniform bars and Specker sequences.
The disadvantage of our approach here is that it relies uponLemma 4.1; however, this result lends itself to reasoning in a way familiar to classical mathematicians and is less steeped in constructive nuance than Bishop’s lemma.
4.1
Fan-Theoretic Equivalents of AS
¬Our exploration of the relationship betweenSpeck[0,1]and the various anti-fan-theorems will shed
light upon a corresponding relationship between the non-Specker propertyAS¬and certainweak fan theorems. We start down this path by observing the following result of Berger and Bridges
[BB07, p. 198] about our embeddingFfromSection 3.1.
Lemma 4.1: BISH ` If(zn)n>1is a sequence of real numbers that is eventually bounded away fromF(α)for eachα∈2N+
, then(zn)is eventually bounded away from each point of[0, 1]. This allows us to prove the following result.
Proposition 4.2: BISH+anti-FTFull ` Speck[0,1].
Proof. Assuminganti-FTFull, letBbe a nonuniform bar: that is, for eachn, there exists a pathun of lengthnthat missesB. Define a sequence(zn)n>1in[0, 1]byzn =F(un). We will show that (zn)is a Specker sequence.
Fix any infinite pathα≡(an)n>1. SinceBis a bar, we can findN1∈N+such thatαN1∈B. To illustrate the next part of the proof, suppose thataN1 =0 (the caseaN1 =1 is similar). Letkbe the greatest integer less thanN1such thatak=1 (if no suchkexists, the desired result follows from cases(i)and(ii)of the following argument). Define infinite pathsβ≡(bn)n>1andγ≡(cn)n>1 by: bn = an ifn < k, 0 ifn=k, 1 ifn > k; and cn= an ifn < N1, 1 ifn=N1, 0 ifn > N1
(as illustrated inFigure 4.1). Now, findN2,N3∈N+such thatβN
2∈BandγN3∈B, and define
αN1
α
β γ
Figure 4.1:The relationship between the pathsα,βandγ
integermwith 16m6N1such that[un]m6=ambutun(m−1) =α(m−1)(we know that such anmexists because|un|=n>N>N1butun(N1)6=α(N1)∈B). Four cases arise:
(i) Case 1:am=0 andm=N1.
αN1
γ un
Note that
F(α)6F(γ) =F(a1,a2, . . . ,aN1−1, 1, 0, 0, . . .).
SinceγN3∈Bandunis not blocked byB, we have
F(un)>F(a1,a2, . . . ,aN1−1, 1, 0, 0, . . . , 0, 1
↑
N3
, 0, 0, . . .); hence we deduce thatF(un) −F(α)>2−N3 >2−N.
(ii) Case 2:αm=0 andm < N1.
αN1
un
In this case we have:
F(un)>F(a1,a2, . . . ,am−1, 1, 0, 0, . . .), andF(α)6F(a1,a2, . . . ,am−1, 0, 1, 1, . . . , 1, 0 ↑ N1 , 1, 1, . . .) =F(a1,a2, . . . ,am−1, 0, 1, 1, . . . , 1, 1 ↑ N1 , 0, 0, . . .). Hence: F(un) −F(α)>2−m− 2−(m+1)+2−(m+2)+· · ·+2−N1=2−N1 >2−N.
(iii) Case 3:am=1 andm=k, wherekis, as before, the greatest integer less thanN1such thatak =1. αN1 β un We have F(α)>F(β) =F(a1,a2, . . . ,ak−1, 0, 1, 1, . . .).
SinceβN2∈Bandunis not blocked byB, F(un)6F(a1,a2, . . . ,ak−1, 0, 1, 1, . . . , 1, 0 ↑ N2 , 1, 1, . . .); henceF(α) −F(un)>2−N2 >2−N.
(iv) Case 4:am=1 andm < k.
αN1 un We have: F(un)6F(a1,a2, . . . ,am−1, 0, 1, 1, . . .) =F(a1,a2, . . . ,am−1, 1, 0, 0, . . .), andF(α)>F(a1,a2, . . . ,am−1, 1, 0, 0, . . . , 0, 1 ↑ N1−1 , 0, 0, . . .); henceF(α) −F(un)>2−(N1−1)>2−N1 >2−N.
In all cases, we have|F(α) −zn|>2−N, and a similar argument holds whenaN1 =1. That is,(zn) is eventually bounded away fromF(α)for eachα∈2N+
. It follows byLemma 4.1that(zn)is a
Specker sequence in[0, 1].
We now strengthen this result by showing thatanti-FTFullandSpeck[0,1]are in fact equivalent
overBISH, and furthermore,anti-FTcalso falls into their equivalence class [BDMJ12].
Lemma 4.3: BISH ` If(zn)n>1 is a Specker sequence in[0, 1], then there exists a sequence (un)n>1in 2∗ such that(F(un))n>1 is a Specker sequence, and|zn −F(un)| < 2−n for each
Proof. Observe that thedyadic rational numbers— those of the formF(u)for someu∈2∗— are
dense in[0, 1]. Hence we see that for eachn∈N+, there existsu
n∈2∗with|zn−F(un)|<2−n. Define a sequence(yn)n>1by yn = F(un). Since(zn) is a Specker sequence we can, given
x∈[0, 1], pickN∈N+such that|z
n−x|>2−N+1for alln>N. Then for suchn, we have
yn−x > x−zn − zn−yn >2−N+1−2−n >2−N+1−2−N =2−N.
Hence(yn)is a Specker sequence.
Proposition 4.4: BISH+Speck[0,1] ` anti-FTc.
Proof. Let(zn)n>1be a Specker sequence in[0, 1]. In view ofLemma 4.3, we may assume that for eachn∈N+there exists a pathu
n ∈2∗such thatzn=F(un). Appending zeroes as necessary, we may further assume that|un|>nfor eachn. Following a proof of Berger and Bridges [BB07, pp. 200–201], we see that D≡ u∈2∗: F(u) −z|u| >2−|u|+1
is detachable (since the numbersznare rational), and that
B≡u∈2∗: (∀v)
u∗v∈D
is a c-bar for2∗. Given anyn∈N+, suppose thatu
n(n)is blocked byB. Thenun(n)∈B(since c-subsets are closed under extensions), and so
F(un(n)) −zn >2−n+1>2−n > F(un(n)) −F(un) = F(un(n)) −zn ,
Corollary 4.5: BISH ` The following are equivalent. (i) Speck[0,1]. (ii) anti-FTc. (iii) anti-FTΠ01cl. (iv) anti-FTΠ01. (v) anti-FTFull.
Proof. Proposition 4.4shows that(i)=⇒ (ii), andProposition 4.2shows that(v)=⇒(i). To
complete the proof, observe that all c-bars areΠ01cl-bars, which are themselvesΠ01-bars, which are
themselves bars; hence we have(ii)=⇒(iii)=⇒(iv)=⇒(v).
Contraposing these equivalences, we arrive at our fan-theoretic equivalents ofAS¬; namely,weak fan theoremsof the form
FT¬¬? For every ?-barBfor2∗, it is impossible thatBbe nonuniform.
Corollary 4.6: BISH ` The following are equivalent.
(i) AS¬. (ii) FT¬¬c . (iii) FT¬¬Π0 1cl. (iv) FT¬¬Π0 1 . (v) FT¬¬Full.
Structurally, these fan theorems are similar to — though not the same as2— double-negated variations
uponFTcthroughFTFull(hence the notation). For non-decidable properties, such a double-negation-
like construction represents a loss of information: in this case, the lost information corresponds to
the structural distinctions between these weak fan theorems. This may be whyFT¬¬∆ seemsnotto fall into the equivalence class ofCorollary 4.6: since it refers to a property that is in part decidable, its weakening ofFT∆represents a lesser degree of information loss.