The second weak anti-Specker property we will consider is thenon-Specker propertyAS¬X. This is
just the denial ofSpeckand states, for a metric spaceX:
AS¬X If(zn)n>1is a sequence inX, then it is impossible for(zn)to be eventually bounded away from each point ofX.
It is straightforward to show that this is equivalent to the following property, whereX∪{ξ}is a
one-point extension ofX(as inASandASltd):
If(zn)n>1is a sequence inX∪{ξ}that is eventually bounded away from each point of
X, then
(∀i)¬¬(∃n > i)zn=ξ
.
Where the limited and full anti-Specker properties depended upon the choice of the parent spaceZ,
formulations of the non-Specker property are fully determined by the choice ofX. Accordingly, not
Proposition 3.13: BISH ` The following are equivalent.
(i) AS¬[0,1].
(ii) AS¬
2N+.
(iii) AS¬Xfor each inhabited compact spaceX.
Proof. As inProposition 3.1,(iii)=⇒(i)is trivial and(i)=⇒(ii)follows fromProposition 3.2.
For(ii)=⇒(iii), again letXbe a compact metric space, and useTheorem 1.4of [BR87, p. 106]
to find a continuous mapping of2N+
ontoX. Lemma 3of [BD10, p. 436] now gives the desired
result.
Notice that we may also prove the implication(i)=⇒(ii)using the (somewhat less elegant) original
construction from page25. For, suppose there exists a Specker sequence(αn)n>1in2N
+
. As before, define the sequence(xn)n>1in[0, 1]∪{2}by
xn= F(un) ifun=α|un|(|un|), and 2 otherwise.
Now consider the sequence(yn)n>1defined by
yn= xn ifxn ∈[0, 1], and yn−1 otherwise,
wherey0 = 0. It is not hard to show that(yn) is a Specker sequence in[0, 1]: fix any point
x ∈ [0, 1]. Then there existsN ∈ N+ andδ > 0 such that|x
n −x| > δfor alln >N[BB07, p. 200]. Consider all the pathsumwith|um| = |uN|+1 (where(un)n>1is again the one-one
enumeration(λ, 0, 1, 00, 01, 10, 11, 000, . . .)of2∗): we must haveuM = α|uM|(|uM|)for one of
these. ThenyM=xMand thus, for allm>M, there exists somen>Msuch thatym=xn. But
M > N, so it follows that|ym−x|> δfor allm>M.
So we see again that the existence of a Specker sequence in2N+
entails the existence of a Specker sequence in[0, 1]; contraposing, we obtain the desired result.
InChapter 4, we will study a number of principles that are equivalent to the Specker property
Speck[0,1]. This will allow us to identify, by contraposition, severalnegativeprinciples that are
equivalent toAS¬: most notably, we will see inCorollary 4.5thatAS¬falls into an equivalence
class with severalweak fan theorems. It would be illuminating to identify apositiveresult equivalent
toAS¬; however, it is not clear how such an equivalence could be found.
The non-Specker property is also significant in that it is the strongest anti-Specker property that we have been able to show followsdirectlyfromWLPO[BDMJ12]. While we know already that
all of the anti-Specker properties we will be interested in follow fromWLPO5(and, indeed,LLPO,
due to a result of Diener [Die13]), direct proofs enable us to better understand preciselyhowthese
principles interact with each other — and in this case, some concrete payoff comes as soon as
Proposition 3.22.
We begin this proof by extendingWLPOto predicates onN+×N+(as opposed to binary sequences,
which correspond to predicates onN+).
Lemma 3.14: BISH+WLPO ` IfPis a decidable predicate onN+×N+, then
(∀i)¬¬(∃m)P(i,m)∨ ¬¬(∃i)(∀m)¬P(i,m).
Proof. For eachi∈N+, define a binary sequence(λ(i)
m)m>1such that
λ(mi)=0 =⇒ ¬P(i,m), and
λ(mi)=1 =⇒ P(i,m).
WLPOallows us to decide whether or notλ(i) = 0; accordingly, we can define another binary
sequence(µi)i>1so that µi=0 =⇒ ¬(∀m) λ(mi)=0, and µi=1 =⇒ (∀m) λ(mi)=0 . Now applyWLPOto(µi)to obtain the following two cases:
5For, overBISH:WLPO=⇒FT
v Ifµi=0 for alli∈N+, then (∀i)¬(∀m)¬P(i,m), and so(∀i)¬¬(∃m)P(i,m). v Otherwise, we have ¬(∀i)¬(∀m)¬P(i,m), and so¬¬(∃i)(∀m)¬P(i,m).
We now turn to our main result. The notation #Xrefers to the cardinality of the setX.
Proposition 3.15: BISH+WLPO ` AS¬.
Proof. Let(zn)n>1be a Specker sequence in[0, 1]. We construct, inductively, a sequence(In)n>0
of intervals such that for eachn, two properties hold.
(i) |In|=2−n, andIn⊂In−1ifn>1;
(ii) (∀i)¬¬(∃m)h#j6m: zj∈In >i
i
.
Most of the work of this proof revolves around property(ii), which is a weakening of the claim that the number of terms of(zn)inInis unbounded.
Start the induction by settingI0 = [0, 1], which clearly satisfies(i). Furthermore,zj ∈I0for all
j∈N+, so given anyi∈N+, choosingm
>iyields #{j6m: zj∈I0}>i. That is,
(∀i)(∃m)h#j6m: zj∈I0 >i
i
and property(ii)follows for this base case.
Now fix anyk ∈ Nand suppose that we have constructed an intervalIk = [ak,bk]with the relevant properties. Denote byξkthe midpoint ofIk, and letHLandHRbe the left and right closed halves ofIk, respectively. We will make our choice ofIk+1from these halves: clearly, no matter
Since(zn)is a Specker sequence, it is eventually bounded away fromξkandbk: that is, there exist
N∈N+andδ >0 such that|z
n−ξk|> δand|zn−bk|> δfor alln>N. Hence for anyn>N, we can decide whether or notzn ∈HR. This means that the predicate
P(i,m)≡#j: N6j6N+m∧zj∈HR >i
onN+×N+is decidable, so byLemma 3.14, we can distinguish between the following two cases:
v In the case
(3.1) (∀i)¬¬(∃m)h#j: N6j6N+m∧zj∈HR >i
i
, setIk+1=HR. Fix anyi∈N+and assume that
¬(∃m)h#j6m: zj∈Ik+1 >i i , (3.2) and so(∀m)¬h#j6N+m: zj∈Ik+1 >i i . Then since #j: N6j6N+m∧zj∈Ik+1 >i =⇒ #j6N+m: zj∈Ik+1 >i
for allm, we have
(∀m)¬h#j: N6j6N+m∧zj∈Ik+1 >i
i
.
But this contradicts (3.1); hence we obtain¬(3.2), and sinceiwas arbitrary, property(ii)
follows forIk+1.
v In the case
(3.3) ¬¬(∃i)(∀m)h#j: N6j6N+m∧zj∈HR < i
i
it is impossible for there to be infinitely manymwithzm∈HR. SetIk+1=HL, and assume both (∃i)(∀m)h#{j: N6j6N+m∧zj∈HR}< i i (3.4) and(∃i)¬(∃m)h#j6m: zj∈Ik+1 >i i . (3.5)
Accordingly, construct numbersi1,i2∈N+such that
(∀m)h#{j: N6j6N+m∧zj∈HR}< i1 i (3.6) and(∀m)h#j6m: zj∈Ik+1 < i2 i . (3.7)
We can weaken (3.7) to obtain
(3.8) (∀m)h#j: N6j6N+m∧zj∈Ik+1 < i2 i
.
Now from our choice ofNwe see that, for alln > N, we havezn ∈ Ik only if either
zn ∈HL≡Ik+1orzn∈HR. Hence we can combine (3.6) and (3.8) to obtain: (3.9) (∀m)h#j: N6j6N+m∧zj∈Ik < i1+i2
i
. But it follows from induction assumption(ii)that
¬(∀m)h#j6m: zj∈Ik < i1+i2+ (N−1) i , and since #j: N6j6N+m∧zj∈Ik < i1+i2 =⇒ #j6N+m: zj∈Ik < i1+i2+ (N−1) =⇒ #j6m: zj∈Ik < i1+i2+ (N−1) for allm, we have
¬(∀m)h#j: N6j6N+m∧zj∈Ik < i1+i2
i
which contradicts (3.9). So we have (3.5)=⇒¬(3.4). But (3.3)≡¬¬(3.4); hence we conclude
¬(3.5), and property(ii)follows forIk+1.
This completes the construction of(In). Now, it follows from property(i)thatTn>1Inconsists of a single point — call itξ. PickN∈N+andδ >0 such that|z
n−ξ|> δfor alln>N, and choose
kfor which|Ik|< 12δ. Now for alln>N, we have
zn−ξk > zn−ξ − ξ−ξk > δ− 12δ= 12δ
(where, as earlier,ξkdenotes the midpoint of the intervalIk). Thereforezn ∈/ Ik. It then follows that
¬(∃m)h#j6m: zj∈Ik >N+1
i
,
which contradicts property(ii)ofIkin the casei≡N+1.