We now comment on the hypothesis that S1 is weakly located at 0. That
hypothesis is used to prove, in Corollary41, thatρ(0, Sr)>0 for each r >0,
and hence, in Proposition43, thatθ−1is pointwise continuous onG(from which
the uniform continuity ofθ−1 on compact sets follows). It is easy to see that
the pointwise continuity ofθ−1 at 0 isequivalent to the weak locatedness ofS 1
at 0. Without the latter hypothesis, but still making use of Proposition43, we can prove the following two propositions.
Proposition 46 Let θ be a continuous isomorphism of R onto a complete
abelian groupG. Thenθ−1 is sequentially continuous onG.
Proof. It clearly suffices to prove that θ−1 is sequentially continuous at 0.
Accordingly, let θ(tn) → 0 in G, and let ε > 0. By Ishihara’s tricks, either
|tn| < ε eventually or else |tn| > ε/2 infinitely often. In the latter case, by
Lemma 6.6.9 of [12], LPO holds; so S1, being separable, is located in G. It
follows from Proposition 43 that θ−1 is pointwise continuous at 0, which is
absurd, since|tn|> ε/2 infinitely often. Thus, in fact,|tn|< εeventually. Since
ε >0 is arbitrary, we conclude thatθ−1 is sequentially continuous at 0.
Proposition 47 Let θ be a continuous, one-one homomorphism ofR onto a
complete (metric) abelian group(G, ρ), and letr >0. Thenθ[−r, r]is compact, and the restriction ofθ−1 toθ[−r, r]is uniformly continuous.
Proof. Let 0< ε < r. Sinceθ is uniformly continuous on the totally bounded intervals (ε, r] and [r, ε), the set
S ={θ(t) :ε <|t|6r},
is totally bounded. Construct a strictly decreasing sequence (δn)n>1of positive
numbers converging to 0 such that for eachn, ifx∈Gand ρ(0, x)< δn, then
ρ(0, nx)<2−n. Construct an increasing binary sequence (λ
n)n>1 such that
λn= 0⇒ρ(0, S)< δn,
λn= 1⇒ρ(0, S)> δn+1.
If λn = 0, choose tn such that ε < |tn| 6 r and ρ(0, θ(tn)) < δn; setting
xn =θ(ntn), we haveρ(0, xn) <2−n. If λn = 1−λn−1, settm =tn−1 and
xm =xm−1 for all m > n. Then (xn)n>1 is a Cauchy sequence in G and so
converges to a limitx∞∈G. Letτ=θ−1(x∞), and compute a positive integer
N >|τ|/ε. Suppose thatλN = 0. Ifλm= 1−λm−1for somem > N, thenε <
|tm−1| 6r and θ(τ) =θ((m−1)tm−1). Since θ is one-one, τ = (m−1)tm−1
and therefore
|τ|= (m−1)|tm−1|>(m−1)ε>N ε,
a contradiction. Hence λn = 0 for all n > N and therefore for all n. Thus
|ntn|> nε→ ∞, but θ(ntn)→0. This contradicts Proposition46. Hence the
caseλN = 0 is ruled out, and we have λN = 1. Thus ρ(0, S)>0. It follows
that if|t|6rand ρ(0, θ(t))< δN+1, thenθ(t)∈/ S and therefore|t|6ε.
We can now compute α > 0 such that if t ∈ [−2r,2r] and ρ(0, θ(t)) < α, then|t|< ε. Corollary 8shows that K=θ[−r, r] is compact; so the mapping (x, y) Ã y−x is uniformly continuous on K ×K. Pick δ > 0 such that if x, x0, y, y0 ∈ K and ρ((x, y),(x0, y0)) < δ, then ρ(y−x, y0−x0) < α. If
t, t0∈[−r, r] and ρ(θ(t), θ(t0))< δ, then
ρ((θ(t), θ(t)),(θ(t0), θ(t))) < δ
and so
ρ(0, θ(t−t0)) =ρ(0, θ(t)−θ(t0))< α.
Since|t−t0|62r, it follows that|t−t0|< ε. Sinceε >0 is arbitrary, the proof
is complete.
In [23] (Theorem 4), Ishihara showed that the statement
Every sequentially continuous mapping of a separable metric space into a metric space is pointwise continuous
is equivalent, inBISH, toBD-N. SinceBD-Nholds in the classical, intuition- istic, and recursive models ofBISH(again see [23]), it follows from Proposition 46and the remark preceding it that the statement
(*) For each continuous isomorphism ofRonto a complete abelian group, the image of{t∈R:|t|>1}is weakly located at 0
holds in each of these models. This precludes our obtaining a Brouwerian or a recursive counterexample to (*). Also, in these models, Theorem 2 holds without additional hypothesis.
Propositions46and47enable us to show that the weak locatedness hypothesis in Theorem36can be derived if we assume a form of local path connectedness for the groupG. Note that, unlike the weak locatedness ofS1 at 0, this is an
intrinsic property of the groupG, independent of the isomorphismθ:R→G. Recall that a metric spaceX ispath connected if for allx, x0 ∈X there exists a
(uniformly continuous) pathγ: [0,1]→X withγ(0) =xandγ(1) =x0.
Lemma 48 Let X be a path connected metric space, andf a sequentially con-
tinuous mapping ofX intoR. Let a, b∈X andt∈Rsatisfy f(a)< t < f(b). Then for eachε >0, there exists x∈X such that |f(x)−t|< ε.
Proof. There exists a uniformly continuous mapping γ: [0,1]→X such that
γ(0) =aandγ(1) =b. Thenf◦γis a sequentially continuous mapping of [0,1] into Rsuch that f ◦γ(0)< t < f ◦γ(1). It follows from a well-known result (see Exercise 17 on page 58 of [12]) that for eachε >0, there exists ζ ∈[0,1] such that|f(γ(ζ))−t|< ε.
Letabe a point of a metric spaceX. We say thatXisball-locally path connected ataif there existsτ >0 such thatB(a, r) is path connected for all but countably manyr∈(0, τ).
Proposition 49 Let θ be a continuous, one-one homomorphism ofR onto a
complete (metric) abelian groupGthat is ball-locally connected at0. Then θ−1
is pointwise continuous.
Proof. By Proposition 42, it is enough to prove that θ−1 is continuous at 0.
Sinceθis injective, we can computer >0 such that 0< r < 1
2min{ρ(0, θ(1)), ρ(0, θ(−1))} andB(0, r) is path connected. We show that
θ−1¡B(0, r)¢⊂[−1,1]. (9)
Note thatρ¡θ(1), B(0, r)¢> randρ¡θ(−1), B(0, r)¢> r. Now computeα >0 such that if t, t0 ∈ [−1,1] and |t−t0| < α, then ρ(θ(t), θ(t0)) < r. Given
is sequentially continuous onG, it follows from Lemma48 there existst0 such
that 1−α < t0<1 andθ(t0)∈B(0, r). By our choice ofα,
ρ¡θ(1), B(0, r)¢6ρ(θ(1), θ(t0))< r,
a contradiction. The other possibility, t < −1, is ruled out similarly. We conclude thatt∈[−1,1]. Hence (9) holds.
Givenε >0 and using Lemma 47, now choose δ >0 such that if t, t0 ∈[−1,1]
and ρ(θ(t), θ(t0)) < δ, then |t−t0| < ε. If ρ(0, x) < min{δ, r} in G, then
x∈ B(0, r)⊂θ[−1,1], so ¯¯θ−1(x)¯¯< ε. Sinceε >0 is arbitrary, we conclude
that θ−1 is pointwise continuous at 0. Reference to Proposition 42 completes
the proof.
Corollary 50 Letθbe a continuous, one-one homomorphism ofRonto a com-
plete (metric) abelian group(G, ρ)that is ball-locally path connected at 0. Then
ρ(0, S1)>0 andS1 is weakly located at0.
Proof. By Proposition 49, there existsδ >0 such that ifρ(0, θ(t))< δ, then
|t|<1. Thus if|t|>1, thenρ(0, θ(t))>δ, from which the desired conclusions follow.
We now return to the condition “ρ(0, S1)>0” that, as we pointed out earlier,
is vital for the proof of our main theorem. First, we give a result that is worth including for the sake of completeness of exposition.
If, under the hypotheses of Theorem36, the groupGis locally compact, then, as a consequence of Corollary41 and our next proposition, we can prove that
S1 is located.
Proposition 51 Letθbe a continuous homomorphism ofRonto a locally com-
pact abelian groupG such thatρ(0, S1)>0, where Sr is defined at (8). Then
Sr is located inGfor eachr >0.
Proof. Fixr, Rsuch that
B =B(0, R+ 1)∩G
is compact and contains the (totally) bounded setθ[−r−1, r+ 1]. Let 0< ε <
1, and chooseδ∈(0,1/2) such that for all t, t0 ∈[−r−1, r+ 1], if|t−t0|6δ,
then ρ(θ(t), θ(t0))< ε/2. In view of Lemma 38, there exists c > 0 such that
ρ(0, Sδ)> c. Now, since Gis locally compact, the mapping (t, x)Ã θ(t) +x
is uniformly continuous on the compact subset [−r, r]×B ofR×G, so we can chooseα∈(0, ε/2) such that ify, z ∈B andρ(y, z)< α, then
Let t ∈ [−r, r], and let |t0| > r+δ be such that θ(t0) ∈ B. Suppose that
ρ(θ(t), θ(t0))< α. Then−t∈[−r, r], so
ρ(0, θ(t0−t)) =ρ(θ(t−t)), θ(t0−t))< c,
which contradicts our choice ofc, since|t0−t|>δ. Thus
ρ(y, θ[−r, r])>α (y∈Sr+δ∩B). (10)
Construct a finiteα/2-approximation {ξ1, . . . , ξm} to B, and write{1, . . . , m}
as a union of setsP, Q such that
i∈P ⇒ρ(ξi, θ[−r, r])>0,
i∈Q⇒ρ(ξi, θ[−r, r])<
α
2.
Consider t such that θ(t) ∈ Sr∩B. There exists y ∈ Sr+δ ∩B such that
ρ(θ(t), y) < ε/2. Indeed, either |t| > r+δ, when we take y = θ(t), or else
|t| < r+ 1−δ; in the latter case, since θ[−r−1, r+ 1] ⊂B, it follows from our choice ofδthat settingy=θ
³
t+ t |t|δ
´
fulfils our requirements. Pickisuch thatρ(y, ξi)< α/2. Ifi∈Q, then ρ(y, θ[−r, r])6ρ(y, ξi) +ρ(ξi, θ[−r, r]) <α 2 + α 2 =α,
which contradicts (10). Hence i /∈ Q and therefore i ∈ P. It follows that
{ξi:i∈P} is a finitely enumerableε-approximation to Sr∩B. Sinceε >0 is
arbitrary, it follows thatSr∩B is totally bounded. Hence Sr is locally totally
bounded and therefore (by Proposition 2.2.18 of [12]) located.
In the notation of Theorem36, can we prove, without assuming local compact- ness of our groupG, that if it is noncompact, thenS1is located? (We have just
shown that we can whenGis locally compact andρ(0, S1)>0.) The answer is
‘no’: the statement
If θ is a continuous, one-one homomorphism of Ronto a complete abelian groupG, such thatρ(0, S1)>0, thenS1is located in G
implies LPO. To prove this, consider any binary sequence (an)n>1 with at
most one term equal to 1. Construct a continuous injectionθ:R→Cwith the following properties:
I if eithert61 oran= 0 for alln6t, thenθ(t) = (2t,0);
I ifan= 1 andn < t6n+ 1, thenθ(t) = (2n, t−n);
Figure 4: What does the image ofRunderθlook like?
Thenθis a continuous isomorphism betweenRand the groupG=θ(R), taken with the Euclidean metric and the addition defined by
θ(t) +θ(t0) =θ(t+t0) (t, t0 ∈R).
To prove thatGis complete, we first observe
(a) that ifaN = 1, then Gis the closure inCof the set
((−∞, N)× {0})∪ {(N, y) : 06y61} ∪((−∞, N]× {1}) and so is complete; and
(b) that if Imθ(t)<1, then Reθ(t)>2t−2.
Let (tn)n>1 be any sequence in R such that (θ(tn))n>1 is a Cauchy sequence
in G. Then z = limn→∞θ(tn) exists in C. Either Imz > 0 or Imz < 1. In
the first case, computing in turn positive integersN, msuch that Imθ(tN)>0
andtN < m, we see that an = 1 for some n6m; so, by observation (a), Gis
complete. In the case Imz <1, compute a positive integerM > 1
2Rez. There
exists κsuch that Reθ(tk) <2M and Imθ(tk)<1 for all k >κ. For such a
value ofk, suppose thattk > M+1. Then Reθ(tk)<2tk−2; so, by observation
(b), Imθ(t)>1, a contradiction. Hence tk6M + 1 for allk>κ. Ifan= 0 for
alln6M + 1, then z= lim k→∞,k>κθ(tk) = limk→∞(2tk,0), so t= 1 2Rez= limk→∞tk 6M+ 1
andz= (2t,0) =θ(t). If, on the other hand,an= 1 for somen6M + 1, then
Gis complete. This completes the proof of completeness forG. It is clear thatρ(0, S1)>1. Now suppose that
s= inf{ρ(θ(0), θ(t)) :|t|>1}
exists. Either s > 1 and therefore an = 0 for all n, or else s < 2. In the
latter case we can chooset with |t|>1 andρ(θ(0), θ(t))<2. Ift <−1, then
ν > t. Ifan= 0 for all n6ν, thenρ(0, θ(t)) = 2t>2, which is absurd. Hence
an= 1 for somen6ν.
Note that in this example, the groupGis ball-locally path connected at 0: for 0< r <1,the set
B(0, r) = [−r, r]× {0}
is actually connected. However,B(0,1) is not path connected if there existsn
withan= 1. Moreover, if Gis locally compact, then eitheran = 0 for allnor
else there existsnsuch thatan= 1. For there exists a compact subset KofG
such that
∀t∈R(ρ(θ(0), θ(t))<2⇒θ(t)∈K) .
Pickt1, . . . , tmsuch that{θ(t1), . . . , θ(tm)}is a 1/2-approximation toK. Either
Imθ(tj)<1 for all j, in which casean = 0 for alln; or else there existsj such
Chapter 5
Open problems
5
Open problems
In this thesis we have presented a constructive consideration of the classical theorem
Theorem 1Letθbe a continuous homomorphism of Ronto a compact (metric) abelian group G. Then there exists τ >0 such that θ(τ) = 0.
and its contrapositive
Theorem 2Let θ be a continuous, one-one homomorphism of Ronto a com- plete (metric) abelian group G. Then Gis noncompact.
within the framework of Bishop’s constructive mathematics. However, our con- structive versions of Theorems 1 and 2 each involve extra hypothesis. The natural question to ask is then: Can we do better, or are Theorems 1 and 2 fundamentally non-constructive?
In particular, is the assumption thatθ(0,∞) is open in G necessary in Theo- rem 18? The next corollary, which follows directly from Theorem 18and the discussion following it, helps clarify this question.
Corollary 52 Letθbe a continuous homomorphism ofRonto a compact (met-
ric) abelian groupG. Then the following are equivalent.
(1) θ is periodic.
(2) θ(0,∞)is open.
(3) There exists a sequence (Rn)n>1 of real numbers strictly increasing to
infinity such thatθ[−Rn, Rn]is weakly bilocated at0 for each n.
(4) There existst >0 such that G=θ[0, t].
Proposition 12 shows that LPO implies condition (3) of the above corollary; whence Theorem1holds inBISH+LPO. Moreover, in any Brouwerian coun- terexample to Theorem 1 we cannot have θ(0,∞) open. Such a group and homomorphism, suggested by Hannes Diener as a potential Brouwerian coun- terexample to Theorem 1, is given by θ : R → G, where θ(t) = aeiat for
is open in G if and only if a = 0 or a 6= 0. To see this, let δ > 0 be such that B(θ(1), δ) ⊂ θ(0,∞). If |a| > 0 we are done, so we may assume that
|a|< δ/2. Thenθ(−1)∈B(θ(1), δ)⊂θ(0,∞), so there exists t >0 such that
θ(−1) =θ(t); that is, such that ae−ia=aeiat. Either |a(t+ 1)| >0, in which
casea6= 0, or|a(t+ 1)|<2π. In the latter case if a6= 0, then eia(t+1)6= 1—a
contradiction. Hence¬(a6= 0) and thereforea= 0.
However, Hannes showed, with the next result, that this is not the group we are after.
Proposition 53 Let a ∈ R and define θ : R → G by θ(t) = aeiat. Then
G=θ(R), with the metric induced fromC, is complete if and only ifa= 0or
a6= 0.
Proof. Without loss of generality we may assume that|a| <1. Construct an increasing binary sequence (λn)n>1such that
λn = 0 ⇒ |a|< 1
n; λn = 1 ⇒ |a|> 1
n+ 1.
Ifλn = 0, setxn =θ(0) =a; if λn = 1−λn−1, setxk =θ(π/|a|) =−afor all
k>n. Then (xn)n>1 is a Cauchy sequence inG. IfGis complete, then there
existst∞∈Rsuch thatxn →θ(t∞) asn→ ∞. Pick an integer N such that
t∞ < πN. Either λN = 1 and a6= 0, orλN = 0 and |a|<1/N. In the latter
case, if λm = 1 for some m > N, then t∞ = π/|a| > πN—a contradiction.
Henceλn= 0 for alln, soa= 0.
The problem of finding the minimal period of a homomorphism is a little more complex. We have shown that, under the hypothesis of Theorem18, we cannot hope in general to find the minimal period ofθ, but thatτmin exists whenθis
differentiable andG satisfies the twin tangent ball condition. So, can we find more general conditions under which the minimal period exists? Our first task might be to generalise Lemma33, and hence Theorem32, ton-dimensions. With this in mind, we define ann-dimensional closed curveto be a one-one uniformly continuous mappingf :S1→Rn with a uniformly continuous inverse. A curve
G in Rn is said to satisfy the encircled tangent ball condition if there exists
ν >0 such that for eachx∈Gand ally∈(f0(x))⊥
∩∂B(x, ν) we have
B(y, ν)∩G={x}.
We might then hope to prove
Conjecture 54 Let Gbe a differentiable n-dimensional closed curve that sat-
isfies the encircled tangent ball condition. Then there existsβ >0 such that if
r∈ (0, β], w ∈Rn,0 6t
1 < t < t2 62π, and
°
°f¡eitk¢−w°°6r (k= 1,2),
in which case we would have
Conjecture 55 12Letθbe a differentiable homomorphism fromRonto a non-
trivial compact abelian groupGthat is contained inRnand satisfies the encircled tangent ball condition. Ifθis periodic, then θ has a minimal period.
Our Brouwerian example establishing that we cannot always construct the minimal period of a periodic homomorphism involved a subset, Ga, of the 4-
dimensional space S1×C; this can, however, easily be recast as a subset of
the 3-dimensional space S1×R. So, can we perhaps construct the minimal
period ofθwheneverGis contained in a 2-dimensional space? With the abelian groups Ga1,...,an, introduced in (7), in mind we might suggest the next con- jecture. Note that 2m = |P({p
1, . . . , pm})| is the number of distinct positive
divisors ofM =p1· · ·pm.
Conjecture 56 Let θ be a continuous periodic homomorphism of R onto a
compact (metric) abelian group G. If G is contained inRn, then there exists a subset {τ1, . . . , τk} of Rsuch that k62n−2 and θ(τmax) = 0, where τmax =
max{τ1, . . . , τk}.
With Theorem2we are slightly closer to having the complete picture. We have shown thatBD-Nis sufficient to prove Theorem2and, consequently, that this Theorem holds inINTandRUSSin addition toCLASS. The major remaining questions are then: Can we prove Theorem2 withinBISH? and if not, then does Theorem2 imply BD-N? If this latter case holds, then can we establish Theorem2in the special case that our abelian group is locally compact? Theorem 36 shows that in any Brouwerian counterexample to Theorem 2 we cannot haveS1 weakly located at 0. Forgetting for the moment that Theorem
2holds inBISH+BD-N, we might try definingG=θ(R) whereθ(t) =atfor somea∈Rsuch that¬(a= 0). However, not surprisingly, Proposition3shows that the completeness of this set is equivalent to Markov’s principle13. Another
possible approach would be to let (an)n>1 be a binary sequence with at most