Continuous homomorphisms of R onto a compact group

Math. Log. Quart. 56, No. 2, 191 – 197 (2010) / DOI 10.1002/malq.200810048

Continuous homomorphisms of R onto a compact group

Douglas Bridges∗ and Matthew Hendtlass∗∗

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

Received 23 December 2008, revised 8 June 2009, accepted 10 June 2009Published online 18 March 2010

Key words Constructive mathematics, compact group, periodic.MSC (2000) 03F60, 22C05, 26E40

It is shown within Bishop’s constructive mathematics that, under one extra, classically automatic, hypothesis,a continuous homomorphism from R onto a compact metric abelian group is periodic, but that the existence ofthe minimum value of the period is not derivable.

c© 2010 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

In this paper, which is written within the framework of Bishop’s constructive mathematics (BISH),1) weconsider a partial abstraction of the well-known classical result

COP Every compact orbit of a dynamical system is periodic [12].

The abstraction is the following:

Theorem 1 Let θ be a continuous homomorphism of R onto a compact (metric) abelian group G, such that

T0 ≡ {θ(t) : t > 0}.

is open. Then θ is periodic, in the sense that there exists τ > 0 such that θ(τ) = 0.

Here we use standard additive notation for abelian group operations. By a metric abelian group we meanan abelian group G equipped with a metric ρ, such that the mapping (x, y) � y − x is pointwise continuousat (0, 0) ∈ G × G, and uniformly continuous on compact subsets of G × G. The mappings x � −x and(x, y) � x + y are then pointwise continuous throughout their domains, and uniformly continuous on compactsubsets of their domains. Moreover, for each positive integer n, the mapping x � nx is continuous at 0. Notethat if G is locally compact – that is, every bounded subset of G is contained in a compact subset of G – then thepointwise continuity of the mapping (x, y) � y− x is a consequence of its uniform continuity on compact sets.

We say that a homomorphism θ of the abelian group R into a metric abelian group G is continuous if it isuniformly continuous on each compact (or, equivalently, on each bounded) subset of R.

Although we know of no reference for Theorem 1 in the literature, we believe that the following argument,based on the standard classical one used to prove COP, would be the natural one for the classical mathematicianto use. Under the hypotheses of Theorem 1, first observe that for each positive integer n, since θ is continuous,the closed subset

Cn ≡ {θ(t) : t ∈ R, t � n}

of G is compact. Since C1 ⊃ C2 ⊃ · · · , the set⋂

n�1 Cn is nonempty; from which it is not hard to deducethat each Cn is dense in G. On the other hand, by the continuity of θ, the sets θ[−n,n] are compact and henceclosed. Since G is complete and equals the union of the sets θ[−n,n] (n � 1), it follows from the Baire categorytheorem that the interior of θ[−N,N] is nonempty for some N. We can therefore find t1, t2 such that |t1| � N,t2 � N+ 1, and θ(t1) = θ(t2). Setting τ = t2 − t1, with very little more work we see that θ(τ) = 0.

∗ Corresponding author: e-mail: [email protected]∗∗ e-mail: [email protected]) That is, mathematics with intuitionistic logic and an appropriate set-theoretic foundation such as those in [1, 6, 10, 13]. For more on

BISH and other varieties of constructive mathematics, see [4, 5, 8, 9].

c© 2010 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

192 D. Bridges and M. Hendtlass: Homomorphisms onto a compact group

Where, then, are the constructive flaws in this argument? There are several, beginning with the compactnessof Cn: for that, it is not enough, constructively, that Cn be a closed subset of the compact space G; it must alsobe located in G, in the sense that

ρ(x,Cn) ≡ inf{ρ(x, y) : y ∈ Cn}

exists for each x ∈ G (see [9, Chapter 2]). The next constructive flaw in the proof lies in its claim that⋂

n�1 Cn

is nonempty. Actually, we would want the set to be inhabited, in the sense that we can construct points thatbelong to it; this is a stronger property than the mere denial of emptiness. In BISH, the statement

Every descending sequence of compact subsets of a metric space has inhabited intersection

is equivalent2) to the essentially nonconstructive lesser limited principle of omniscience,

LLPO For each binary sequence (an)n�1 with at most one term equal to 1,either an = 0 for all even n or else an = 0 for all odd n.

A further flaw in the classical argument under scrutiny is the claim that θ[−n,n] is compact, a claim based on thepreservation of compactness by uniformly continuous mappings: it is well-known that if, for each real number a,the image of [−1, 1] under the mapping x � ax is compact, then LLPO is derivable in BISH.

We are almost there with our criticism of the classical proof of Theorem 1. The final constructive problemarises in the application of Baire’s category theorem: although the intersection of a sequence of dense open setsin a complete metric space is, as classically, dense, the classical contrapositive version of Baire’s theorem – theversion applied above – does not hold in BISH without some quite strong extra hypotheses; see [8, Chapter 2]and also the paper [11].3)

We conclude that the constructive problems in proving Theorem 1 are nontrivial. As we now show, the trickis to set things up for an application of Baire’s theorem in the “dense open sets” version. Here, then, is the proofof Theorem 1.

P r o o f. Under the hypotheses of that theorem, for each r > 0 let

Tr ≡ {θ(t) : t > r}.

We first prove that ρ(0, Tr) = 0. Given ε > 0, we compute δ > 0 such that if y, y′ ∈ G and ρ(y, y′) < δ, thenρ(0, y−y′) < ε; this is possible in view of [5, page 400, Proposition (1.2)] . Now pick t1, . . . , tm in R such that{θ(t1), . . . , θ(tm)} is a δ-approximation to G, and let

t > max{t1, . . . , tm}+ r.

There exists i � m such that ρ(θ(t), θ(ti)) < δ and therefore ρ(0, θ(t − ti)) < ε; moreover, t − ti > r. Sinceε > 0 is arbitrary, we conclude that ρ(0, Tr) = 0.

Next, consider any t ∈ R and ε > 0. By [5, page 400, Proposition (1.2)] , there exists δ > 0 such that ify, y′ ∈ G and ρ(0, y − y′) < δ, then ρ(y, y′) < ε. The first part of the proof enables us to construct t′ suchthat t′ > r + |t| and ρ(0, θ(t′)) < δ. Then ρ(θ(t), θ(t′ + t)) < ε, where t′ + t > r. Since t ∈ R and ε > 0 arearbitrary, we conclude that Tr is dense in G.

To prove that Tr is open, fix t > r and compute ε > 0 such that B(θ(t− r), ε) ⊂ T0. There exists δ > 0 suchthat if x, y ∈ G and ρ(x, y) < δ, then ρ(x− θ(r), y− θ(r)) < ε. If t′ ∈ R and ρ(θ(t′), θ(t)) < δ, then

ρ(θ(t′ − r), θ(t− r)) < ε

and so θ(t′ − r) ∈ T0; whence there exists t′′ > 0 such that θ(t′ − r) = θ(t′′) and thus θ(t′) = θ(t′′ + r) ∈ Tr.Therefore B(θ(t), δ) ⊂ Tr, and so Tr is open.

We now see that (Tn)n�1 is a sequence of dense, open subsets of G. Applying Baire’s theorem, we seethat

⋂n�1 Tn is dense in G and so, in particular, contains θ(t0) for some t0 ∈ R. Pick a positive integer

N > |t0|. Then θ(t0) ∈ TN, so there exists t′0 such that |t′0| > N and θ(t0) = θ(t′0). Then t0 − t′0 �= 0 andθ(t0 − t′0) = 0.

2) Even with the additional hypothesis that there is at most one point in the intersection, the existence of such a point is equivalent, overBISH, to a version of Brouwer’s fan theorem; see [3].

3) Richman has recently obtained a constructive proof of the “closed sets” form of Baire’s theorem under relatively weak hypotheses [14].

c© 2010 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim www.mlq-journal.org

Math. Log. Quart. 56, No. 2 (2010) / www.mlq-journal.org 193

In connection with our proof, it behooves us to comment on the hypothesis that T0 be open. First, we observethat if θ is periodic, then for each t ∈ R there exists t′ > 0 such that θ(t) = θ(t′), so T0 = G and is open.Now let θ be a continuous homomorphism of R onto any metric abelian group, and assume that T0 is not open.Let θ(t) = 0 and suppose that t �= 0. Then θ is periodic, so T0 is open, a contradiction. Hence ¬(t �= 0) andtherefore t = 0. This shows that θ is one-one. On the other hand, we see from the first part4) of the proof ofTheorem 1 that

{θ(t) : |t| � 1}

is dense and therefore located in G. It follows from [7, Proposition 21] that θ−1 is uniformly continuous on G;whence T0 is open in G. This contradiction shows, constructively, that T0 cannot fail to be open; classically, itfollows that it must be open.

The next major question is: Under the hypotheses of Theorem 1, when θ is nontrivial (that is, there existst > 0 such that θ(t) �= 0), can we find the minimum period τ of θ? In other words, can we find τ0 > 0 such thatθ(τ0) = 0 and θ(t) �= 0 whenever 0 < |t| < τ0? Classically, once we have found τ > 0 as in Theorem 1, weknow that τ0 exists and has the form τ/n for some positive integer n. To make constructive progress, we derivesome lemmas, the first three of which are elementary though both nontrivial and useful.

Lemma 2 If y > x > 0 and ε > 0, then there exists a positive integer N such that either Nx < y < (N+1)xor |y−Nx| < ε.

P r o o f. Pick a positive integer ν such that νx > y− ε. Either (ν− 1)x < y and we take N = ν− 1, or else(ν − 1)x > y − ε. In the later case, either (ν − 2)x < y and we take N = ν − 2, or else (ν − 2)x > y − ε.Carrying on in this way for at most ν− 1 steps, we are guaranteed to find the desired positive integer N.

Lemma 3 If y > x > 0, then there exists a positive integer N such that |Nx− y| < x.

P r o o f. Apply Lemma 2 with ε = x.

Lemma 4 Let 0 < τ < r < t. Then there exists a nonnegative integer N such that 0 < t−Nτ < r.

P r o o f. By Lemma 2, there exists a positive integer N such that

either Nτ < t < (N+ 1)τ or |t−Nτ| < min{τ, r− τ}.

In the former case, 0 < t−Nτ < τ; in the latter, −τ < t−Nτ < r− τ, so 0 < t− (N− 1)τ < r.

Lemma 5 Let θ be a continuous homomorphism of R onto a nontrivial compact abelian group G. Then thereexists t0 > 0 such that θ(t) �= 0 whenever 0 < |t| < t0.

P r o o f. Fix t1 > 0 such that θ(t1) �= 0, and choose t0 ∈ (0, t1) such that ρ(0, θ(t)) >1

2ρ(0, θ(t1)) when-

ever |t− t1| < t0. Let 0 < t < t0. By Lemma 4, there exists a nonnegative integerN such that 0 < t1 −Nt < t0.

Since |Nt− t1| < t0, we have ρ(0,Nθ(t)) = ρ(0, θ(Nt)) >1

2ρ(0, θ(t1)); whence Nθ(t) �= 0 and therefore

θ(t) �= 0. If, on the other hand, −t0 < t < 0, then, by the foregoing, θ(−t) �= 0 and therefore again θ(t) �= 0.

Lemma 6 Let θ be a continuous homomorphism of R onto a nontrivial compact abelian group G, letθ(τ) = 0, where τ > 0, and let t0 be as in Lemma 5. If |t− τ| < t0, then θ(t) �= 0.

P r o o f. By Lemma 5, θ(t0 − τ) �= 0 and therefore θ(t0) �= θ(τ) = 0.

4) That part does not use the hypothesis that T0 is open.

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Lemma 7 In the notation of Lemma 5, the restriction of θ to the interval [0, t0) is one-one.

P r o o f. Let t1, t2 be distinct points of [0, t0). Then 0 < |t1 − t2| < t0, so, by Lemma 5,

θ(t1) − θ(t2) = θ(t1 − t2) �= 0

and therefore θ(t1) �= θ(t2).

Proposition 8 Let θ be a continuous homomorphism of R onto a nontrivial compact abelian group G, andlet τ be as in the conclusion of Theorem 1. Then there exists a positive integer N such that θ(τ/n) �= 0 for alln > N.

P r o o f. We need only compute, firstly, t0 > 0 as in Lemma 5, and then a positive integer N such thatτ/N < t0.

With N as in this last proposition, and donning, for the moment, a classical-logical hat, we now see that theminimum period of θ in Theorem 1 is τ/n for some positive integer n � N. Doffing that hat again, we providea Brouwerian example that knowing that our continuous homomorphism is periodic does not, of itself, enable usto compute the minimum period.

In order to do this, for each a ∈ R we define

(1) Ga ≡ {(e2πit, aeπit) : t ∈ R} ⊂ S1 × C,

where S1 = {z ∈ C : |z| = 1}. Then

(e2πit, aeπit) + (e2πit′ , aeπit′) ≡ (e2πi(t+t′), aeπi(t+t′))

defines an addition operation that turns Ga into an abelian group with identity (1, a), and

(2) θa(t) ≡ (e2πit, aeπit)

is a continuous homomorphism of R onto Ga. Moreover, θa(2) = 0, and θ(0,∞) equals Ga and is thereforeopen.

Lemma 9 The following are equivalent:(i) The group Ga is compact for each a ∈ R.

(ii) LLPO.

P r o o f. Assuming (i), let (an)n�1 be a binary sequence with at most one nonzero term, and define

a =∑

∞

n=0 2−nan.

If ak = 0 for all k � n, set tn = n; if an = 1, set tj = n for all j � n. For all m,n we have

ρ(θa(tm), θa(tn)) = a|eπitm − eπitn | � 2a.

If ε > 0, then either 2a < ε, in which case ρ(θa(tn), θa(tm)) < ε for all m and n, or else a > 0. In thelatter case, pick a positive integer N such that 2−N < a; then a = 2−ν for some ν < N, so tm = tn = ν, andtherefore ρ(θa(tm), θa(tn)) = 0 < ε, for all m,n � ν. Since ε > 0 is arbitrary, it follows that (θa(tn))n�1 isa Cauchy sequence in Ga. Using (i), we can find t ∈ R such that θa(tn) → θa(t) as n → ∞. Choose a positiveinteger κ such that κ � t < κ+ 2. If ρ(t, {κ, κ+ 1}) > 0, then

1 �= e2πit = limn→∞ e2πitn = 1,

which is absurd; we conclude that either t = κ or t = κ+ 1. Now suppose that ak = 1 for some k �≡ t (mod 2).Then a �= 0, tn = k for all n � k, and eπi(k−t) = −1; so

limn→∞ θ(tn) = θ(k) = (e2πik, aeπik) �= (e2πit, aeπit) = θ(t),

another contradiction. Hence ak = 0 for all k not congruent to κ. Thus (i) implies LLPO.



Conversely, assuming (ii), fix a in R and let (tn)n�1 be a sequence in R such that (θa(tn))n�1 is a Cauchysequence in Ga and hence in S1 × C; then (e2πitn)n�1 is a Cauchy sequence in S1. Since S1 and S1 × C arecomplete spaces, there exist t ∈ R and z ≡ (z1, z2) ∈ S1 × C such that

max{|e2πitn − e2πit|, ρ(θa(tn), z)} → 0 as n → ∞.

By LLPO, either ρ(z, θa(t)) � ρ(z, θa(t+ 1)), in which case we take s ≡ t, or else

ρ(z, θa(t+ 1)) � ρ(z, θa(t)) = ρ(z, θa(t+ 2)),

when we take s ≡ t+ 1. Thus in either case,

(3) ρ(z, θa(s)) � ρ(z, θa(s+ 1)).

Note that

(∗) if s = t, then eπi(s+1) = −eπit = −eπis;and if s = t+ 1, then eπis = −eπit and eπi(s+1) = eπi(t+2) = eπit.

We show that z = θa(s) ∈ Ga. To that end, suppose that z �= θa(s). Since

e2πis = e2πit = limn→∞ e2πitn = z1,

we see that

aeπis �= z2 = limn→∞ aeπitn

and therefore that a �= 0. Compute a positive integer N such that for all n � N,

ρ(θa(tn), z) < δ ≡ 1

2min{|a|, |a|2, ρ(θa(s), z)}, and |e2πitn − e2πit| <

(δ

a

)2

.

Noting (*), we see that for such n,

|eπitn − eπis||eπitn − eπi(s+1)| = |eπitn − eπit||eπitn + eπit|

= |e2πitn − e2πit| <

(δ

a

)2

.(4)

If |eπitn − eπis| < δ/|a|, then (note that (δ/|a|)2 < δ)

ρ(θa(s), z) � ρ(θa(tn), θa(s)) + ρ(θa(tn), z)

< max{

(δ

a

)2

, |a||eπitn − eπis|}+ δ

< 2δ � ρ(θa(s), z),

which is absurd. Hence, by (4), |aeπitn − aeπi(s+1)| < δ. Since e2πi(s+1) = e2πis = e2πit, it follows that

ρ(θa(tn), θa(s+ 1)) < δ � 1

2ρ(θa(s), z)

for all n � N. Letting n → ∞, we obtain ρ(z, θa(s + 1)) � 1

2ρ(θa(s), z). It follows from this and (3) that

ρ(θa(s), z) = 0 and therefore z = θa(s), a contradiction. We now conclude that z does, after all, equal θa(s),and therefore that Ga is complete. Thus LLPO implies (i).



We now prove that if, in Theorem 1, we can always find the smallest positive τ such that θa(t + τ) = θa(t)for all t, then LLPO implies the weak limited principle of omniscience,

WLPOFor each binary sequence (an)n�1, either an = 0 for all nor it is impossible that an = 0 for all n.

Assuming LLPO, we see from Lemma 9 that for each a ∈ R, the group Ga is complete; since Ga is totallybounded, it is therefore compact. Suppose that

τmin = inf{t > 0 : θa(t) = (1, a)}

exists. Either τmin < 2 or τmin > 1. In the first case, if a �= 0, then τmin = 2, a contradiction; whence a = 0. Inthe case τmin > 1, we cannot have (1,−a) = θa(1) = θa(0) = (1, a), so ¬(a = 0). Since a ∈ R is arbitrary,we have proved that

(∀x ∈ R)(x = 0∨ ¬(x = 0)),

which is constructively equivalent to WLPO.Since LLPO is provably weaker than WLPO (see [2]), we conclude from the foregoing analysis that under

the hypotheses of Theorem 1, we cannot derive the conclusion that there exists a minimum positive value of τsuch that θ(τ) = 0.

Finally, we consider what may initially appear to be a ridiculously trivial question: if a continuous homomor-phism of R onto a metric abelian group is periodic, is the group compact?

Proposition 10 Let θ be a continuous homomorphism of R onto a metric abelian group G, and τ a positivenumber such that θ(τ) = 0. Then θ[0, τ] is dense in G, θ[0, 2τ] = G, and G is totally bounded.

P r o o f. Given ε > 0, compute δ ∈ (0, τ) such that if t, t′ ∈ [−τ, τ] and |t− t′| < δ, then ρ(θ(t), θ(t′)) < ε.Fix t ∈ R. Either |t| < δ, in which case ρ(0, θ(t)) < ε, or else t �= 0. In the latter case, taking t > 0 forillustration, we compute an integer n � 0 such that nτ � t < (n + 2)τ. Then either |t − (n + 1)τ| < δ < τ ort �= (n+ 1)τ. In the first case,

ρ(0, θ(t)) = ρ(0, θ(t) − (n+ 1)θ(τ)) = ρ(0, θ(t− (n+ 1)τ)) < ε.

In the second case, either t ∈ [nτ, (n+ 1)τ) or t ∈ ((n+ 1)τ, (n+ 2)τ], so there exists ν ∈ {n,n+ 1} such thatt− ντ ∈ [0, τ]; since

ρ(0, θ(t)) = ρ(0, θ(t− ντ) + νθ(τ)) = ρ(0, θ(t− ντ)),

we have θ(t) ∈ θ[0, τ]. Since ε and t are arbitrary, θ[0, τ] is dense in G. The uniform continuity of θ on [0, τ]ensures that θ[0, τ], and hence G, is totally bounded. Finally, given t ∈ R, and computing an integer N such thatt−Nτ ∈ [0, 2τ], we have θ(t) = θ(t−Nτ). Hence G = θ[0, 2τ].

Classically, under the hypotheses of Proposition 10 we would conclude that G is compact. Constructively, wecannot do so: for, referring to the definitions (1) and (2), we see from Lemma 9 that if, for each a ∈ R, the groupGa is compact and hence complete, then LLPO is derivable in BISH.

A concluding comment: our investigation of continuous group homomorphisms arose from Arno Berger ask-ing whether the Baire category theorem was essential for the proof of COP, whose (partial) abstraction to thecontext of abelian groups we have analysed constructively in this paper and [7]. We still do not know the answerto Berger’s question, but it is interesting that the proof of our direct constructive analogue of the (abstracted)periodicity theorem uses the “dense open sets” version of Baire’s theorem, whereas our proof of a contrapositiveversion, in [7], uses a constructively inequivalent “union of closed sets” version.

Acknowledgements The authors thank Hannes Diener for stimulating conversations about the problems addressed in thispaper, and Fred Richman for pointing out an error in our original proof of Theorem 1. They also thank the University ofCanterbury and the New Zealand Institute of Mathematics and its Applications, each for supporting Hendtlass by a Master’sScholarship; and the Royal Society of New Zealand, for supporting Bridges’s research by Marsden Award UOC0502.



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Continuous homomorphisms of R onto a compact group