Subsum Sets: Intervals, Cantor Sets, and Cantorvals

SUBSUM SETS: INTER V ALS, CANTOR SETS, AND CANTOR V ALS ZBIGNIEW NITECKI Abstract. Giv en a sequence { x i } ∞ i =1 con verging to z ero, we consider the set Σ  { x i } ∞ i =1  of n umbers which are sums of (infinite, fini te, or empty) subse- quences of { x i } ∞ i =1 . When the original sequence is not absolutely summable, Σ  { x i } ∞ i =1  is an un bounded closed inte rv al which includes zero. When it is absolutely summable Σ  { x i } ∞ i =1  is one of the following: a finite union of (non trivial) compac t in terv als, a Can tor s et, or a “symmetric Can torv al”. Recently , while trying to think up some challenging pr oblems for m y under grad- uate Real Analysis s tuden ts, I stumbled onto an elementary and, I think, natural question on which I was unaw are of any literature. One of the most counterin tuitive facts in the ele ment ar y theor y o f series is that, even if a sequence of r eal num b ers { x i } c onv erges to zero (that is, it is a null sequence ), the co rresp onding series P ∞ i =1 x i might diverge. The example of this which most of us encounter first is the harmonic se quenc e 1 k , which co nv erges to zero, but whose s um, sur prisingly , diverges: P ∞ k =1 1 k = ∞ . How ever, if we throw aw ay enough o f these terms—for example, if we thr ow aw ay all r ecipro cals o f num- ber s which are not p ow ers o f t wo—w e end up with a se quence whose cor resp onding series do es co nv erge. W e will call such a sequence a summable su bse quenc e of o ur original sequence, and its sum a subsum , of our (orig inal) sequence. Then we might ask ab out the set o f all p ossible subs ums of a g iven sequence (as suming al- wa ys that the origina l se quence go es to zer o): is it an interv al, a finite union of int erv als... or something more complica ted? This turns out to b e a challenging question: I set out trying to answ er it and came to a n um b er of in teres ting conclusions, but w as unable to giv e a satisfactory general description of such sets on my own. H ow ever, a c omment by Micha l Misiur ewicz led me by chance to a 198 8 pap er by J. A. Guthrie and J. E. Nymann [ 8 ], which gives a complete top olog ical de scription of subsum sets as well as a review of some earlier work on the problem. 1 After writing up what I had found, I came across the pap er o f Rafe Jones [ 10 ], which cov ers some o f the s ame material, but see ms Date : No vem ber 7, 2021. 2010 Mathematics Subje ct Classific ation. 40A05, 11B05. Key wor ds and phr ases. subsum set, absolutely summable, conditionally summ able, Cantor set, iterated function system, Can torv al. I would like to thank Aaron Brown, Keith Burns, Bil l Dunham, Richard K en yon, Micha l Misiurewicz, Don Plant e, Jim Propp, Charles Pugh, and Mariusz Urbanski for useful co nv ersations in the course of preparing this paper. 1 I wa s gratified to d iscov er that the terminology I had ad opted in my musings o n t he sub ject is almost iden tical to that used in most earl ier writing on the sub j ect. The one substant ial exception is the w ord “Can torv al”, coined by tw o Brazilians in [ 14 ], whic h evok es for me the Sam ba on F at T uesda y in Rio.. . 1 2 ZBIGNIEW NITECKI unaw are of the definitiv e result of [ 8 ]. How ever, it go es b eyond the a ssumption that the sequence c onv erges to zero. A t the end o f the present pa per , I will sketc h some of the extensions suggested in Jones’ pa p er, as well a s the further e xtensions in the work o f Mor´ an [ 16 , 17 ] re ferenced there. Our story inv olves an int ere sting in terplay betw een standard topics on sequences and series, s ome elementary n umber theor y , and the to po logy o f subsets of the line , which pr ovides a n app ea ling “extra topic” for underg raduate analys is studen ts. Most of our discuss ion will focus on p ositive null sequence s, which can b e studied using geometric ideas. Howev er, w e shall see tow ard the end of this pap er ( § 6 ) how the description of al l subsum sets ca n be r educed to the corresp onding description of subsum s ets for p ositive null sequences. 1. Positive, Non-Summable Sequences W e can think of the har monic sequence as an infinite collection o f domino es of successively s horter lengths: the k th domino has length 1 k . The fact that the series diverges means that if w e put them all e nd-to-end, w e will fill out a whole half-line. Now supp os e we are given a p ositive real num b er r . Can we find a co llection o f domino es from this set which ex actly fill up an interv al o f length r ? W ell, we know the lengths of the do mino es conv erge monotonica lly to ze ro, so except for the first few, they are all shorter than a ny s pec ified fraction o f r . This means that w e can, b y star ting far enough do wn the line, fit a string of any sp e cifie d finite numb er of suc c essive domino es inside the int erv al. If w e sta rt with the n th domino a nd fit in as many successive domino es as we can (starting from the n th ), then the fir st do mino that “p o kes out” will certainly be shor ter than 1 n . In fact, if we hav e manag ed to squeeze in k do mino es (starting from the n th ) but cannot fit the next one in, then the one that po kes out has leng th 1 n + k . This means that the ones we c an fit fill a n in terv al that is shorter than r —but its leng th plus 1 n + k is mor e than r . It follows tha t after we have squeezed in k successive domino es starting from the n th , we are left with an unfilled gap which is sho rter than 1 n + k . Now, we lo ok for more domino es, to fill this gap. W e sta rt further down our list of do mino es, finding a set o f k ′ successive ones , sta rting with the ( n ′ ) th (where n ′ ≫ n + k ), that fill o ut our g ap—except for a new, sma ller gap of length less than 1 n ′ + k ′ . And we c ontin ue. With a little more car e, w e ca n choose our star ting po int at each stag e so that the size of the gap is cut to less than half its cur rent v alue with eac h new filling. When we a re all do ne, we hav e created a subsequence of our domino es whose combined tota l length is exactly r . Let’s lo ok back at what we did. W e didn’t rea lly use any sp ecia l prop erties of the harmonic sequence in this construction, other than the fact that the lengths of the dominoes go to zero, but th eir sum diverges (to infinity). So w e ha ve a theorem: Theorem 1. If { x i } is a p ositive nul l se quenc e for which P ∞ i =1 x i = ∞ , then every r > 0 is the sum of some su bse qu enc e of { x i } . Actually , ther e is one minor technical point we need to note here. When thinking ab out the harmo nic sequence, w e did ta ke a dv antage of the fact that it is decreasing. In ge neral, the seq uence we a re lo o king at migh t be presented in an order which is not decrea sing. F o rtunately , for a sequence o f p ositive num b ers, the sum (of the series) is not changed by rearr anging their order. (This was noted b y Dirichlet in 1837 [ 5 , p. 315] without explicit pro o f; a pro o f can b e found in many basic SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 3 analysis b o oks, fo r ex ample [ 21 , Thm. 3.56, p. 6 8]. 2 ). Intuitiv ely , the total le ngth of a co llection of domino es set end-to-end is not c hanged if we set them do wn in a different order . This mea ns we can work with them in a non-incr e asing order: x k +1 ≤ x k for every k . This will b e an implicit as sumption in a ll of o ur reasoning , at least while lo oking at p ositive sequences: Standing Assumption: When de aling with p ositive se quen c es, we assume (without loss of gener ality) that the given se qu enc e is non-incr e asing: x i +1 ≤ x i for all i = 1 , 2 , . . . . 2. Positive S ummable Sequences OK, so we have a nswered our question for a sequence of p ositive num b ers g oing to zero whose sum diver ges . What ab o ut if the sum c onver ges ? W e start with t wo examples. First, co nsider the seq uence of (p os itive integer) p ow ers of 1 2 x i = 1 2 i , i = 1 , 2 , . . . which s ums to ∞ X i =1 x i = 1 / 2 1 − 1 / 2 = 1 . W e can a gain picture our sequence as a colle ction of domino es (the i th has le ngth  1 2  i ); clearly , since al l of them place d end-to-end fill a n interv al of length 1, any sub collection will fill a sho rter in terv al; that is, any subsum b elongs to the in terv al [0 , 1 ]. Now, expressing a num be r in [0 , 1] as a sum of (distinct) p owers of 1 2 is the same as giving its binar y or b ase 2 expansion: to be more precise, a binary sequence ξ = { ξ i } ∞ i =1 (each ξ i is 0 or 1) co rresp onds to the num b er x ( ξ ) = ∞ X i =1 ξ i 2 i . Every num be r b etw ee n 0 a nd 1 has a binary expansio n, so the subsum set in this case 3 is an in terv al with e ndpo int s 0 and 1. Now, co nsider the sequence of powers o f 1 3 x i = 1 3 i , i = 1 , 2 , . . . which s ums to ∞ X i =1 x i = 1 / 3 1 − 1 / 3 = 1 2 . 2 The basic i dea is that the partial sums f or an y ordering are themselv es a strictly increasing sequence, and any particular partial sum f or one ordering can b e brac ket ed b et ween tw o partial sums of an y other particular order, so the t wo limits are the same. 3 W e shall see later that this needs some clarification: see Definition 2 . 4 ZBIGNIEW NITECKI As b efore, an y subsum b elongs to the int erv al  0 , 1 2  . But on clo ser insp ection, it bec omes clea r that not every point in this interv al o cc urs as a subsum. F or example, any subs um which do es not inv olve the first term, 1 3 , is a t most eq ual to X 1 := X i> 1 1 3 i = 1 / 9 1 − 1 / 3 = 1 6 and hence belo ngs to the int erv al J 0 :=  0 , 1 6  whereas any subsum which do es inv olve the first term b elongs to J 1 :=  1 3 , 1 2  . Note tha t J 1 is the translate o f J 0 by x 1 = 1 3 , and the set of subsums is actually contained in the union o f tw o disjoint closed interv als C 1 := J 0 ∪ J 1 . That is, dis tinguishing subsums accor ding to whether they do or don’t inv olve the first term of the sequence brea ks the set of all subsums into tw o pieces, the s econd a tra nslate of the fir st. When we take account o f all the p ossibilities for which of the first two terms of the sequence o ccur in a given subsum, we find that the se t of subsums is con tained in the union of four disjoint c losed in terv a ls–tw o s ubin terv als of J 0 and tw o subinterv als of J 1 . Of cours e, we can contin ue this pro cess . A subsequence of { x i } ca n b e sp ecified using the sequence ξ = ξ 1 ξ 2 · · · of zero es and ones defined b y (1) ξ k = ( 1 if x k is included in the subse quence , 0 if it is not . The sum co rresp onding to this subsequence is then (2) x ξ := ∞ X k =1 ξ k · x k . F o r our particula r example, this reads x ξ = ∞ X k =1 ξ k 3 k which is a ba se three expa nsion for x ξ . The interv als J 00 , J 01 , J 10 and J 11 result fro m so rting the subsum set acco rding to which of the fir st tw o terms o f the sequence { x i } = { x i } ∞ i =1 are included in a given subsum–that is, acco rding to the initial “word” o f leng th 2 in the defining sequence ξ . In general, we can pars e any subsum into an initial finite sum, x ξ 1 ··· ξ n determined by the initial “word” ξ 1 · · · ξ n of length n , and the rest of the s um, which is a subsum of the sequence { x i } ∞ i = n +1 obtained by omitting the first n terms of { x i } ∞ i =1 . Let us infor mally 4 denote the subsum set of a sequenc e { x i } ∞ i =1 by 4 A formal definition will be given shortly in Definition 2 . SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 5 Σ( { x i } ∞ i =1 ), and wr ite Σ ( n ) := Σ  { x i } ∞ i = n +1  for the s et of subsums whic h do not involv e the first n ter ms x 1 , . . . , x n . Then the collection of all subs ums w hose defining seq uence ξ ha s initial word ξ 1 · · · ξ n can b e written x ξ 1 ··· ξ n + Σ ( n ) ; letting the initial word of length n ra nge ov er all the p ossible n -tuples of zero es and ones, we fill up o ur subsum set: (3) Σ( { x i } ∞ i =1 ) = [ ξ 1 ··· ξ n ∈{ 0 , 1 } n  x ξ 1 ··· ξ n + Σ ( n )  . As b efore, Σ ( n ) is contained in the clo sed interv al 5 J 0 n = [0 , X n ] where X n is the highest sum in Σ ( n ) X n = X k>n x k and it follo ws that (for each fixed n ) our whole subsum set is contained in the union of 2 n closed interv als C n = [ ξ 1 ··· ξ n ∈{ 0 , 1 } n J ξ 1 ··· ξ n where J ξ 1 ··· ξ n := x ξ 1 ··· ξ n + J 0 n = [ x ξ 1 ··· ξ n + 0 , x ξ 1 ··· ξ n + X n ] . In our example, X n = ∞ X k = n +1 1 3 k = 1 / 3 n +1 1 − 1 3 = 1 2 · 3 n so J 0 n =  0 , 1 2 · 3 n  . Having fixed an initial word of length n , we hav e t wo p os sibilities for the next, ( n + 1) st ent ry in ξ : either ξ n +1 = 0 or ξ n +1 = 1. This mea ns tha t each interv al J ξ 1 ··· ξ n of C n contains t wo subinterv als as so ciated to initia l words of length n + 1 in ξ : J ξ 1 ··· ξ n 0 = x ξ 1 ··· ξ n + [0 , X n +1 ] and J ξ 1 ··· ξ n 1 = x ξ 1 ··· ξ n + 1 3 n +1 + [0 , X n +1 ] where X n +1 = 1 2 · 3 n +1 . 5 0 n denotes the w ord of length n consisting of all zero es. 6 ZBIGNIEW NITECKI The impor tant thing to notice is that these t wo subinterv als hav e the sa me leng th, X n +1 , and the sec ond is a tra nslate o f the firs t b y an amount grea ter than X n +1 . This means they ar e disjoin t. Lo oking a bit more closely , we note that the fir st subint erv al starts at the left endp oint o f J ξ 1 ··· ξ n while the s econd en ds a t its right endpo int. Thus, pa ssing from the unio n C n of interv als deter mined by words of length n to the union C n +1 of those determined by words o f length n + 1 , each comp onent interv al o f C n acquires a gap in its middle, separating tw o subinterv als which are co mpo nents of C n +1 . In fact, since X n +1 = 1 3 X n , this gap is precisely the “ middle thir d” o f each component. Hence w e are ca rrying o ut the cons truction of the middle- third Cantor set, ex cept that we start from the in terv a l  0 , 1 2  instead of [0 , 1 ]. In this wa y , when we pas s to the intersection C ∞ = ∞ \ n =1 C n we obtain a version o f the s tandard Cantor set, but scale d down by a fa ctor of a half. The argument ab ov e shows that the subsum set of the sequence o f p owers of 1 3 is a Cantor s et. Ho wev er, the construction of the sets C n and C ∞ applies to any po sitive summable null sequence, with the pr oviso that in genera l, the in terv als J ξ 1 ··· ξ n need not b e disjoint—so o ur final se t C ∞ need not b e a Cantor set. In fact, for the p ow ers of 1 2 , w e hav e X n = 1 2 n , and the int erv als J ξ 1 ··· ξ n abut, so C n = [0 , 1] for all n (and hence for “ n = ∞ ”). As we shall see, ev en more complicated b ehavior is p ossible which mixes overlap a nd disjointness. In genera l, though, the pro cedure we ha ve outlined pro duces the compact set C ∞ , which is g uaranteed to contain our subsum s et. B ut c ertainly at each finite s tage, the set C n contains more than Σ( { x i } ∞ i =1 ). So, what abo ut the intersection?—do es Σ( { x i } ∞ i =1 ) e qual C ∞ , or is it a prop er subse t? The a nswer to this hinges o n what we mean by a “subsequence”. Usually a “subsequence” of a n infinite seque nce is understo o d to itself b e infinite. If we use this no tion in our definition of subsums, w e exclude any num b er given as a finite sum of p ow ers of 1 3 —that is, we exclude the left endp oint of each of our interv als J ξ 1 ··· ξ k . The resulting set is a bit awkward to describ e. So we follow a conv ention going back to S. Kakey a (whose 191 4 pap er [ 11 ] is the first one I am aware of on this topic) and include finite subsequences, as well a s the empty seque nce (whose sum we take to b e zer o), in our formal definition of the subsum set. Definition 2. The subsum set of a nul l se quenc e x 1 , x 2 , · · · → 0 is the c ol le ction Σ( { x i } ∞ i =1 ) of al l numb ers of t he form x ξ := ∞ X k =1 ξ k · x k , wher e ξ = ξ 1 ξ 2 · · · SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 7 is any se quenc e of zer o es and ones for which the s u bse qu enc e { ξ i · x i } ∞ i =1 is sum- mable. 6 This definition simply co difies the idea tha t we take s ums of infinite, finite, and empty subseq uences of { x i } ∞ i =1 . Note that a finite subsum corre sp onds to a sequence ξ which is even tua lly all zero es; we s hall often o mit the “tail o f zero es ” when sp ecifying the sequence ξ in such a situation. Now, w e hav e constructed a nested sequence of compact sets C n , each containing our subsum set; it follows th at Σ( { x i } ∞ i =1 ) is con tained in the compact set C ∞ . F ur- thermore, C n consists of interv als of length X n , each having nonempty intersection with Σ( { x i } ∞ i =1 ) (for ex ample its endp oints), whic h means that all p oints o f C n are within distance X n of the set Σ( { x i } ∞ i =1 ). Since we have a ssumed our sequence is summable, its “ tails” X n m ust conv erge to zero, which implies tha t C ∞ is the closur e of Σ( { x i } ∞ i =1 ). The cons truction of C ∞ automatically implies several proper ties: • Since X n > 0, ev ery comp onent of C n is an interv al, and hence it has no iso- lated p oints—it is p erfe ct . This pro per ty p ersis ts under nested in terse ction, so C ∞ is a per fect set. • C n is a union o f clo sed in terv als J ξ 1 ··· ξ n of leng th X n ; in particular , ea ch po int of C n is within distance X n of at leas t one right endp oint and at least one left endpo int of some J ξ 1 ··· ξ n . Since X n → 0, this means the right ( r esp . left) endp oints of the v ar ious int erv als J ξ 1 ··· ξ n are dense in C ∞ . In the case of p ow ers of 1 3 , w e hav e a n explicit homeomor phism between the subsum set Σ  1 3 k  ∞ k =1  and the middle-third Can tor set, telling us that this subsum set is compac t, a nd hence e quals C ∞ . T o go b eyond this exa mple, w e need to show that Σ( { x i } ∞ i =1 ) is clo sed in g eneral. This was done in [ 11 ] by a dir ect arg ument, but we can finesse the general ca se using the example and a sneaky tr ick. F o r our example (p ow ers of 1 3 ), the sequence ξ for a particular subsum is a base 3 expa nsion of that subsum, s o p oints of the Cant or s et are in one-to-o ne corres po ndence with sequences ξ of zero es and ones. F urthermor e, this mapping is a homeomorphism (p oints with expansio ns that agr ee for a long time are clo se to ea ch other, and vice-versa). But half of this also applies to a general subs um set: for any sequence { x i } , two subsums whose defining sequences ξ agree for at least n places b elong to the s ame interv al J ξ 1 ··· ξ n , which is an interv al of leng th X n . And that length, which is by definition a “ tail” of a conv ergent series, g o es to zer o. Thu s, the ma pping taking a p oint of the (middle-third) Cantor set to its defining sequence ξ and then to the p oint x ξ in our s ubsum set corresp o nding to the same sequence is co nt inuous. Since it is also onto, we hav e exhibited a gener al subsum set Σ( { x i } ∞ i =1 ) as a contin uo us imag e of a compact set—hence it is also compact. F r om this we can conclude tha t C ∞ = Σ( { x i } ∞ i =1 ) . Hence Σ( { x i } ∞ i =1 ) has the prop erties no ted ab ov e for C ∞ : it is a perfect set, and (since the left ( r esp . righ t) endp oint of any J ξ 1 ··· ξ n is the sum of a finite ( r esp . infinite) subsequence), b oth kinds of sums a re dense in Σ( { x i } ∞ i =1 ). 6 In the cont ext of this section, where we hav e assumed the original sequence is p ositive and summable, every s ubsequence is summable. 8 ZBIGNIEW NITECKI Σ( { x i } ∞ i =1 ) also has so me sy mmetry prop erties. W e hav e already seen (fixing n ) that C n is a union of sets J ξ 1 ··· ξ n which are just tr anslates of e ach other; this mea ns that for ea ch fixed n the sets Σ( { x i } ∞ i =1 ) ∩ J ξ 1 ··· ξ n are homeomor phic. Another symmetry is the reflection a bo ut the midpo int , given b y (4) x 7→ X 0 − x. T o see this pa rticular symmetr y , note tha t when x is a subs um o f our sequence defined b y the sequence ξ of 0’s and 1’s, then X 0 − x is defined b y the sequence ˜ ξ i , where ˜ ξ i = 1 − ξ i —that is, X 0 − x is the sum of all the terms not included in the sum defining x . W e summarize 7 these ge neral observ ations in the following theo rem: Theorem 3. F or every summable, p ositive nu l l se quenc e x 1 , x 2 , · · · → 0 with sum ∞ X k =0 x i = X 0 , the subsum set Σ( { x i } ∞ i =1 ) is a p erfe ct set with c onvex hul l [0 , X 0 ] which is symmet- ric under the r efle ct ion x 7→ X 0 − x. F urthermor e, the c ol le ction of al l sums of finite subse quenc es (as wel l as the c ol le ction of al l sums of infinite subse quenc es) is dense in Σ( { x i } ∞ i =1 ) . The fac t that Σ( { x i } ∞ i =1 ) is p erfect was proved by Shoichi Kakeya in 1914 [ 11 ] and independently by Hans Hornich in 19 41 [ 9 ] 8 The reflection symmetry of subsum sets w as noted by Hornich, as well as b y Joseph Nymann and Ricardo Saenz in [ 18 ]. 3. Terms vs. T ails: Subsum sets of geometric and p -series In the exa mples studied so far, we ha ve o bserved tw o extre mes of be havior. F or the pow ers of 1 3 , the interv als J ξ 1 ··· ξ n for an y fixed n a re disjoint, and in the limit we obtain a Cantor set as C ∞ . But for p owers of 1 2 , thes e interv als touch, as a r esult of which all the sets C n are the same, a nd C ∞ is an interv al. T o understand the basis of these phenomena in g eneral, we exa mine the recursive step in the construction of C ∞ . When w e g o from C n − 1 to C n , each interv al J ξ (for a fixed ( n − 1)-word ξ ) is replaced by the union of tw o subinterv als, cor resp onding to the n -words ξ − = ξ 0 and ξ + = ξ 1 obtained by app ending 0 ( r esp . 1) to ξ . Bo th of these subinterv als hav e length equal to the n th tail X n , and the seco nd is the translate o f the first by the n th term x n . F urthermore, the r ight ( r esp . left) endpoint of J ξ is the same as the right ( r esp . left) endp oint o f J ξ − ( r esp . J ξ + ). Thus we ca n distinguish tw o cases: T erm exceeds T ail : If x n > X n , the tw o in terv als are disjoint, so J ξ in C n is replac ed by a disjoint union of subinterv als in C n +1 ; that is, J ξ breaks int o the disjoint union o f J ξ − and J ξ + , leaving a “gap” of size x n − X n in the middle. T ail b ounds T erm: If x n ≤ X n , the tw o in terv als shar e at least one p o int , so their union equals J ξ . 7 No pun int ended. 8 A 1948 pap er by P . Kesav a Menon [ 15 ] addresses simi lar issues, but I find it confusing to determine j ust what i s b eing prov ed. SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 9 Note that with n fixed, the interv als J ξ ob ey the same r ule for every n -word. Also remember that they a ll hav e length X n , which go es to zero (since X n is the tail of a co nv ergent s eries). Note a lso that in the first case, C n +1 is obtained fr om C n by deleting an interv al of length x n +1 − X n +1 from each of its 2 n comp onents. Since these a ll hav e length X n , the total length of C n +1 is 2 n { X n − ( x n +1 − X n +1 ) } = 2 n +1 X n +1 . This shows Theorem 4 . Supp ose { x i } is a summable se quenc e of p ositive r e al nu m b ers. Then (1) If x n > X n (i.e., the t erm ex c e e ds the t ail) for every n , then for e ach n , C n is the disjoint union of the 2 n close d intervals J ξ as ξ r anges over the wor ds of length n . It fol lows that C ∞ = Σ( { x i } ∞ i =1 ) is a Cantor set whose L eb esgue me asur e is lim n →∞ 2 k X k . (2) If x n ≤ X n (i.e., the tail b ounds the term) for every n , then for e ach n , C n = C 0 = [0 , X 0 ] , so C ∞ = Σ( { x i } ∞ i =1 ) is the interval [0 , X 0 ] . These prop er ties were es tablished by Hornich [ 9 ]. Kakeya [ 11 ] noted the sec ond prop erty (in fact that the tail alw ays b ounds the term if a nd only if the s ubsum set is an interv al—cf. our Lemma 8 and Prop ositio n 9 ). F o r a ge ometric sequence with first ter m a a nd ra tio 9 ρ ∈ (0 , 1), we know that x n = aρ n − 1 and X n = aρ n 1 − ρ so X n x n = ρ 1 − ρ which is at least 1 for ρ ≥ 1 2 and strictly less than 1 for 0 < ρ < 1 2 . This imm ediately gives us a descr iption of Σ( { x i } ∞ i =1 ) for a ny p o sitive g eometric seq uence. 10 Corollary 5. If  x i = aρ i − 1  is a ge ometric se quenc e with initial term a > 0 and r atio ρ ∈ (0 , 1) , t hen Σ( { x i } ∞ i =1 ) is (1) a Cantor set of me asur e zer o for 0 < ρ < 1 2 (2) t he interval h 0 , X 0 = a 1 − ρ i for 1 2 ≤ ρ < 1 . Theorem 4 tells us what happ ens when only one of the tw o p ossible rela tions betw een the terms and the tails oc curs. What about if both o ccur, but o ne of them o ccurs eventual ly ? 11 As an example, co nsider the sequence star ting with 2 and then follow ed b y the powers of 1 2 . W e already know that the seq uence s tarting from the seco nd term 9 (that is, a geometric sequence whose terms are p ositive and tend to zero) 10 Jones [ 10 , Prop. 3.3] gives a kind of extension of the first case of this corollar y , in the spi rit of the ratio test for conv ergence. 11 A prop erty is said to hold ev entual ly for a sequence if there i s some pl ace K in the sequence so that the pr operty holds for al l later terms—or equiv alently , if the prop erty fail s to hold for at most a finite num b er of terms. 10 ZBIGNIEW NITECKI ( i.e. , , just the p owers of 1 2 ) has subsum set [0 , 1], and it follows from Equatio n ( 3 ) that the full subsum set is C ∞ = [0 , 1] ∪ (2 + [0 , 1]) = [0 , 1] ∪ [2 , 3] . These tw o interv als a re dis joint b ecause the first term, x 1 = 2 , is gr eater than the first tail, X 1 = 1. In gener al, if the tail b ounds the term a fter the N th place x k ≤ X k for k > N then Theor em 2 a pplied to the sequence sta rting after p o sition N tells us that Σ ( N ) = [0 , X N ] and then Equation ( 3 ) tells us tha t Σ( { x i } ∞ i =1 ) is the union of 2 N closed interv als, which means, a llowing for some overlaps b et ween them, that it is the disjoint unio n of at most 2 N int erv als. F urther more, if the term exceeds the tail for al l of the first K places x k > X k for k = 1 , 2 , ..., K then the interv als J ξ 1 ··· ξ n are all dis joint, so C K consists of 2 K disjoint in terv a ls. So in this cas e C ∞ = C N has at le ast 2 K comp onents. Summarizing, we ha ve Prop ositi on 6. Supp ose { x i } is a p ositive, su mmable nul l se quen c e. (1) If the tail b ounds t he term eventual ly, then C ∞ is a finite u nion of close d intervals. (2) In p articular, if t he tail b oun ds the term for al l k > N t hen C ∞ = C N c onsists of at most 2 N disjoint close d intervals. (3) If in addition the term exc e e ds the t ail for k = 1 , . . . , K , t hen C ∞ c onsists of at le ast 2 K disjoint close d intervals. As an e xample, consider the p -se quenc e x k = 1 k p where p > 1 is a fixed expo nent. The pr ecise v alue of the n th tail X n is hard to determine, but we can take adv antage of the standar d pro of of summability (that is, the integral test) to estimate it a nd so tr y to chec k which terms exceed the asso ciated tails a nd which tails b ound the terms . F r om Figure 1 we obtain the estimates Z ∞ n +1 dx x p < ∞ X k = n +1 1 k p < Z ∞ n dx x p . Carrying out the integration on either s ide, we hav e (5) ( n + 1) 1 − p p − 1 < X n < n 1 − p p − 1 . Thu s we can guar antee tha t the n th tail exceeds the n th term x n < X n whenever 1 n p ≤ ( n + 1) 1 − p p − 1 . SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 11 1 k p 1 ( k +1) p 1 k p 1 ( k +1) p y = 1 x p Figure 1. Integral T es t for p -ser ies a condition which can b e rewritten p − 1 ≤ ( n + 1)  n n + 1  p . Fixing p > 1, the frac tion on the right conv erg es to 1, while the firs t factor go es to infinit y , so (for a given exp onent p ) the the n th tail b ounds the n th term eventual ly . W e leav e it to the rea der to check that the function f p ( x ) = ( x + 1)  x x + 1  p is strictly incr easing. How e ver, the condition x n > X n is guara nteed to ho ld whenever n 1 − p p − 1 ≤ 1 n p or p − 1 ≥ n : the n th term exc e e ds the n th tail at le ast for 12 n ≤ K := ⌊ p ⌋ − 1. W e then have Corollary 7. The su bsu m set of a summable p - se qu enc e is a finite union of disjoint close d intervals. The n umb er of these intervals is b etwe en 2 K and 2 N , wher e • K is the highest inte ger less than or e qu al to p − 1 , and • N is the le ast inte ger such that p − 1 ≤ ( N + 1)  N N + 1  p . 12 ⌊ p ⌋ denotes the highest i nt eger ≤ p 12 ZBIGNIEW NITECKI Prop os ition 6 takes ca re o f sequences fo r which the tail ev entually b ounds the term. The situation is mor e complicated when the term event ually exceeds the tail, but not immediately . If at some stage the term exceeds the tail, it is still true that each of the interv als J ξ 1 ··· ξ n will split in to tw o subinterv als separ ated by a “g ap”. How ever, if the tail bo unded the term at some previous s tage, w e can no longer assume that the in terv als which are splitting ar e disjoint: in pr inciple the “g ap” cr eated when one of them splits can b e cov ered over by par t of another one, so that (at lea st as far as this part of the set is c oncerned) no new gap is crea ted in C n +1 . An example o f this phenomenon is the sequence 2 5 , 9 25 , 12 125 , 54 625 , . . . defined by 13 x 2 k = 9 · 6 k − 1 5 2 k x 2 k +1 = 2 · 6 k 5 2 k +1 . (6) This sequence is summable, with X 0 = ∞ X k =0 2 · 6 k 5 2 k +1 + ∞ X k =1 9 · 6 k − 1 5 2 k = 1 and the fir st four tails are X 1 = 3 5 , X 2 = 6 25 , X 3 = 18 125 , X 4 = 36 625 . In general, the even -num b ered tails exceed the corre sp onding terms , but the o dd - nu mbered terms exce ed the corr esp onding tails . In particular , in the passa ge from C 2 n to C 2 n +1 , each in terv al br eaks into tw o ov erlapping interv als, s o C 2 n +1 = C 2 n . How e ver, the pass age from an o dd-num ber ed set to an even-n umbered se t is more complicated: it is still true that ea ch o f the interv als J ξ is repla ced by tw o dis- joint subin terv a ls, but the “gap” this pro duces is so metimes covered by one of the subint erv als coming fro m a different J ξ ′ . F or example, C 3 has three comp onents: J 000 ∪ J 001 =  0 , 30 125  , J 010 ∪ J 011 ∪ J 100 ∪ J 101 =  45 125 , 80 125  , J 110 ∪ J 111 =  95 125 , 1  . A straig ht forward but tedious calculation shows that each of the tw o end comp o- nent s br eaks into three comp onents–for example the left co mpo nent b eco mes  J 0000 =  0 , 36 625  ∪  J 0001 ∪ J 0010 =  54 625 , 96 625  ∪  J 0000 =  114 625 , 150 625  –but the middle comp onent remains unchanged. Nonetheless, the co mpo nents o f C n do app ear to keep breaking up into subinterv als, sugge sting that at the end there will b e infinitely many comp onents to Σ ( { x i } ∞ i =1 ). In fact, this turns out to be true. T o see wh y , w e need to study what happ ens at the far left of C n when the term exceeds the tail. Every in terv al J ξ ( ξ a word of length n ) is a translate to the r ight of the leftmost int erv al J 0 n by x ξ . Since we have ass umed the sequence is non-incr easing, the shortest of these translations is given by x n . Thus for any n the interv al [0 , x n ) int ers ects J 0 n but is dis joint from all the o ther interv als J ξ ( ξ a word of le ngth n ) making up C n . Supp os e now that the n th term exceeds the n th tail ( x n > X n ), so tha t [0 , x n − 1 ] brea ks into tw o subint erv als, [0 , X n ] and [ x n , x n + X n ]. The only 13 W e shall see ho w this mysterious sequence was created in § 5 . SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 13 word ξ of length n − 1 for which J ξ int ers ects [0 , x n − 1 ] is ξ = 0 n . Thus, the “g ap” ( X n , x n ) introduced into J 0 n when it brea ks up into J 0 n 0=0 n +1 and J 0 n 1 bec omes a gap in C n . A simila r ar gument applies when x ξ is the left endp oint of so me comp onent of C n − 1 ; the easiest wa y to s ee this is to transla te the who le picture using x ξ and note that to the left o f x ξ there is a la rger “gap” coming from some earlier separ ation. Finally , w e can use the symmetr y of C ∞ under x 7→ X 0 − x establis hed in Theorem 3 to draw the same conclusion for the right endp oints of components of C n . This giv es us Lemma 8. Supp ose { x i } is a p ositive, non-incr e asing su mmable se quenc e. 14 If at stage n the term exc e e ds the tail x n > X n and [ a, b ] is a c omp onent of C n − 1 , then [ a, a + X n ] and [ b − X n , b ] ar e disjoint c omp onents of C n . Prop ositi on 9. Supp ose { x i } is a p ositive, non- incr e asing summable se quenc e. Then Σ( { x i } ∞ i =1 ) has (1) infi n itely many c omp onents if the term exc e e ds the tail infi nitely often; (2) at le ast 2 N c omp onents if t he term exc e e ds the t ail N times. Note that ( 2 ) ge neralizes the low er b ound given in Pro po sition 6 ( 3 ). In par ticular, Corollary 10. The subsu m set of a p ositive, non-incr e asing su mmable se qu en c e { x i } is a fin ite union of intervals if and only if the tail eventual ly exc e e ds the term. Prop os ition 9 ( 1 ) and Cor ollary 10 strong ly sug gest tha t a subs um set is either a finite union o f closed interv als or a C antor set. How ever, to show t hat a subsum set is a Cantor set, we need to show not only that it has infinitely many comp onents, but also that it has empt y interior, or equiv alently , that every component is a sing le po int. The following obser v atio n, which follows fro m Lemma 8 , suggests that this might b e true: Remark 11. S upp ose { x i } is a p ositive, non-incr e asing summable se quenc e. If the term ex c e e ds the tail infinitely often, then e ach endp oint of every c omp onent of e ach C n c onstitu tes a one-p oint c omp onent of Σ( { x i } ∞ i =1 ) = C ∞ . Kakey a [ 11 ] conjectured tha t the subsum set is a Ca nt or s et if and o nly if the term exc eeds the tail infinitely o ften. Initially , I had the sa me intu ition. How ever, it turns out tha t there exist subsum sets with infinitely man y comp o nent s but nonempty interior. W e shall study some exa mples in the nex t section. 4. Cantor v als The following example was analyzed by Guthrie and N ymann in [ 8 ] in th e pro cess of characterizing the r ange of an arbitra ry finite mea sure. Consider the p ositive 14 Ev ery p ositive sequence can b e rewritten i n non-increasing order without changing the sub- sum set. How ever, the sequence of tails—and hen ce, presumably , the times when the term excee ds the ta il— is ce rtainly affecte d by suc h a reordering. It is critical for our argumen t that the sequence be given i n non-increasing order before this condition is che ck ed. 14 ZBIGNIEW NITECKI decreasing summable s equence 3 4 , 2 4 , 3 16 , 2 16 , . . . that is, x 2 k − 1 = 3 4 k x 2 k = 2 4 k . The tails of this seque nce are X 2 k = 5 3 · 4 k , k = 0 , 1 , . . . X 2 k − 1 = 11 3 · 4 k , k = 1 , . . . . Since 3 < 11 3 and 2 > 5 3 , w e see that every even-n umbered term exc eeds the corres po nding tail, so Σ( { x i } ∞ i =1 ) has infinitely many components, b y Pr op osition 9 . Guthrie and Nymann show that the subsum set co nt ains the in terv al  3 4 , 1  , but their argument (and example) can b e seen as a s p ecial c ase of a num be r-theoretic argument shown me by Rick K eny on, in the c ontext of a n example he sent me befo re I ran across [ 8 ], namely 6 4 , 1 4 , 6 16 , 1 16 , . . . or x 2 k − 1 = 6 4 k x 2 k = 1 4 k . W e note that this or der, while it ma kes tra nsparent the genera ting formulas for the sequence, is not monotone: for e xample, x 2 = 1 4 = 4 16 < 6 16 = x 3 , and x 4 = 1 16 = 2 2 > x 5 = 6 64 = 3 32 . F o r the record, the no n-increasing order is 6 4 , 6 16 , 1 4 , 6 64 , 1 16 , 6 256 , 1 64 , . . . . The r eader can verify that the term exceeds the tail infinitely often. The following arg umen t, sugg ested b y Keny on [ 12 , § 2], gives a wa y to genera te many examples with nonempty interior and, pr esumably , infinitely many co mp onents (including the Guthrie -Nymann one). 15 The key observ ation (in the case of Ke ny o n’s example ab ov e) is tha t every con- gruence cla ss mo d 4 ca n b e obtained as a sum of the “digits” 6 a nd 1, since 6 ≡ 2 mo d 4 and 6 + 1 = 7 ≡ 3 mod 4 . Thu s the se t o f sums of K eny o n’s sequence is the set of all re als which can b e expres sed as “g eneralized ba se 4 expansions” us ing the “digits” 0 , 1 , 6 and 7: Σ( { x i } ∞ i =1 ) = ( ∞ X i =1 a i 4 i | a i ∈ { 0 , 1 , 6 , 7 } ) . Prop ositi on 12 (R. Keny on) . Supp ose we ar e given n ∈ N and n inte gers d 0 , d 1 , . . . , d n − 1 such that d i ≡ j mo d n. Then the set of “ gener alize d b ase n ex p ansions” using these “digits” S = ( ∞ X i =1 a i n i | a i ∈ { d 0 , . . . , d n − 1 } ) has nonempty interior. 15 [ 10 , p. 515] gives another example, which he attributes to Dan V elleman, very muc h in the same spir it. SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 15 Pr o of. The first step is to co nfirm the somewhat optimistic intuition that, since the digits include representativ es o f all the c ongruence classes mo d n , the finite sums of the for m k X i =1 a i n i , a i ∈ { d 0 , . . . , d n − 1 } should, b y analogy with t he standa rd cas e d j = j , ha ve fractiona l parts that include all rational n umber s of the form a n k . The “obvious” rea soning w e migh t expect do e s not apply: for example, 1 4 + 2 4 2 = 6 16 while 1 4 + 6 4 2 = 10 16 ; the difference is not an int eger e ven though 6 = 2 mod 4. How ever, it is tr ue that differ ent ex pressions of this form have differ ent fractional pa rts. T o see this, supp ose we hav e t wo such sums with the same frac tional part: a 1 n + a 2 n 2 + · · · + a k n k = b 1 n + b 2 n 2 + · · · + b k n k + N (where each a i and b i is one of our “digits” d 0 , . . . , d n − 1 , and N ∈ N ). W e can rewrite this a s a 1 − b 1 n 1 + a 2 − b 2 n 2 + . . . a k − b k n k = N and multiply both sides by n k : n k − 1 ( a 1 − b 1 ) + n k − 2 ( a 2 − b 2 ) + · · · + n ( a k − 1 − b k − 1 ) + ( a k − b k ) = n k N . T a king the cong ruence clas s of b oth sides mo d n , we get a k − b k ≡ 0 mo d n . But since the p oss ible dig its b elong to different congr uence classes mo d n , w e m ust have a k = b k . Thu s by induction o n k , a i = b i for i = 1 . 2 , . . . , k . Now, for a g iven (fixed) k , there ar e n k sums of the form k X i =1 a i n i as well as n k fractions of the form a n k with 0 ≤ a < n k . Hence by the pigeo nhole principle, cong ruence mo d n gener ates a bijection b etw een the t wo sets, confirming our int uition. The se cond step is then to reinterpret this s tatement to say that the in teger translates of S cov er the whole real line [ k ∈ Z ( k + S ) = R . Finally , we in vok e the Bair e Cate gory The or em , which in our cont ext says that if a co untable union of sets equa ls R then at lea st one o f them ha s no nempt y interior [ 1 ]. 16 F r om this we conclude that for a t least one integer k , ( k + S ) has non-empt y int erio r—but since it is a translate of S , the same is true of S .  16 This was Baire’s do ctoral dissertation; see Dunham’s highly r eadable account in [ 6 , pp. 184- 191]. A more general version of this (inv olving complete metric spaces), is prov ed i n many basic analysis texts; for example, see [ 19 , Thm. 4.31, pp. 243-5] or [ 21 , Pr ob. 16, p. 40], 16 ZBIGNIEW NITECKI Having established the existence of subsum sets with infinitely m any comp onents but non-empty interior, we should try to under stand b etter the structure o f thes e sets. Suppo se a subsum set Σ( { x i } ∞ i =1 ) has infinitely man y comp onents but non-empty int erio r. F or each n , we can write Σ( { x i } ∞ i =1 ) as the union of 2 n translates of the set Σ ( n ) . Invoking the Ba ire Category Theor em again (this time in its weaker f or m, inv o lving a finite union) we conclude that one, and hence all, of these tra nslates has non- empt y int erio r. In particula r, ea ch interv al J ξ 1 ··· ξ n contains a subinterv al of Σ( { x i } ∞ i =1 ). This means that every p oint of Σ( { x i } ∞ i =1 ) is within distance X n of some s ubint erv al of Σ( { x i } ∞ i =1 ). Since X n → 0, the subinterv als (in particular the non-trivial comp onents) of Σ( { x i } ∞ i =1 ) a re dense. A t the same time, Re mark 11 tells us that the trivial ( i.e. , one- p oint) compo nent s of Σ( { x i } ∞ i =1 ) a re als o dense , in the sense that every endp oint of a no n-trivial comp onent is an accumulation po int of trivial comp onents. In addition to Guthrie a nd Nymann [ 8 ], such sets were studied by Mendes and Oliveira [ 14 ], in connection with the s tructure of arithmetic s ums of Cantor sets (motiv ated by the study of bifurcatio n phenomena in dynamical systems). They dubbed them Cantorvals . In their context, three v arie ties of Cantorv als can arise, but b ecause o f the symmetry of subsum sets, the only kind that arises in o ur context is what they call an M -Cantorval . I prefer the more descriptive term symmetric Cantorval . F or mally: Definition 13. A symmetric Cantorval is a nonempty c omp act subset S of the r e al line su ch that (1) S is the closur e of its interior (i.e., t he nontrivial c omp onents ar e dense) (2) Both endp oints of any nontrivial c omp onent of S ar e ac cumulation p oints of trivial (i.e., one-p oint) c omp onents of S . The remar ks ab ove es tablish a full top olo gical class ification o f subsum sets for summable p ositive sequence s, prov en by Guthrie and Nymann (with differe n t ter- minology) in [ 8 ]: Theorem 1 4 (Guthrie-Nymann) . The subsum set of a p ositive summable se quenc e is one of the fol lowing: (1) a finite un ion of (disjoint) close d intervals; (2) a Cantor set; (3) a symmetric Cantorval. Each of the first t wo categ ories in Theorem 14 provides a list of possible top olo g- ical types : in the first ca se, the nu mber of co mpo nent s determines the top o logical t yp e, while in the s econd, a ll Cantor sets are ho meomorphic, by a well-known theo- rem (see for example [ 19 , pp. 1 03-4 ]). It turns out that all (s ymmetric) Cant or v als are also homeomorphic. This w as pr ov ed in [ 8 ] and stated without explicit pr o of in [ 14 ]. Prop ositi on 15. Any two symmet ric Cantorvals ar e home omorphic. Pr o of. Given t wo Can torv als C and C ′ , fir st identif y the lo ngest comp onent of each; if there is s ome ambiguit y (b ecause several comp onents hav e the same max imal length), then pick the leftmost o ne. There is a unique affine, or der-pres erving homeomorphism b etw een them. Note that by definition there ar e o ther co mpo nents of C ( r esp . C ′ ) on either side of the chosen o ne. In pa rticular, its complement is contained in tw o disjoint SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 17 int erv als, one to the right and one to the left, and the par t of each Cantorv al in each of these interv als is again a Can torv al. Th us, we can apply the same algorithm to pair the longest no nt rivia l compo nent to the left ( r esp . right) of the chosen one in C with the corresp o nding one in C ′ . Contin uing in this w ay , we get an order- preserving corres po ndence b etw een the no n-trivial comp onents of C and those of C ′ , and a n o rder-pre serving homeomorphis m b etw e en co rresp onding co mp onents. But this means we hav e an or der-pres erving co ntin uo us mapping from the (dense) int erio r of C onto the interior of C ′ . This uniquely extends to a homeomorphis m from all of C onto all of C ′ .  Guthrie and Nymann po int out that a mo del symmetric Cantorv al c an b e con- structed b y following the standard co nstruction of the middle-third Ca ntor set (removing the middle third o f each comp onent at a given stag e) but then g oing back a nd “filling in” the g aps at every other stage. 5. Bi-Geometric Sequences W e saw in § 3 that the subsum set of a geo metric sequence is either an interv al or a Ca n tor set, b ecause the relation be t ween the term and the tail is alwa ys the sa me. W e can construct exa mples which exhibit a ny pa rticular patter n o f alternation betw een the t wo p ossible relations by lo oking at the se quence of sets C n in a differ ent w ay , in terms of ra tios. T o b e precis e, given a sequence { x i } of terms, let us lo ok at the asso c iated sequence of tails, { X i } , and for each index i , cons ider the pr op ortion o f X i taken up by x i +1 : ρ i := x i +1 X i or equiv ale nt ly (7) x i +1 = ρ i · X i Then, since X i = x i +1 + X i +1 , we have (8) X i +1 = (1 − ρ i ) · X i . Conv ersely , the sequence of r atios { ρ i } together with the total sum X 0 determines the sequence { x i } recurs ively , via the initia l condition x 1 = ρ 0 X 0 and the r elation x i +1 = ρ i  ρ − 1 i − 1 − 1  x i . Equiv a lently , x i can b e given by an ex plicit formula: (9) x i = ρ i Y j =0 ,...,i − 1 (1 − ρ j ) X 0 . The initial (tota l) sum X 0 is simply a sca ling factor, so to determine what k ind of set o ccurs w e ca n assume that the total sum is X 0 = 1. Now, at ea ch stage, the term and tail are determined from the previo us ta il by ( 7 ) and ( 8 ), from which it is easy to see that 18 ZBIGNIEW NITECKI • the sequence { x i } is non-increasing if a nd only if for every i (10) ρ i ≤ ρ i − 1 1 − ρ i − 1 ; • the n th term exceeds the n th tail ( x n > X n ) if a nd only if 17 ρ n − 1 < 1 2 and (equiv alently) • the n th tail b ounds the n th term ( x n ≤ X n ) if a nd only if ρ n − 1 ≥ 1 2 . So one w ay to crea te a seq uence for whic h both p os sibilities oc cur infinitely often is to pick tw o ratios , (11) 0 < α < 1 2 < β < 1 , and to set ρ i = ( α for even i, β for o dd i. This leads to the se quence x 2 k = β (1 − α ) k (1 − β ) k − 1 x 2 k +1 = α (1 − α ) k (1 − β ) k . (12) A seq uence defined in this w ay spiritually resem bles a geometric sequence, except that it inv olves tw o distinct r atios, so w e migh t refer to it as a bi-geome tric sequence . 18 W e have seen three examples of bi-geometric sequences earlier in t his pap er. The sequence defined b y E quation ( 6 ) was constructed so that α = 2 5 , β = 3 5 while b oth the Guthrie-Nymann and Keny on examples have α = 9 20 , β = 6 11 . The first o bserv a tion above says that, in o rder to hav e a non-increa sing sequence { x i } , we also need α and β to satisfy α ≤ β 1 − β (13) β ≤ α 1 − α . (14) 17 In view of Theorem 1 , by picking an increasing sequence of ratios con verging to 1 2 at an appropriate rate, we can create sequences who se subsum set is a Cant or set of an y desired Leb esgue measure m < 1. 18 An ob vious generalization of this i dea, which could b e call ed a multi-ge ometric se quence , is one where the sequence of ratios ρ i is p erio dic; we could refer to a sequence for which ρ i + m = ρ i for some fixed m > 0 and all i as an m -g e ometri c se quenc e . W e shall deal only with bi-geometric sequences i n this pap er. SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 19 Note that, since we r equire β > 1 2 , E quation ( 14 ) puts further limitations on the po ssible v alues of α : 1 2 < β ≤ α 1 − α forces 1 − α < 2 α or (15) α > 1 3 . By contrast, Equation ( 13 ) puts no further restrictions on β . W e can tr y to analyze the subsum set of a bi- geometric sequence by using the idea of an iter ate d function syst em ([ 3 ], [ 7 ]). Supp ose we hav e a sequence defined in ter ms of tw o parameters 1 3 < α < 1 2 < β < 1 , sub ject to ( 13 ) and ( 14 ), by Equation ( 12 ). The sets C 0 and C 1 are the same, and we can describ e the set C 2 in terms of the set C 0 = [0 , 1] as the unio n o f four interv als J ij , i , j ∈ { 0 , 1 } , each of length X 2 = (1 − α )(1 − β ), with resp ective endpo ints x 00 = 0 x 10 = x 1 = α x 01 = x 2 = β (1 − α ) x 11 = x 1 + x 2 = α + β − αβ . Each of these interv als can be o btained from the basic in terv a l C 0 = [0 , 1] by sca ling and translation; sp ecifically , we can define four affine functions, all with the sa me scaling factor λ = (1 − α )(1 − β ) : ϕ 00 ( x ) = λx ϕ 01 ( x ) = x 2 + λx = β (1 − α ) + λx ϕ 10 ( x ) = x 10 + λx = α + λx ϕ 11 ( x ) = x 1 + x 2 + λx α + (1 − α ) β + λx. Then it is easy to see that, in terms of our ea rlier notation, J ξ i ξ j = ϕ ij ( C 0 ) . But o ur r ecursive relations for x k and X k rep eat ev ery t wo steps, a nd hence we get recursive definitio ns of the sets C k and J ξ : C 2 k = C 2 k − 1 = 1 [ i,j =0 ϕ ij ( C 2 k − 1 = C 2 k − 2 ) ; 20 ZBIGNIEW NITECKI more spe cifically , for e ach word ξ o f length 2 k in zero es and ones, if its initial 2 k − 2-word is ˜ ξ and last tw o entries are i, j ∈ { 0 , 1 } , then J ξ = ˜ ξij = ϕ ij  ˜ ξ  . The v arious ov erlaps b etw een images of ϕ 01 and ϕ 10 make it difficult to carr y out a car eful analys is of the sets C n in g eneral. How ever, one easy obser v ation allows us to c onclude in certain ca ses that the set C ∞ is a Cantor set. At each s tage, the set-mapping C 2 k 7→ C 2 k +2 = ϕ 00 ( C 2 k ) ∪ ϕ 01 ( C 2 k ) ∪ ϕ 10 ( C 2 k ) ∪ ϕ 11 ( C 2 k ) first sca les C 2 k by the fa ctor λ = (1 − α )(1 − β ), duplicates four copies of the sc aled version, then lays them down (with some ov erlap). Ignor ing the ov erlap, we ca n assert that the total of the lengths of the interv als making up C 2 k +2 is less than 4 λ times the cor resp onding measur e for C 2 k . In particula r, the longes t int erv al in C 2 k will hav e length at most (4 λ ) k . This allows us to formulate Remark 1 6. Supp ose { x i } is a bi-ge ometric se quenc e with r atios 0 < α < 1 2 < β < 1 satisfying Equation ( 14 ) . If (16) λ := (1 − α )(1 − β ) < 1 4 , then Σ( { x i } ∞ i =1 ) is a Cantor set. This s hows in par ticular that our fir st e xample yields a Cantor set, since λ = 6 25 < 1 4 . By contrast, the Guthrie- Nymann a nd Keny on examples b oth hav e λ = 11 20 · 5 11 = 1 4 . In Fig ure 2 we have sketched the para meter spa ce ( α, β ) ∈ [0 , 1] × [0 , 1 ] for bi- geometric seq uences. Our discussion ab ove concerned the upp er- left q uarter of this square,  0 , 1 2  ×  1 2 , 1  , characterized b y the ineq ualities ( 11 ), but by interc hanging the roles of α and β where necess ary we can e xtend it to the who le squar e. The hatched areas are excluded b y the requir ement tha t the s equence { x i } b e non- decreasing (Equation ( 13 ) a nd ( 14 )). The upp er gr ay area is where Equatio n ( 16 ) holds, guara nt eeing that Σ( { x i } ∞ i =1 ) is a Ca n tor set. Note that the tw o exa mples of Ca nt or v als (Guthrie-Nymann and Ken yon) b oth cor resp ond to a p o int on the bo undary of this regio n, where λ = 1 4 . The lower gray area is where b oth α and β a re a t mo st eq ual to 1 2 , whic h means the tail a lwa ys b ounds the term—so C n = [0 , 1 ] for a ll n . This leav es the t wo white reg ions (lab eled w ith a question marks) where one r atio is a t most 1 2 while the other is greater than 1 2 , where o ur analysis so far cannot completely determine the top olog y of the subsum set; howev er, we do know that in this regio n the subsum set has infinitely many compo nents, so for each bi-geometric sequence coming fro m pa rameters in this interv al, the subsum set is either a Cantor set or a symmetric Cantorv al. How ever we have not developed a test to distinguish, in general, which pos sibility a par ticular exa mple exhibits. In fact, I do n’t know if there are Cantorv al exa mples with λ > 1 4 , or, in the o ther direction, if there are any bi-g eometric sequences with pa rameters in the white r egion which yield Cantor sets. SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 21 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 0 0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 α β α = β 1 − β β = α 1 − α ? ? Σ = [0 , 1] λ < 1 4 : Σ i s Can tor • Can torv al examples x 2 > x 1 x 3 > x 2 β = 3 − 4 α 4 − 4 α Figure 2. Bi-geometr ic sequences 6. Sequences with V ar ying Sign W e tur n no w to the g eneral case, when some terms ar e p o sitive while others are negative. Her e we take adv antage of another observ ation, g iven b y Riema nn in [ 20 ] but a ttributed by him to Dir ichlet: let us separate out the p ositive terms o f x i as  x + i  and the nega tive terms as  x − i  . Since the terms of each of thes e t wo sums hav e constant sig n, we c an define X x + i = X + ∈ [0 , ∞ ] X x − i = X − ∈ [ −∞ , 0 ] . W e can distinguish three p oss ible config urations: • If b o th X − and X + are finite, the seque nce is absolutel y summable ( P ∞ i =0 | x i | converges), b ecaus e ∞ X i =0 | x i | ≤   X −   + X + . 22 ZBIGNIEW NITECKI Recall tha t a s a consequence e very reordering of the seq uence sums to the same (finite) num b er. • If b oth X − and X + are infinite, the sequence is conditionally summable . It is a standard fact (attributed to Riemann) that if a series con verges while the co rresp onding serie s of abso lute v alues diverges, then by rearra nging the or der o f the terms w e can get a series summing to any real n umber, or diverging to either + ∞ or −∞ . Riemann’s informal pro of of this fact [ 20 , § 3] rests o n the o bserv a tion that in this case b oth X − and X + are infinite. • If one is finite and the other infinite, we will call the sequence unconditi on- ally unsumm able . In this case , every reo rdering g ives rise to a div erg ent series; for exa mple, if X + = ∞ a nd X − is finite, then a partial s um o f p o s- itive terms can b e ma de ar bitrarily larg e, while including negative terms as well can at worst low er this sum by | X − | , so any rear rangement diverges to ∞ . In the absolutely s ummable case, Kakeya [ 11 ] stated without pro o f that Σ( { x i } ∞ i =1 ) equals the interv al [ X − , X + ] if a nd only if all the tails b ound the s ums for the se- quence of absolute v alues {| x k |} . Hor nich [ 9 ] took this further: again assuming that the seq uence is a bsolutely summable (so b o th X ± are finite), and given a subse- quence { y i } of our sequence, consider the tra nslated sum of its abs olute v a lues X | y i | + X − = X   y + i   + X   y − i   + X − = X y + i + X   y − i   − X k   x − k   where the last summand is the sum o f the absolute v alues of al l the ne gative terms of the origina l sequence. If we co mb ine the last t wo sums, the terms y i in the sub- sequence get cancelled, leaving the sum of all the negative terms whic h are exclude d from the s ubsequence. This o f cour se is ano ther subsum of our sequence. F urther- more, every subsum of the full s equence can b e expre ssed in this wa y , w hich shows that the subsum set of the (absolutely summable) sequence { x i } is the translate by X − of the subsum set of the sequence {| x i |} of absolute v alues. Prop ositi on 17 (Hornich) . If { x i } is an absolutely summable se quenc e, t hen Σ( { x i } ∞ i =1 ) = Σ( {| x k |} ∞ k =1 ) + X − . This means that the criteria we gav e in Theorem 4 , Pr op osition 6 and Co rol- lary 10 can be applied to the (p ositive) sequence of absolute v alues to determine the top ology of the subsum set of the or iginal, v ariable sign but absolutely summable sequence. Finally , if our orig inal sequence is not abso lutely summable, w e can easily sp ecify the subsum s et. In this case we know that at least one of X ± is infinite. W e con- centrate on the case X + infinite; the other case is analogous. Since the subsequence of positive terms is not summable, by Theorem 1 Σ  x + i  = [0 , ∞ ): we can obtain any p ositive num b er as the sum of a s ubsequence of p ositive terms. If X − is finite, we can obtain any num b er in [ X − , ∞ ) b y adding a p ositive num b er to X − ; if it is infinite, we can obtain any ne gative num be r as the sum of some subsequence of ne gative terms—so Σ( { x i } ∞ i =1 ) = R in this case. With a little a buse o f notatio n and sne aky reinterpretation, we can fo rmulate a general characterization of all subsum sets. The abuse o f notatio n is that we will a llow closed interv al nota tion with one or bo th endp oints infinite; it will be unders to o d that in s uch a case the square bracket at that end should b e replaced by a round pa renthesis. SUBSUM SETS: INTER V ALS, CANTOR S ETS, AND CANTOR V ALS 23 The sneaky reinterpretation is simply this: if a p ositive s equence is not summa- ble, then every “tail” is infinite, so bo unds a ny term. With these t weaks, we c an state a genera l result, extending Theorem 14 : Theorem 18. Given a nul l s e quenc e x k → 0 , let X + (r esp. X − ) b e the (p ossi- bly infinite) sum of al l the p ositive (r esp. ne gative) terms. Then the subsum set Σ( { x i } ∞ i =1 ) is a close d, p erfe ct set whose c onvex hul l is the int erval [ X − , X + ] , and which is symmetric with r esp e ct to r efle ction acr oss the midp oint of this interval. F urthermor e, denote the se quenc e of absolute values of our terms by a k = | x k | , k = 1 , 2 , . . . and its tails by A k = X i>k a k . Then: (1) If the tail b ounds t he t erm a k ≤ A k for al l k > K , and the numb er of t erms which exc e e d the tail a k > A k is N , t hen Σ( { x i } ∞ i =1 ) is the u nion of b etwe en 2 N and 2 K disjoint close d intervals. (2) If the term exc e e ds t he tail infi nitely often, then Σ( { x i } ∞ i =1 ) is either a Cantor set or a symmetric Cantorval. In p art icular, if the term always exc e e ds the tail, then Σ( { x i } ∞ i =1 ) is a Cantor set. 7. Generaliza tions W e comment briefly on tw o extensio ns of the material discuss ed in this pap er. First, Ra fe Jones [ 10 ] consider s non-null real sequences. Several new phenomena are po ssible in this context. If the sequence conv erg es to a nonzer o limit, then its subsum set is a countable, unbo unded set; in fact, [ 10 , P rop. 4.1] any sequence po ssessing no null subsequences has a countable subs um set. Jones notes [ 1 0 , p. 514] that in general the subsum set of a non-null seq uence need not b e clos ed (for example, Σ  n +1 n  has 1 as an accumulation point, but do es not contain it). In genera l, the subsum set of any seq uence is either meag er ( i.e. , of first Bair e category , and hence totally dis connected), o r else its interior is a dense subset [ 10 , Theorem 3.1 ]. If it is neither coun table nor a n unbounded interv al, then it c onsists of a countable union o f translates o f some null subsequence [ 10 , Pr op. 3.2]. A sec ond extension, referenced by Jones, is the work of Manuel Mor´ an [ 16 , 17 ] which considers subsum s ets of sequences in higher dimensio ns, in particular of co mplex sequences , under an assumption (“quick convergence”) analo gous to our “terms exceed tails” conditio n. In this context, Mor´ an studies the Hausdorff dimension of the fractal sets generated b y f amilies sequences obtained from analytic functions. 24 ZBIGNIEW NITECKI References [1] Ren´ e Baire. Sur les fonctions des variables r ´ eel le s . Impri merie Bernardoni de C. Reb esc hini & Ci e, 1899. 4 [2] Roger Bak er, Charles Christenson, and Henry O rde (translators). Bernhar d Riemann Col- le ct e d Pap ers . Kendrick Pr ess, 2004. 20 [3] Mi c hael Barnsl ey . F r actals Everywher e . Academic Pr ess, 1988. Second Edition, Morgan Kuf- mann 1993 (Hardback) , 2000 (P ap erback ). 5 [4] Garrett Birkhoff, editor. A Sour ce Bo ok i n Classic al Analysis . Harv ard Univ ersity Press, 1973. 20 [5] P . G. L. Di richlet. Bew eis des Satzes, dass jede unbegrenzte arithmetische Progression, deren erstes Glied und Di fferenz ganze Zahlen ohne gemeinsc haftlichen F actor sind, unendlich viele Primzahlen en th¨ alt. 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