Journal of the Egyptian Mathematical Society 25 (2017) 326–330
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Journal of the Egyptian Mathematical Society
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Original Article
Some existence results on Cantor sets
∗ Borys Álvarez-Samaniego a, , Wilson P. Álvarez-Samaniego a, Jonathan Ortiz-Castro b a Núcleo de Investigadores Científicos, Facultad de Ingeniería, Ciencias Físicas y Matemática, Universidad Central del Ecuador (UCE), Quito, Ecuador b Facultad de Ciencias, Escuela Politécnica Nacional (EPN), Quito, Ecuador
a r t i c l e i n f o a b s t r a c t
Article history: The existence of two different Cantor sets, one of them contained in the set of Liouville numbers and the Received 2 November 2016 other one inside the set of Diophantine numbers, is proved. Finally, a necessary and sufficient condition
Revised 25 January 2017 for the existence of a Cantor set contained in a subset of the real line is given. Accepted 11 February 2017 Available online 18 March 2017 ©2017 Egyptian Mathematical Society. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license.
MSC: ( http://creativecommons.org/licenses/by-nc-nd/4.0/ ) 54A05 54B05 54A25
Keywords: Cantor set Liouville numbers Diophantine numbers
1. Introduction that q > 1 and p 1 First, we will introduce some basic topological concepts. 0 < α − < . (1.1) q q n Definition 1.1. A nowhere dense set X in a topological space is a If β ∈ R is not a Liouville number, we say that β is a Diophan- set whose closure has empty interior, i.e. int ( X ) = ∅ . tine number. The sets of Liouville and Diophantine numbers are, Definition 1.2. A nonempty set C ⊂ R is a Cantor set if C respectively, denoted by L and D . is nowhere dense and perfect (i.e. C = C , where C := { p ∈ Joseph Liouville, by giving two different proofs, showed the ex- R ; p is an accumulation point of C} is the derived set of C ). istence of transcendental numbers for the first time in 1844 [1,2] . Definition 1.3. A condensation point t of a subset A of a topolog- Later, in 1851 [3] more detailed versions of these proofs were ical space, is any point t , such that every open neighborhood of t given. It was also shown in [3] that the real number contains uncountably many points of A . + ∞ 1 , We denote by Q the set of rational numbers. The symbol Z is 10 k ! k =1 used to denote the set of integers, Z + = { 1 , 2 , . . .} denotes the pos- itive integers and N = { 0 , 1 , 2 , . . .} is the set of all natural numbers. which was already mentioned in [1] , is a transcendental number. The cardinality of a set B is denoted by | B | . We denote by OR , the It is worth noting that the techniques used in [1–3] allow proving class of all ordinal numbers. Moreover, represents the set of all that all Liouville numbers are transcendental. countable ordinal numbers. Some general properties of the set of Liouville numbers are: L is a null set under the Lebesgue measure (i.e. λ(L ) = 0 ), it is α ∈ R \ Q α Definition 1.4. Let . We say that is a Liouville number a dense G δ set in the real line, L is an uncountable set and, ∈ N , = = , if for all n there exist integer numbers p pn and q qn such more specifically, it has the cardinality of the continuum. Since the Lebesgue measure of the Diophantine numbers is infinity, it is an uncountable set. By using Theorem 3.1 below, there exists a Cantor ∗ Corresponding author. set in D , and since any uncountable closed set in R has the cardi- E-mail addresses: [email protected] , [email protected] , balvarez@ D impa.br (B. Álvarez-Samaniego), [email protected] (W.P. Álvarez- nality of the continuum, it follows that has also the cardinality Samaniego), [email protected] (J. Ortiz-Castro). of the continuum. Moreover, in view of the density of Q in R , the http://dx.doi.org/10.1016/j.joems.2017.02.002 1110-256X/© 2017 Egyptian Mathematical Society. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license. ( http://creativecommons.org/licenses/by-nc-nd/4.0/ ) B. Álvarez-Samaniego et al. / Journal of the Egyptian Mathematical Society 25 (2017) 326–330 327 set of Diophantine numbers is also dense in the real line. In addi- Proof. Let f be the function given by tion, D is a set of first category, i.e. it can be written as a countable f : A −→ { 0 , 1 } N union of nowhere dense subsets of R . x −→ y, Now, we would like to prove an interesting result, which will N only be mentioned in this section, related to the fact that the prop- where y = (y n ) n ∈ N ∈ { 0 , 1 } is defined by erty of being a Cantor set is preserved by homeomorphisms when , = , the homeomorphic image of its domain is a closed subset of R . 1 if x2 n +1 1 y n = 0 , if x 2 n = 1 , Proposition 1.1. Let A, B ⊂ R . Suppose that f : A −→ B is a homeo- morphism and B is a closed subset of R . If C ⊂ A is a Cantor set, then for all n ∈ N . From the definition of function f , one gets that f is N N f ( C ) ⊂ B is also a Cantor set. bijective. Then, | A | = | { 0 , 1 } | . Since c = | R| = | { 0 , 1 } | , we con- clude that A has the cardinality of the continuum. Proof. From the fact that C ⊂ A is a perfect set, we get that C is closed subset of R . Moreover, since f −1 is continuous, we have that Using (2.1) , we define the set ( f −1 ) −1 (C) = f (C) is a closed set in B , and since B is a closed sub- + ∞ x − set of R , it follows that f ( C ) is closed in R . Now, we claim that = n 1 = ( ) ∈ . S : x xn n ∈ N A (2.2) ⊂ ∈ ε > ∈ 10 n! f(C) (f(C)) . In fact, let y f(C) and 0. So, there is x C such n =1 that y = f (x ) . Since f is continuous at x , there exists δ > 0 such that The following lemma will be used in the proof of Proposition 2.1 .
= ( ) ∈ {− , , } for all z ∈ A, | z − x | < δ ⇒ | f (z) − f (x ) | < ε. (1.2) Lemma 2.2. Let z zi i ∈ Z + be a sequence such that zi 1 0 1 for all i ∈ Z + . If ∈ = , ∈ ⊂ < Considering that x C C there exists z0 C A such that 0 + ∞ | z − x | < δ, and by (1.2) we deduce that | f (z ) − f (x ) | < ε. Fur- z 0 0 i = 0 , = i ! thermore, the injectivity of f implies that f(x) f(z0 ). Then, f(z0 ) 10 i =1 ∈ ( B ( y , ε) ࢨ{ y }) ∩ f ( C ). In consequence, y ∈ ( f ( C )) . Hence, f ( C ) is a = ∈ Z + perfect set in R . On the other hand, since C = ∅ , we see that then zi 0 for all i . f ( C ) = ∅ . Finally, we will show that int ( f (C)) = ∅ . We suppose, by Proof. Since contradiction, that there exists y ∈ int( f ( C )). Then, there is r > 0 + ∞ ( − , + ) ⊂ ( ) ( − , + ) such that y r y r f C . Since y r y r is a connected z 1 z i −1 − = , set, we have that f ((y − r, y + r)) ⊂ C ⊂ A is also a connected i ! 10 = 10 set. Let us take u, v ∈ f −1 ((y − r, y + r)) such that u < v . Then, i 2 ( , ) ⊂ −1 (( − , + )) ⊂ = ∅ u v f y r y r C. Thus, int(C) , which is a contra- we see that diction. Hence, int ( f (C)) = ∅ . Since f ( C ) is a nonempty perfect and + ∞ + ∞ + ∞ z z | z | 1 nowhere dense set in R , we conclude that f ( C ) is a Cantor set. 1 = i ≤ i ≤ 10 10 i ! 10 i ! 10 i ! i =2 i =2 i =2 It is worth mentioning that “every uncountable G δ or F σ set + ∞ 1 in a Polish space contains a homeomorphic copy of the Cantor 1 2 1 < = 10 = . space” [4]. In addition, Alexandroff [5] showed that every uncount- i − 1 = 10 1 90 able Borel-measurable set contains a perfect set. By using these i 2! 10 last facts, one can also obtain some of the results of this paper. | | < 1 < = Thus, z1 9 1. Hence, we conclude that z1 0. However, in Section 2 , we will mainly proceed in a different way, Now, we suppose, by induction, that for n ∈ Z + such that n ≥ by constructing an uncountable perfect and nowhere dense subset = ∈ { , , . . . , − } 2, we have that zk 0 for k 1 2 n 1 . From the fact that of the Liouville numbers. It deserves remark that Bendixson’s The- + ∞ z z orem, which states that every closed subset of the real line can be − n = i , represented as a disjoint union of a perfect set and a countable set, 10 n ! 10 i ! i = n +1 is used in the proofs of the main results given in Sections 3 and 4 . we get This paper is organized as follows. In Section 2 , the existence of a Cantor set in the set of Liouville numbers is proved. Section 3 is + ∞ + ∞ + ∞ z z | z | 1 devoted to show the existence of a Cantor set inside the set of n = i ≤ i ≤ n ! 10 10 i! 10 i! 10 i! Diophantine real numbers. Finally, in Section 4, a necessary and i = n +1 i = n +1 i = n +1 sufficient condition for the existence of a Cantor set contained in a + ∞ 1 1 (n +1)! 10 1 subset of R is given. < = 10 = · . ( + ) 10 i 1 − 1 9 10 n 1 ! i =(n +1)! 10 2. Existence of a Cantor set contained in the Liouville numbers Then,
n ! We begin this section showing the existence of an uncount- 10 10 10 n !(1 −(n +1)) | z n | < · = · 10 able closed set, S , contained in L . Then, we prove that S is a per- 9 10 (n +1)! 9 λ(L ) = , λ fect set. Finally, since S is a closed set and 0 where is 10 − ( ) 10 − 1 = · 10 n n! ≤ · 10 4 = < 1 , the Lebesgue measure on the real line, we conclude that S is also 9 9 90 0 0 nowhere dense. where we have used the fact that −n (n !) ≤−4 for n ∈ { 2 , 3 , . . .} . First, let us consider the following set Hence, | z n | < 1, and thus we conclude that z n = 0 . N A = { x = (x n ) n ∈ N ∈ { 0 , 1 } : x 2 n + x 2 n +1 = 1 , ∀ n ∈ N } . (2.1) The next proposition shows that the set S has the cardinality of The next result concerns the cardinality of set A . the continuum.
Lemma 2.1. The set A, given in (2.1) , has the cardinality of the con- Proposition 2.1. Let S and A be the sets respectively defined by tinuum. (2.2) and (2.1) . Then, | S| = | A | = c. 328 B. Álvarez-Samaniego et al. / Journal of the Egyptian Mathematical Society 25 (2017) 326–330
+ ∞ + ∞ Proof. Let g be the function given by 1 1 ≤ ε + < ε + k k −→ i! i g : A S = + 10 =( + ) 10 + ∞ i k 2 i k 2 ! xn −1 (k +2)! x = (x ) ∈ N −→ . 0 . 1 10 1 10 1 n n n ! = ε + = · + · 10 k n =1 1 − 0 . 1 9 10 (k +2)! 9 10 (k +2)! 20 1 By the definition of function g, we see that g is surjective. More- = · , over, by Lemma 2.2 , it follows that g is injective. Therefore, we con- 9 10 (k +2)! clude that g is bijective. Hence, | S| = | A | = c. where in the second inequality on the right-hand side of the im- plication above, we have used (2.5) . Thus, for all m, n ∈ N , The subsequent result states that S is closed. ( + ) k 1 ! m n 20 10 Proposition 2.2. The set S ⊂ R , given in (2.2) , is a closed subset of m, n ≥ N ⇒ | x − x | < · k k k ( k +2 )! R . 9 10 20 1 20 (2.6) = · < < 1 ∈ R ( n ) ( k +1 )! ( k +1 ) 4 Proof. Let y be such that there is a sequence x n ∈ N in S with 9 10 9 · 10
n = ∈ N , ⇒ m = n. lim x y. So, for every n we can write xk xk n → + ∞
∈ n = + ∞ It follows from (2.6) that there is xk {0, 1} such that lim x x n n → + ∞ k n i −1 x = , x . By the principle of finite induction, we conclude that for all 10 i ! k i =1 l ∈ N , where (x n ) ∈ A . Since (x n ) ∈ N is a convergent sequence, we have n = ∈ { , } . i i ∈ N n lim x : xl 0 1 (2.7) n → + ∞ l ( n ) ε = 1 > , that x n ∈ N is a Cauchy sequence. Thus, for 0 90 0 there is Moreover, it follows from the last expression that for all l ∈ N , N 0 ∈ N such that for all m, n ∈ N , ∈ N ∈ N , there exists Nl such that for all n + ∞ m n x − x m n i −1 i −1 ≥ ⇒ n = . m, n ≥ N 0 ⇒ | x − x | = < ε 0 . n Nl xl xl (2.8) 10 i ! i =1 = ( ) ∈ Claim 1: x : xi i ∈ N A. Then, for m, n ∈ N such that m, n ≥ N , we get Let i ∈ N . We see that for all n ∈ N , 0 n n + ∞ + ∞ + = . m n m − n m − n x2 i x2 i +1 1 | x − x | x − x − x − x − 0 0 ≤ i 1 i 1 + i 1 i 1 10 10 i ! 10 i ! By using ( 2.7 ) into the last expression we get i =1 i =2 + ∞ + ∞ + ∞ x 2 i + x 2 i +1 = 1 . (2.9) | m − n | x − x − 1 1 < ε + i 1 i 1 ≤ ε + < ε + N 0 0 0 By ( 2.7 ), x ∈ 2 , and using ( 2.9 ), we conclude that x ∈ A. i! i! i = 10 = 10 = 10 + ∞ xi −1 i 2 i 2 i 2! Claim 2: y = . i =1 10 i! 2 ∞ 0 . 1 1 1 1 ε > 1 < + ∞ , = (ε) ∈ In fact, let 0. Since i =1 i ! there is a a = ε 0 + = + = . 10 1 − 0 . 1 90 90 45 { 2 , 3 , . . .} such that
n = m ∈ n = + ∞ Therefore, x x . Hence, there is x0 {0, 1} such that lim x 0 0 n → + ∞ 0 1 + < ε. (2.10) x 0 . We proceed now by induction on k . Let k ∈ Z . Let us suppose i ! 10 ∈ { , . . . , − } , n = ∈ { , } i = a that for all j 0 k 1 there exist lim x : x j 0 1 . Us- n → + ∞ j n By ( 2.8 ), there exists P = P (ε) ∈ N , such that for all n ∈ N , ing again the fact that (x ) n ∈ N is a Cauchy sequence, we see that
ε = 10 · 1 > , ∈ N , ∈ N , n for k ( + ) 0 there is Nk −1 such that for m n n ≥ P ⇒ x = x ∈ { 0 , 1 } , ∀ k ∈ { 0 , 1 , . . . , a − 2 } . (2.11) 9 10 k 2 ! k k + ∞ So, for all n ∈ N , x m − x n m n i −1 i −1 m, n ≥ N − ⇒ | x − x | = < ε . (2.3) + ∞ + ∞ + ∞ k 1 i ! k n 10 x − x − x − i =1 n ≥ P ⇒ x n − i 1 = i 1 − i 1 i ! i ! i ! 10 10 10 ∈ { , , . . . , − } , n = ∈ i =1 i =1 i =1 Since for every j 0 1 k 1 there exist lim x x j n → + ∞ j − + ∞ { , } ∈ N ∈ N , a 1 n − n − 0 1 we have that there is Nj such that for all n x − xi −1 x − xi −1 = i 1 − i 1 n i ! i ! n ≥ N ⇒ x = x . (2.4) 10 10 j j j i =1 i = a + ∞ Let N := max{ N , N , . . . , N , N } ∈ N . Thus, for all m, n ∈ N , n − k 0 1 k −1 k −1 x − xi −1 = i 1 + ∞ i! m − n = 10 x − x − i a , ≥ ⇒ | m − n| = i 1 i 1 < ε , m n Nk x x k (2.5) + ∞ + ∞ i! | n − | = + 10 x − xi −1 1 i k 1 ≤ i 1 ≤ < ε, 10 i ! 10 i ! where in the last inequality we have used (2.3) and (2.4) . Then, for i = a i = a all m, n ∈ N , (2.12) | m − n | x x where in the third equality above we have used (2.11) and in the last , ≥ ⇒ k k m n Nk 10 (k +1)! inequality we have used (2.10). Then, + ∞ + ∞ + ∞ x m − x n x m − x n x − i −1 i −1 i −1 i −1 n = i 1 . ≤ + lim x i ! i ! n → + ∞ i! 10 10 = 10 i = k +1 i = k +2 i 1 + ∞ m n By the uniqueness of the limit in the real line, we conclude that y = | x − x | i −1 i −1 + ∞ xi −1 < ε + = . k 10 i ! i 1 10 i! i = k +2 B. Álvarez-Samaniego et al. / Journal of the Egyptian Mathematical Society 25 (2017) 326–330 329
∈ + ∞ xi −1 Using Claims 1 and 2, we obtain that y S. Hence, S is a closed ∈ S , there is (x ) ∈ N ∈ A such that a = . For all i ∈ N , we n n i =1 10 i! R subset of . define By using a standard proof, we now show the following propo- 1 − x i , if (2 N − i )(2 N + 1 − i ) = 0 , y i := (2.16) sition. x i , otherwise. N Proposition 2.3. The set S, given in (2.2) , is contained in the set of Since (x n ) n ∈ N ∈ { 0 , 1 } , it follows directly from (2.16) that ( ) ∈ { , } N ∈ N , + = Liouville numbers, more precisely yn n ∈ N 0 1 . We will now show that for all i y2 i y2 i +1 = , + = − + − = + − 1. In fact, if i N then y2 i y2 i +1 1 x2 i 1 x2 i +1 1 1 ⊂ ( , ) ∩ L ⊂ L . S 0 1 (2.13) = = + = + = 1 1. On the other hand, if i N, then y2 i y2 i +1 x2 i x2 i +1 1. ( ) ∈ N ∈ , ∈ ( ) ∈ = Thus, yn n A and therefore Proof. Let y S. Thus, there exists xn n ∈ N A such that y + ∞ x − i 1 + ∞ i =1 i ! . Then, y − 10 = i 1 ∈ . b : S (2.17) + ∞ + ∞ + ∞ i! = 10 x − 1 1 1 i 1 0 < y = i 1 < < = < 1 . 10 i ! 10 i ! 10 i 9 In addition, i =1 i =1 i =1 + ∞ + ∞ + ∞ + ∈ Z , ∈ Z x − y − x − − y − We now consider n . We define qn pn as follows < | − | = i 1 − i 1 = i 1 i 1 0 a b 10 i! 10 i! 10 i! n i =1 i =1 i =1 x − = n! > = · k 1 . qn : 10 1 and pn : qn k! + ∞ = 10 x − y x − y x + − y + k 1 = i i = 2N 2N + 2N 1 2N 1 10 (i +1)! 10 (2 N+1)! 10 (2 N+2)! Therefore, = i 0 + ∞ + ∞ x 2 N − 1 + x 2 N x 2 N+1 − 1 + x 2 N+1 p n x − 1 = + − = k 1 < ( + ) ( + ) y 10 2N 1 ! 10 2N 2 ! q k! k! n = + 10 = + 10 k n 1 k n 1 2 x − 1 2 x + − 1 | 2 x − 1 | | 2 x + − 1 | = 2N + 2N 1 ≤ 2N + 2N 1 + ∞ + ∞ 1 1 1 10 (2 N+1)! 10 (2 N+2)! 10 (2 N+1)! 10 (2 N+2)! < = · 10 k 10 (n +1)! 10 k 1 1 1 1 k =(n +1)! k =0 = + ≤ + 10 (2 N+1)! 10 (2 N+2)! 10 (2 N)! 10 (2 N)! 1 10 10 = · < 2 10 (n +1)! 9 10 (n +1)! ≤ < ε, 10 2 N 10 n ! 1 1 1 ∈ ≤ = = = . where the last equality is a consequence of the fact that for all z (n +1)! n !(n +1) −n ! n ! ·n n 10 10 10 qn { 0 , 1 } , | 2 z − 1 | = 1 . This concludes the proof of the proposition. For the sake of completeness, we will show here that y ∈ R \ Q . We now proceed to prove the key theorem of this section. In fact, we suppose, by contradiction, that there are p, q ∈ Z + such p + ∞ xi −1 that = y = . Since y ∈ (0, 1), we see that 0 < p < q . Theorem 2.1. The set S, given in (2.2) , is a Cantor set contained in q i =1 10 i! Then, p ∈ { 1 , 2 , . . . , q − 1 } . Moreover, there is m ∈ Z + such that the set of Liouville numbers. · − q < 10 m! m 1. (2.14) Proof. Let S be the set given by (2.2) . By Propositions 2.1, 2.3 and 2.4 , S is an uncountable perfect set contained in L . Moreover, by p + ∞ xi −1 Furthermore, the expression = is equivalent to q i =1 10 i! Proposition 2.2 , S is a closed subset of R , since λ(L ) = 0 and S ⊂ L , R m + ∞ we have that S is a nowhere dense subset of . We may therefore m ! m ! −k ! m ! xk −1 ⊂ L p · 10 = q · x − 10 + q · 10 · . (2.15) conclude that S is a Cantor set such that S . k 1 10 k ! k =1 k = m +1 Before ending this section, we state an important definition and m ! + ∞ xk −1 + a lemma that we will use in the proof of Proposition 2.5 below. Using (2.15) we see that q · 10 · ∈ Z . Then, k = m +1 10 k! Definition 2.1 (Cantor–Bendixson’s derivative) . Let A be a subset of + ∞ + ∞ x − 1 ≤ · m! · k 1 < · m! · a topological space. For a given ordinal number α ∈ OR , we define, 1 q 10 q 10 α 10 k! 10 k! using transfinite recursion, the αth derivative of A , written A ( ), as k = m +1 k = m +1 + ∞ + ∞ follows: · m ! m ! 1 q 10 1 < q · 10 · = · • (0) = , k (m +1)! k A A 10 10 10 (β+ ) (β) k =(m +1)! k =0 • A 1 = (A ) , for all ordinal β, (λ) (γ ) q 10 q • A = A , for all limit ordinal λ = 0. = · < < 1 , γ <λ 10 m ! ·m 9 10 m ! ·m −1 where in the last inequality we have used (2.14) . Last expression The next lemma and its proof can be found in [6, Lemma 2.1] . shows that the assumption y ∈ Q leads to a contradiction. Hence, + Lemma 2.3. Suppose that n ∈ Z . Let F 1 , F 2 , . . . , F n be closed subsets ⊂ ( , ) ∩ L ⊂ L S 0 1 . of R . Then, for all ordinal number α ∈ OR , we have that The succeeding result says that the set S is equal to its set of (α) n n (α) accumulation points. = . Fk Fk = = Proposition 2.4. The set S, given in (2.2) , is a perfect set, i.e. S = S . k 1 k 1 We close this section with a general topological result on the Proof. Since S is closed, it is enough to show that every element of real line. S is an accumulation point of S . In order to prove the last assertion, let a ∈ S , and ε > 0. We will show that there exists b ∈ S such that Proposition 2.5. Every element of a perfect set C ⊂ R is a condensa-
< | − | < ε ∈ N > 1 · 2 0 a b . We take N satisfying N 2 log10 ε . Since a tion point of C. 330 B. Álvarez-Samaniego et al. / Journal of the Egyptian Mathematical Society 25 (2017) 326–330
Proof. Let C ⊂ R be a perfect set. We suppose, for the sake of con- 4. A necessary and sufficient condition for the existence of a tradiction, that there is a ∈ C such that a is not a condensation Cantor set contained in a subset of the real line point of C . Then, there exists r > 0 such that C ∩ (a − r, a + r) is countable. Thus, C ∩ [ a − r, a + r] is also countable. Since C is a per- We begin this section with a preliminary result. fect set, we have that C is closed. Hence, C ∩ [ a − r, a + r] is a closed Lemma 4.1. Every nonempty perfect set in R contains a Cantor set. and countable subset of R . By Theorem C of Cantor [7] , there exists a countable ordinal number α ∈ such that the αth derivative of Proof. Let P ⊂ R be a nonempty perfect set. There are two cases to (α) C ∩ [ a − r, a + r] is empty, namely, (C ∩ [ a − r, a + r]) = ∅ . More- consider. over, we write • If int (P ) = ∅ , then P is a Cantor set. = ( \ ( − , + )) ( ∩ ( − , + )) C C a r a r C a r a r • If a ∈ int( P ), there is r > 0 such that (a − r, a + r) ⊂ P . Let C be
− r , + r ⊂ (R \ (a − r, a + r)) ∪ (C ∩ [ a − r, a + r]) , the usual triadic Cantor set in the closed interval a 3 a 3 . Then, R \ ( − , + ) = (−∞ , − + , + ∞ ) where a r a r a r] [a r is a perfect set. r r Using Lemma 2.3 , we see that C ⊂ a − , a + ⊂ (a − r, a + r) ⊂ P. 3 3 (α) (α) a ∈ C = C ⊂ [ (R \ (a − r, a + r)) ∪ (C ∩ [ a − r, a + r]) ] (α) (α) = (R \ ( − , + )) ∪ ( ∩ − , + ) a r a r C [a r a r] Finally, we show the purpose of this section. = R \ (a − r, a + r) , Theorem 4.1. X ⊂ R contains a Cantor set if and only if X contains a which is a contradiction. Therefore, every element of C is a con- closed and uncountable subset of R . densation point of C . Proof. Let F ⊂ X be a closed and uncountable subset of the real
3. Existence of a Cantor set contained in the Diophantine line. By Bendixson’s Theorem, there is a perfect and uncount- ⊂ numbers able set P F. By Lemma 4.1, there is a Cantor set K, such that K ⊂ P ⊂ F ⊂ X . Reciprocally, since all Cantor sets are closed and un- First, let us consider the following representation of the set of countable, the theorem is proved. Liouville numbers, Acknowledgments L = U n , n ∈ N The authors would like to thank the anonymous referees for where their valuable comments and suggestions. + ∞ p 1 p p p 1 References U n = − , , + q q n q q q q n q =2 p∈ Z [1] J. Liouville , Sur des classes très étendues de quantités dont la valeur n’est ni is an open and dense set of R , for all n ∈ N . Then, algébrique, ni même réductible àdes irrationnelles algébriques, C. R. Acad. Sci. Paris 18 (1844) 883–885 . c [2] J. Liouville , Nouvelle démonstration d’un théorème sur les irrationnelles al-
D = R \ L = L c = = c = , gébriques inséré dans le compte rendu de la dernière séance, C. R. Acad. Sci. Un Un Dn (3.1) Paris 18 (1844) 910–911 . n ∈ N n ∈ N n ∈ N [3] J. Liouville , Sur des classes très étendues de quantités dont la valeur n’est ni = c ∈ N algébrique, ni même réductible àdes irrationnelles algébriques, J. Math. Pures where Dn Un is a closed and nowhere dense set, for all n . We Appl. 16 (1851) 133–142 . come now to the main result of this section. [4] A.S. Kechris , Classical descriptive set theory, in: Graduate Texts in Mathematics, 156, Springer Verlag, 1995 . ⊂ D Theorem 3.1. There is a set C such that C is a Cantor set. [5] P.S. Alexandroff, Sur la puissance des ensembles mesurables, C. R. Acad. Sci. Paris 162 (1916) 323–325 . λ(L ) = , D Proof. Since 0 is an uncountable set. Using (3.1), we see [6] B. Álvarez-Samaniego , A. Merino , A primitive associated to the Cantor–Bendix- ∈ N that there is a k such that Dk is an uncountable set. Let C be son derivative on the real line, J. Math. Sci. Adv. Appl. 41 (2016) 1–33. [7] G. Cantor , Sur divers théorèmes de la théorie des ensembles de points situés the set of all condensation points of Dk . By Bendixson’s Theorem, dans un espace continu à n dimensions, Acta Math. 2 (1883) 409–414 . = ( \ ) , ࢨ Dk Dk C C where Dk C is countable and C is a perfect set. R We see that C is an uncountable subset of . Also, since Dk is a nowhere dense set, we get that C is also nowhere dense. Hence, C is a Cantor set contained in D .