The Construction of Cantor Sets

By Ng Tze Beng

Cantor sets play an important role in real analysis, particularly in furnishing counter examples and exotic or pathological functions. Most frequently, we meet Cantor set of zero measure but the construction is canonical enough to apply to give Cantor set of positive measure. These are subsets of the closed interval [0,1] and of measure greater or equal to 0 but less than 1.  For the use of Cantor sets see Composition and Riemann Integrability, Lebesgue Integration and Composition and Change of Variable or Subsitution in Riemann Integration.  

The first Cantor set we shall construct is the Cantor set of measure zero. This is the set left over after repeatedly deleting the middle third open intervals.

The Cantor set C₀

We shall start from the closed unit interval [0,1]. At the first stage, we delete the middle open interval (1/3, 2/3) from [0, 1]. We shall enumerate the open intervals to be deleted. We denote (1/3, 2/3) by I(1,1). That is I(1,1) = (1/3,2/3). The first construction gives us the remaining 2 closed intervals

[0, 1] - I(1,1) = [0,1] - (1/3,2/3) = [0, 1/3] È[2/3, 1].

Then at the second stage we delete the middle third open interval from each of the closed intervals. Thus there are 2 open intervals to be deleted and they are

I(2,1) = 1/3 I(1, 1) =(1/9, 2/9) and

I(2, 2) = I(2, 1) + 2/3 , the translation of I(2,1) by 2/3.

Hence we are left with 4 = 2² closed intervals, two in [0, 1/3] and two in [2/3, 1]. By rescaling we see that the middle open third intervals of each of these 4 closed intervals are given by

I(3,1) = 1/3 I(2,1), I(3,2) =1/3 I(2,2) and

I(3, 3) = I(3,1) + 2/3, I(3, 4) = I(3,2) +2/3.

Repeating this construction, at the k-th stage, there are 2^k-1 open intervals, each of length 1/3^k, to be deleted from [0, 1] and for k ³ 3, they are given by

I(k, j) = 1/3 I(k-1, j), j = 1, 2, ¼, 2^k-2 ;

I(k, 2^k-2+ j) = I(k, j) + 2/3, j = 1, 2, ¼, 2^k-2.

Let G be the union of these countable collection of disjoint open intervals,

.

Thus G is open. The Cantor set C₀ is defined to be the complement of G in [0, 1]. That is C₀ = [0, 1] - G and is therefore a closed subset of [0, 1]. Note that at the k-th stage the total length of the intervals to be deleted is 2^k-1 (1/3^k). Therefore, the measure of G, since the intervals in G are all disjoint, is given by

Thus the measure of C₀ is equal to m([0, 1]) - m(G) = 1-1 = 0.

In order to define the Cantor function, we shall consider another representation of the Cantor set C₀ . This will also show that C₀ has the same cardinality as [0, 1] and so it is uncountable. We shall make use of the ternary representation of numbers in [0, 1]. There is a unique way of representing real numbers in (0, 1] by non-terminating ternary expansion.

We shall redefine these real numbers as the supremum of a sequence of rational numbers. Take any 1 ³ x > 0 in [0,1]. Choose the greatest integer a₁ such that a₁/3 < x £. a₁/3 + 1/3. Consider the rational numbers 0/3, 1/3 , 2/3. Then take the largest of these which is < x. That is we choose a₁ in {0,1,2} such that < x £ + . Thus a₁ = 0 Û 0/3 < x £ 1/3, a₁ = 1 Û 1/3 < x £ 2/3 and a₁ = 2 Û 2/3 < x £ 3/3=1. Let d₁ = 0 + . Then d₁ < x £ d₁ + . Now choose integer a₂ in {0,1,2} such that d₁ + < x £ d₁ + . Let now d₂ = d₁ + = 0 + + . We have d₂ < x £ d₂ + . Continuing like this we shall obtain a sequence a₀ , a₁ , a₂ , ¼, of integers, 0 £ a_i £ 2, such that if d_n = d_n-1 + = 0+ + + ¼+ then d_n < x £ d_n + . The symbol 0 × a₁ a₂ a₃ ¼ is called the non-terminating ternary expansion of x . So the set {d₁ , d₂ , d₃ , d₄ ¼ } is bounded above by x. Then the least upper bound or supremum of this set is x. We deduce this as follows. If x ¹ sup{d₁ , d₂ , d₃ , d₄ ¼ } = M, then x > M . Then by the Archimedean property of the set of real numbers, there exists a counting number m such that < x - M and so M + < x. Now take a non-terminating ternary expansion 0 × c₁ c₂ c₃ ¼ for . Let L be the first integer such that c_L ¹ 0. Then . Therefore, M + < M + < x. Hence since M = sup{d₁ , d₂ , d₃ , d₄ ¼ }, d_L + £ M + < x and this contradicts x £ d_L + . Hence x = sup{d₁ , d₂ , d₃ , d₄ ¼ }.

This representation of the real number x in (0, 1] is unique. Suppose two non-terminating ternary expressions 0× a₁ a₂ a₃ ¼ and 0× b₁ b₂ b₃ ¼ are such that for some integer j ³ 0, a_j ¹ b_j and a_i = b_i for i £ j - 1. If 0× a₁ a₂ a₃ ¼ represents x and 0× b₁ b₂ b₃ ¼ represents y we shall show that x ¹ y. Suppose that a_j < b_j . By the hypothesis we have

0 + + + ¼+ + < x £ 0 + + + ¼+ + .

But 0 + + + ¼+ += 0 + + + ¼+ +

£ 0 + + + ¼+ + < y.

Therefore, x < y and so x ¹ y. Similarly if a_j > b_j, we can show that x ¹ y. Thus this way of representing any real number in (0, 1] is unique. Let 0 be represented by the terminating ternary expansion 0.00. Notice that for x = , the corresponding non-terminating expansion is the expression 0.022222¼¼ which is a non-terminating expression with recurring '2'. is represented by 0.122222....... a non-terminating expression with recurring '2'. 1 is represented by 0.22222........ a non-terminating expression with recurring '2'. We are going to change our representation a little. We are going to use some terminating expansion in our representation as follows. If the non-terminating expansion of x consists of exactly one '1' in its expansion, say 0× a₁ a₂ a₃ .... has precisely a_j = 1 and that a_k = 2 for k > j , then we replace it by the terminating expansion 0× b₁ b₂ b₃ ... b_j -1 2 with b_k = a_k for k £ j-1, b_j = 2 and b_k = 0 for k > j, hence this representation has no '1' in it. Thus we are going to use the following convention: if a number x in [0, 1] has two ternary expansion, one with no 1's and one with at least one '1', then it is the one that has no 1's that is to be used. We refer to this as our system of expansion. We are going to use this representation to describe the set G in [0, 1].

x Î I(1,1) = (1/3, 2/3) Û the non terminating expansion of x has a₁ = 1 and a₂ and subsequent a_j are not all equal to 2 Û the system of expansion used for x has a₁ = 1 (because the terminating expansion excluded here has no '1', namely 2/3 which has terminating expansion 0.2 and recurring expansion 0.12 , where the underscore denotes repeating infinitely many times the number underscored. )

Now dividing by 3 has the effect of shifting the expansion by one place to the right and introducing a zero, i.e. it has the effect of moving the ternary point to the left. Thus

x Î I(2,1) = 1/3 I(1,1) = (1/9, 2/9)

Û the system of expansion used for x has a₁ = 0 and a₂ = 1

Therefore, x Î I(2,2) = I(2,1) + 2/3

Û the system of expansion used for x has a₁ = 2 and a₂ = 1.

Hence x Î I(2,1)È I(2,2) Û the system of expansion used for x has the first '1' occurring in the second ternary place.

Then since I(3,1) = 1/3 I(2,1) and I(3,2) =1/3 I(2,2), x Î I(3,1)È I(3,2) Û the system of expansion used for x has a₁ = 0 and has the first '1' occurring in the third ternary place. Since I(3, 3) = I(3,1) + 2/3 and I(3, 4) = I(3,2) +2/3, x Î I(3, 3)È I(3, 4) Û the system of expansion used for x has a₁ = 2 and has the first '1' occurring in the third ternary place. Therefore, x Î I(3,1)È I(3,2)ÈI(3, 3)È I(3, 4) Û the system of expansion used for x has the first '1' occurring in the third ternary place. Thus by induction we see that Û the system of expansion used for x has the first '1' occurring in the k-th ternary place. Therefore, Û the system of expansion used for x has at least one '1'. Hence x Î C₀ = [0, 1] - G Û the system of expansion used for x has no 1's. We have thus proved the following:

Lemma 1. C₀ consists of those numbers in [0, 1] whose representation in the above system of ternary expansion has no 1's. Therefore, we can write for any x in C₀

, where b_k = 0 or 1.

Lemma 2. C₀ is uncountable.

Proof. Define g : C₀ ®[0, 1] by , where . Then since every real number y in [0, 1] has a non-terminating binary representation of the form or y = 0, we can take x in C₀ to be or 0 if y = 0 and so g(x) = y and thus g is surjective. Since [0, 1] is uncountable, C₀ is uncountable.

Next we would like to extend the function to all of [0, 1]. Let I(i, j) be denoted by the open interval (a(i, j), b(i, j)). Then {a(i, j), b(i, j)} Í C₀ . We shall show that g(a(i, j)) = g(b(i, j)). It is easily seen that in our system of ternary representation

• and

• , for some b_k = 0 or 1, k =1,2,¼i-1.

Hence . We define for x in I(i, j), g(x) = g(b(i, j)) = g(a(i, j)) . We shall next show that g is a non decreasing function, that is g is a monotonic increasing function (not necessarily strictly increasing).

Lemma 3. Represent the real numbers in [0, 1] by non-terminating ternary expansions except for 0. In this representation suppose and . Then

1. x = y Û a_i = b_i for all i ³1.

2. x < y Û a₁ < b₁ or there exists integer k ³ 2 such that a_i = b_i for all 1 £ i £ k -1 and a_k < b_k.

Proof. Part 1 follows from the uniqueness of the representation of the numbers either by the non terminating ternary expansion or by the system of representation used above.

Suppose a₁ < b₁. Then . Similarly, if there exists integer k ³ 2 such that a_i = b_i for all 1 £ i £ k -1 and a_k < b_k, then

• .

Now if x < y, then y - x > 0. Then by the Archimedean property of the real number system , there exists an integer k ³ 1 such that . Then

• .

Therefore, . Then a₁ £ b₁ . For if a₁ > b₁, then and so contradicting . Hence either a₁ < b₁ or a₁ = b₁. If a₁ = b₁, then Therefore, we conclude again that a₂ £ b₂. If a₂ < b₂, then we get the conclusion of the Lemma. Because

• , a_i = b_i cannot hold for all 1 £ i £ k. Thus for some j with 1 £ j £ k

a_i = b_i for 1 £ i < j and a_j < b_j . This completes the proof.

Next we shall see how our system of representation also gives the same conclusion. Observe that the convention used in our system of representation deviates from a non-terminating expansion only if the number is zero or has two possible ternary expansions with one of them involving no 1's. Thus if y has a terminating expansion either y = 0 or the terminating expansions has no 1's and ends in the number '2'. Thus if x < y , and y has a terminating ternary expansion,

• , where b_i' is equal to 0 or 2 and b_l' = 2.

Then we can write Thus if , x < y implies that there exists integer k ³ 1 such that a_i = c_i for all 1 £ i £ k -1 and a_k < c_k or a₁ < c₁, if k = 1 . If k £ l -1, then we have a_i = c_i = b_i' for all 1 £ i £ k -1 and a_k < b_k'. If k = l , then we have a_i = b_i' for all 1 £ i £ l -1 and a_l < c_l = 1< 2 = b_l'. If k > l , then we have a_i = b_i' for all 1 £ i £ l -1 and a_l = c_l = 1 < 2 = b_l'. Hence in any case we obtain that x < y implies that there exists integer l ³ k > 1 such that a_i = b_i' for all 1 £ i £ k -1 and a_k < b_k' or a₁ < b₁' .

Now if x has the terminating ternary expansion , we can write it as Hence by what we have just proved, there exists integer l ³ k > 1 such that d_i = b_i' for all 1 £ i £ k -1 and d_k < b_k' or d₁ < b₁' . If p =1, then d₁ = 1 and if d₁ < b₁' , b₁' =2 and l must be greater than 1, otherwise x = y. Thus if p =1, then a₁' = 2 =b₁' and there exists an integer l ³ k > 1 such that a_i = b_i' for all 1 £ i £ k -1 and a_k '< b_k' . We now assume that p > 1. If k < p, then a_i'= d_i = b_i' for all 1 £ i £ k -1 and a_k'= d_k < b_k'. If If k = p, then a_i'= d_i = b_i' for all 1 £ i £ k -1 and d_k = d_p = 1 < b_p'. Hence b_p' = 2 and so a_p'=2 = b_p'. Then it follows that l > p and that there exists q such that p < q £ l and a_i' = b_i' for all 1 £ i £ q -1 and a_q < b_q'. Now we shall see that k £ p. if k > p, then we have a_i'= d_i = b_i' for all 1 £ i £ k -1 and d_k < b_k'. Therefore, a_i'= d_i = b_i' for 1 £ i £ p -1, 1 = d_p = b_p' , hence contradicting that b_p' is even. Hence we have shown that the same conclusion for Lemma 3 is true also for our system of representation of real numbers in [0, 1] by ternary expansion. The converse of the statement is obvious.

We summarise what we have proved in the following lemma.

Lemma 4. Represent the real numbers in [0, 1] by the system of ternary expansion as described above. In this representation suppose and . Then

1. x = y Û a_i' = b_i' for all i ³1.

2. x < y Û a₁' < b₁' or there exists integer k ³ 2 such that a_i' = b_i' for all 1 £ i £ k -1 and a_k' < b_k'.

Now we are ready to investigate the monotonicity of the function g:[0, 1] ®[0, 1]. We have shown that g is a surjective map.

Proposition 5. The map defined previously g:[0, 1] ®[0, 1] is a bounded surjective monotonic increasing map. That is to say x < y Þ g(x) £ g(y).

Proof. We have already seen that g is surjective. It is obviously bounded. Now suppose x and y are in C₀ and that x < y. Then x and y have the representation:

• and , where the a_i's and b_i's are either 0 or 1. By Lemma 4 either

a₁ = 0 and b₁ =1 or there exists integer k ³ 2 such that a_i = b_i for all 1 £ i £ k -1 and a_k = 0 < b_k = 1. Note that and . Hence if a₁ = 0 and b₁ =1, then . If there exists integer k ³ 2 such that a_i = b_i for all 1 £ i £ k -1 and a_k = 0 < b_k = 1, then

• .

Hence if x < y and x and y are in C₀, then g(x) £ g(y). Suppose now that x < y, x Ï C₀ and y is in C₀. Then x Î I(i, j) = (a(i, j), b(i, j)) for some (i, j). Obviously x < y implies that a(i, j) < y and so since both a(i, j) and y are in C₀ by what we have just proved, g(a(i, j)) £ g(y). Therefore, by definition of g, g(x) = g(a(i, j)) £ g(y).

Similarly, if x < y, y Ï C₀ and x is in C₀, then y Î I(i, j) = (a(i, j), b(i, j)) for some (i, j) and x < b(i, j). Again since both x and b(i, j) are in C₀, we have g(x) £ g(b(i, j) ) = g(y).

Suppose now x < y, x, y Ï C₀. Then for some (i, j) and (i', j') , x Î I(i, j), y Î I(i', j'). Then a(i, j) < x < y < b(i', j'). Therefore, since both a(i, j) and b(i', j') are in C₀, g(a(i, j)) £ g(b(i', j') and so g(x) = g(a(i, j)) £ g(b(i', j') = g(y). Hence g is monotonically increasing. This completes the proof.

Next we shall state a general result concerning bounded monotone function whose range is an interval.

Theorem 6. Suppose f : [a, b] ®R is an increasing function. Then f  is continuous if and only if the range of f , J = f ([a, b]) is an interval.

Proof. If f is continuous, then by the Intermediate Value Theorem, the range f ([a, b]) is an interval. Now if f is increasing, then by Theorem 2 of Monotone Function, Function of Bounded Variation, Fundamental Theorem of Calculus, the discontinuity of f can only be jump discontinuity. Suppose the range J is an interval. Suppose f is discontinuous at x = k in (a, b). Then we have

• and .

Note that for any y < k, because f is increasing . Also for any z > k, . Therefore, ( f ( [a, k)È(k, b]))Ç( f (k ^- ), f (k ⁺ )) =Æ. But by assumption the range J is an interval and so ( f (k ^- ), f (k ⁺ )) Í J and so ( f ( [a, k)È(k, b]))Ç( f (k ^- ), f (k ⁺ )) = (J - { f (k)})Ç( f (k ^- ), f (k ⁺ ))¹ Æ. This contradicts

( f ( [a, k)È(k, b]))Ç( f (k ^- ), f (k ⁺ )) =Æ and so we have f (k ^- ) = f (k ⁺ ) = f (k) and so and that means f is continuous at x = k. We can similarly derive a contradiction when k = a or b. If f is not continuous at k = a, then f (a) < f (a⁺) and so ( f (a) , f (a⁺))ÇJ = Æ, contradicting ( f (a) , f (a⁺)) Í J. Thus f must be continuous at x = a. If f is not continuous at k = b, then f (b^-) < f (b) and so ( f (b^-) , f (b))ÇJ = Æ, contradicting ( f (b^-) , f (b)) Í J. Thus f must be continuous at x = b. Thus f cannot have any discontinuity and so it is continuous. This completes the proof.

Proposition 7. The function g:[0, 1] ®[0, 1] is continuous.

Proof. By Proposition 5, the map g is onto and increasing. Since the range [0, 1] is an interval, by Theorem 6, g is continuous.

Next we shall reveal some interesting facts concerning the Cantor set C₀ .

Theorem 8. The Cantor set C₀ is

(1) compact,

(2) no where dense, i.e., it contains no open intervals,

(3) its own boundary points,

(4) perfect, i.e.,it is its own set of accumulation point,

(5) totally disconnected and

(6) between any two points in C₀ , there is an open interval not contained in C₀.

Proof. (1) Since C₀ is closed and bounded, it is compact by the Heine-Borel Theorem.

(2) Let denote the union of the collection of disjoint open intervals deleted after the p-th stage in the construction. Suppose C₀ contains an open interval, then there is a point x in C₀ and a d > 0 such that (x -d , x + d)Í C₀. Hence (x -d , x + d)ÇG = (x -d , x + d)Ç([0, 1] - C₀ ) = Æ. Since (x -d , x + d) and G are disjoint and are contained in [0,1], m(G) + m((x -d , x + d))£ 1. Hence 1 + 2d £ 1 implies d £ 0, contradicting d > 0. Therefore, C₀ cannot contain any open interval and so is nowhere dense. (We can also use m((x -d , x + d)) £ m(C₀) = 0 and so 2d £ 0 contradicting d > 0.)

(3). C₀ is closed, by a characterization of closed set, because C₀ = [0, 1] - G = and each [0, 1] - G_p is closed. Hence the closure of C₀ is C₀ and so its set of cluster points C₀' is contained in C₀. Let x be in [0, 1]. Then for any open set J containing x, say J = U Ç [0, 1], where U is an open interval, J is non empty and contains more than one point and J Ç G ¹ Æ because J Ë [0, 1] - G = C₀ by part (2). This is because if J Í C₀ , then the interior of J, which is a non empty open interval is contained in C₀ contradicting part (2). Thus the closure of G, = [0, 1]. Therefore, the boundary of C₀, .

(4) Since C₀ is closed the set of accumulation points of C₀, C₀' is a subset of C_0. We claim that C₀' = C₀ . Let x Î C₀ . Suppose x Ï C₀' . Then there exists d > 0, such that ( x - d, x + d) Ç C₀ = {x}. Thus ( x - d, x + d)Ç([0, 1] - C₀) =( x - d, x) È (x, x + d). That means (x - d, x) Í ([0, 1] - C₀) = and so

(x - d, x) =

Since the collection is a collection of disjoint open intervals, and (x - d, x) is connected, (x - d, x) Í I(i, j) for some i and j. This is because otherwise (x - d, x) would be a disjoint union of open intervals contradicting that it is connected. Since x Ï I(i, j), x = sup I(i, j) = b(i, j). Similarly,

(x, x+d) =

We deduce as before then that for some p and q , (x, x + d, ) Í I(p, q). Thus because x Ï I(p, q), x = inf I(p, q) = a(p, q). Therefore, a(p, q) = x = b(i, j), contradicting that a(p, q) ¹ b(i, j) by virtue of the definition of I(i, j)'s . Therefore x Î C₀' . Thus C₀' = C₀ .

(5) Since by (2) C₀ does not contain any open interval and since the connected subsets of R are either the singleton sets or the intervals, the only components of C₀ are the singletons {x}, x Î C₀. Therefore, C₀ is totally disconnected.

(6) Suppose x < y and x , y are in C₀. Then (x, y)Ç([0, 1] - C₀) ¹ Æ and is a disjoint union of open intervals and so (x, y) contains at least one open interval.

The Cantor Set C_k

Now we turn our attention towards constructing a Cantor set of positive measure in [0, 1]. Indeed the construction also applies to give a different Cantor set of measure zero. The procedure of deleting the middle portion of each of the remaining closed sets is followed here but using a different specified length. (This idea can be generalised by not requiring the intervals to be deleted to be the middle portion, to give a generalised Cantor set.) Hence the construction is canonical. Let k be any real number with 0 £ k < 1. Let d = (1 - k) > 0. Start with the closed unit interval [0, 1]. Let I(1,1) be the middle open interval of length d/2. Then if we write I(1,1) = (a(1,1), b(1,1)), then a(1,1) = 1/2 - d/4. Then let G₁ = I(1,1) and F₁ = [0,1] - G₁. The first stage is to form F₁ . Then F₁ is the disjoint union of 2 closed intervals each of length equal to a(1,1) = 1/2 - d/4. Let I(2,1) and I(2,2) be respectively the middle open intervals, each of length d/2³, of the 2 closed intervals of F₁ . The open intervals are ordered from the left to the right by the second indices. Let G₂ = I(1,1)ÈI(2,1)ÈI(2,2). Suppose I(2,1) = (a(2,1), b(2,1)). Then a(2,1) = 1/2 a(1,1) - d/2⁴ = 1/2² -d(1/2² - 1/2⁴). The second stage is to form F₂ = [0,1] - G₂ = [0,1] - G₁ - I(2,1)ÈI(2,2). Then F₂ is the disjoint union of 2² closed intervals each of length equaling a(2,1) = 1/2² -d(1/2² - 1/2⁴). Then let I(3, j), j =1,..,2² be respectively the middle portion of length d/2⁵ of the 2² closed intervals again ordered by the second indices, from left to right in the sense that I(3, j) < I(3, k) if and only if j < k if and only if there exist x in I(3, j) and y in I(3, k) such that x < y. Stage three is to remove from F₂ these 2² open intervals, that is to form F₃ = F₂ - È{ I(3, j), j =1,..,2² } = [0, 1] - G₃ , where . Thus F₃ consists of 2³ disjoint closed intervals, each of length equaling a(3,1), where I(3,1) = (a(3,1), b(3,1)). It is easily seen that a(3,1) = 1/2³ -d(1/2³ - 1/2⁶). Let , where I(4, j), j =1,..,2³ are the middle open intervals each of length d/2⁷, of each of the disjoint closed intervals. Continuing in this way, at the n-th stage we have , where I(n, j) = (a(n, j), b(n, j)) is an open interval of length d/2²ⁿ-1 and the I(n, j)'s are ordered from left to right by the second indices according to the natural ordering of elements in the intervals. Note that a(n,1) = 1/2 a(n-1,1)-d(1/2²ⁿ) =1/2ⁿ -d(1/2ⁿ-1/2²ⁿ). We obtain F_n by deleting from F_n-1 the uion of open intervals H_n. Hence F_n = F_n-1- H_n = [0,1] - G_n , where and F_n is a disjoint union of 2ⁿ closed intervals each of length equaling a(n,1) = 1/2ⁿ -d(1/2ⁿ-1/2²ⁿ). To obtained F_n+1 , we shall delete from F_n open intervals each of length equaling d/2 ²ⁿ⁺¹ from the 2ⁿ closed intervals. That is if we let , where I(n+1, j), j =1,..,2ⁿ are these open intervals to be deleted, then F_n+1 = F_n - H_n+1 =[0, 1] - G_n+1 , where . Obviously F_n+1 is a collection of disjoint closed intervals, each of length equal to a(n+1,1). Note also that for each n ³ 1, F_n+1 Í F_n.

Let . Then G is a disjoint union of open intervals. The length or measure of H_k is given by . Thus the measure or length of G,

The Cantor set C_k is defined to be the complement of G in [0, 1]. That is, . Thus the measure of C_k is equal to m([0,1]) - m(G) = 1 - d = k.

Then C_k satisfies the properties stated in Theorem 8.

Theorem 9. The Cantor set C_k defined above for 0 £ k < 1 is uncountable and

(1) compact,

(2) no where dense, i.e., it contains no open intervals,

(3) its own boundary points,

(4) perfect, i.e., is its own set of accumulation point,

(5) totally disconneceted and

(6) between any two points in C_k , there is an open interval not contained in C_k.

Proof. Part (1) and part (3) to part(6) are proved in exactly the same way as in Theorem 8. Part (2) needs a different approach to prove. Suppose C_k contains an open interval say (c, d). Then implies that (c, d)Í F_k for each k ³ 1. Now since , there exists an integer N such that k > N implies that Now we fix a k > N. Note first that (c, d)ÇF_k =(c, d). Since the non-trivial interval (c, d) is connected and F_k is a disjoint union of closed intervals, (c, d) must be contained in one of these closed intervals, each of length . Hence, contradicting Therefore, C_k does not contain an open interval. (The main thrust of the argument is that the F_k consists of disjoint closed intervals, whose maximum length tends to zero as k tends to infinity.) This proves part (2). That C_k is uncountable is a consequence of the following proposition.

Proposition 10. There is a continuous strictly increasing bijective map f : [0, 1] ® [0,1] mapping the Cantor set C_k onto the the Cantor set C₀ defined earlier using the "deleted middle third intervals" construction.

Proof. This is given by Lemma 1 of Composition and Riemann Integrability. Note that the proof given there applies also to the case k = 0.

Since C₀ constructed using the "deleted middle third intervals" construction is uncountable and the function f given by Proposition 10 maps C_k bijectively onto C₀, C_k is also uncountable.

ã Ng Tze Beng 2001