1. Section 3.2: Sequences and Summations

2. Def: A sequence is a function from a subset of the set of integers (usually the set of natural numbers) to a set S. We use the notation a k to denote the image of the integer k. Ex: Consider the sequence 1, 2, 3, 4, 5, 6, … We could specify this sequence as {a k } where a k = k + 1 and the sequence is indexed with the set of natural numbers. That is, a 0 = 1, a 1 = 2, a 2 = 3, … Note the overloaded use of {}; they do not indicate a set here but a sequence. We will be careful to indicate that we are dealing with a sequence when specifying them in this manner to avoid confusion with sets. Ex: The digits in a real number comprise a sequence. Consider .  = 3.14159265… We can consider the digits after the decimal point as a sequence indexed by the set of natural numbers {p k }. So p 0 = 1, p 1 = 4, p 2 = 1, p 3 = 5, p 4 = 9, p 5 = 2, and so on. Any real number can be considered in this manner [before the decimal must be an integer, after the decimal though is a sequence of digits].

3. Ex: 376.1111… is a real number. To the left of the decimal we have an integer, 376, but to the right we have an infinite sequence of digits. This sequence is specified as {a k } where a k = 1 for all k  N. Ex: 1/2 = 0.5 is a real number. To the left of the decimal we have an integer, 0, and to the right we have a sequence of digits. This sequence only has one digit so we could index the sequence by the subset of the integers {0}. And then a 0 = 5. However we can consider a real number to always have a sequence of digits indexed by the set of natural numbers at its end if we wish. If the decimal terminates, we simply extend it with 0s. That is 1/2 = 0.5000… and our sequence of digits is now {a k } where a 0 = 5 and a k = 0 for all k > 0. Ex: Consider the sequence {a k } where a k = 1/(k + 1) for all k  N. Then the terms of this sequence are 1, 1/2, 1/3, 1/4, 1/5, … Note that this is a sequence of rational numbers. We have dealt with numbers up to this point in our examples but remember that the set S from which our elements are chosen is arbitrary.

4. Def: A geometric progression is a sequence of the general form a, ar, ar 2, ar 3, … where the initial term a and the common ratio r are real numbers. Ex: The sequence {b k } with b k = (-1) k indexed by the set of natural numbers is a geometric progression with initial term 1 and common ratio -1. We can see that b 0 = (-1) 0 = 1, b 1 = (-1) 1 = -1, b 2 = (-1) 2 = 1, and so on. So the terms of this sequence are 1, -1, 1, -1,... Ex: The sequence {c k } with c k = 2(5) k indexed by the set of natural numbers is a geometric progression with initial term 2 and common ratio 5. We can see that c 0 = 2(5) 0 = 2, c 1 = 2(5) 1 = 10, c 2 = 2(5) 2 = 50, and so on. So the terms of this sequence are 2, 10, 50, 250,...

5. Def: An arithmetic progression is a sequence of the general form a, a + d, a + 2d, a + 3d, … where the initial term a and the common difference d are real numbers. Ex: The sequence {b k } with b k = 1 + 4k indexed by the set of natural numbers is an arithmetic progression with initial term 1 and common difference 4. We can see that b 0 = 1 + 4*0 = 1, b 1 = 1 + 4*1 = 5, b 2 = 1 + 4*2 = 9, and so on. So the terms of this sequence are 1, 5, 9, 13,... Ex: The sequence {c k } with c k = -1 + (1/2)k indexed by the set of natural numbers is an arithmetic progression with initial term -1 and common difference 1/2. We can see that c 0 = -1 + (1/2)0 = -1, c 1 = -1 + (1/2)1 = -1/2, c 2 = -1 + (1/2)2 = 0, and so on. So the terms of this sequence are -1, -1/2, 0, 1/2, 1,... Def: A string is a sequence which is indexed by a finite subset of the integers. The length of the string is the cardinality of the indexing set. Ex: 101011 is a string of bits (a.k.a. bit string) of length 6.

6. Summations Sometimes when we have a sequence of numbers {a k } we would like to sum the terms of the sequence. That is, we would like to find the sum a 0 + a 1 + a 2 + … We have a special notation for this sum:  a k = a 0 + a 1 + a 2 + … We also would like to specify the sum of the terms of a sequence in a particular range:  n j = m a j = a m + a m+1 + …+ a n. [We sum all a j where m  j  n] In the above notation j is called the index of summation, m is called the lower limit and n is called the upper limit. Ex:  99 j = 0 1/(j + 1) represents the sum of the first 100 terms of our example sequence {a k } where a k = 1/(k + 1).

7. Ex:  5 j = 1 j 2 = 1 2 + 2 2 + 3 2 + 4 2 + 5 2 = 1 + 4 + 9 + 16 + 25 = 55 Identity: Let {a j } be a sequence of real numbers. Then  n j = m a j +  n j = m b j =  n j = m (a j + b j ) To take advantage of the above identity, we often need to shift the index in a summation: Ex:  5 j = 1 j 2 +  4 k = 0 2k =  5 j = 1 j 2 +  5 r = 1 2(r – 1) =  5 j = 1 j 2 + 2(j – 1) =  5 j = 1 (j 2 + 2j – 2) = (1 2 + 2*1 – 2) + (2 2 + 2*2 – 2) + (3 2 + 2*3 – 2) + (4 2 + 2*4 – 2) + (5 2 + 2*5 – 2) = 1 + 6 + 13 + 22 + 33 = 75 Identity: Let a and r be real numbers with r  0. Then  n j = 0 ar j = (ar n+1 – a)/(r – 1) if r  1. [If r = 1, then (n + 1)a is the sum.] There is a table of useful summation formulas (including the above) in the text on page 232. [See also the discussion of how to evaluate nested sums].

8. Cardinality Recall that the cardinality of a set S was defined to be the number of elements in the set S. Two finite sets have the same cardinality if they have the same number of elements. The following definition extends the concept of cardinality. Def: The sets A and B have the same cardinality if and only if there is a one-to-one correspondence from A to B. Remark: Recall that we pointed out the “if” part of this definition for finite sets when we studied functions. That is, we remarked that if the sets A and B have finite cardinality, then a function from A to B can be one-to-one only if |A|  |B|. And a function from A to B can only be onto if |A|  |B|. Taken together, we realized that for finite sets A and B, a function from A to B could be a one-to-one correspondence only if |A| = |B|.

9. So we just verified that the given definition doesn’t contradict our earlier treatment of finite sets. This may not seem like the most natural definition for cardinality at first, but upon further examination it turns out to be very natural. [What do we do when we count?]. When we count the elements of a finite set, we are forming a one-to- one correspondence with the subset {1, 2, …, n} of the set of integers. Once we have formed this one-to-one correspondence, we conclude that the set we counted has cardinality n. Using this notion of cardinality, we can begin to compare the cardinality of infinite sets as well. Some infinite sets seem to be bigger than other infinite sets, but it turns out that they have the same cardinality. Ex: We know that Z +  N. That is N has every positive integer in it, but in addition it has 0 which Z + does not. So it seems that N is ‘bigger’ than Z +. But consider f: N  Z + defined by f(n) = n + 1. This function f is a one-to-one correspondence between N and Z +. So |N| = |Z + |.

10. Def: A set that has finite cardinality or has the same cardinality as the positive integers is called countable. A set that is not countable is called uncountable. Ex: We just saw that the set of natural numbers is countable. Remark: An infinite set S is countable if and only if it is possible to list all of the elements of S in a sequence (indexed by Z + ). Remember that a sequence of elements from S indexed by Z + is a function f: Z +  S. The requirement that it is possible to list all of the elements of S in a sequence means that there exists such a function f from Z + to S which is onto. And if we can have such an onto function, we can make sure it is also one-to-one by keeping only the first occurrence of each element of S in the sequence. Ex: The set of integers is countable because we can list all of the integers in a sequence: 0, -1, 1, -2, 2, -3, 3, -4, … The bijection is f(x) = (x – 1)/2 for odd x, and f(x) = -x/2 for even x.

11. Ex: The set Z  Z is countable. We use the technique of exhaustively listing the members of this set in a sequence. Once again, the order of listing is important. [Graph] (0, 0), (1, 0), (1, 1), (0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1), (2, -1), (2, 0), (2, 1), (2, 2), (1, 2), (0, 2), (-1, 2), (-2, 2), (-2, 1), (-2, 0), … This is not the only possibility. For another example, we could have organized our listing by distance from the origin: 0: (0, 0) 1: (1, 0), (0, 1), (-1, 0), (0, -1) 2: (2, 0), (1, 1), (0, 2), (-1, 1), (-2, 0), (-1, -1), (0, -2), (1, -1) … Theorem: A subset of a countable set is countable. Corollary: The set of rational numbers is countable.

12. Test 2 In class, Wednesday, April 2 –Send me email to arrange to take the test on an earlier date if you can not take it on April 2 Sections 1.8, 2.4, 3.1, 3.2, 3.3, and 3.4 HW5, HW6 (due 3-24), HW7(will be due 3-31)

13. Theorem: The set of real numbers is uncountable. Remark: Up to this point we have only proven sets to be countable. To prove that an infinite set is countable, we only had to find a bijection between it and the set of positive integers. This sort of thing is an existence proof. However, to prove that a set is uncountable, we must show that there does not exist any bijection between it and the set of positive integers. This is significantly harder. Before we proceed with the proof, recall what a real number is. All real numbers take the form x.d 1 d 2 d 3 … where x is an integer (hence finite) and the thing to the right of the decimal point is a sequence of digits indexed by the set of positive integers. Remember that even real numbers with a decimal that terminates (like 5.725) can still be thought of in this manner. We just tack on 0’s at the end (5.725000…).

14. Proof: Assume, to the contrary, that the set of real numbers is countable. Then the subset of the real numbers B = {x | 0 < x < 1} is also countable (since a subset of a countable set is countable). So there exists a bijection from the set of positive integers to B (call it f). We don’t know anything about f other than it is a bijection from Z + to B. f(1) = 0.d 11 d 12 d 13 …, f(2) = 0.d 21 d 22 d 33 …, f(3) = 0.d 31 d 32 d 33 …, etc. Since f is a bijection (and hence onto) then every real number in B must appear somewhere on this list (as f(n) for some positive integer n). Consider the member 0.d 1 d 2 d 3 … of B where each digit d i = 4 if d ii  4 and d i = 5 if d ii = 4. This member of B can not appear as f(n) for any positive integer n since the nth digit (d n ) of this number differs from the nth digit (d nn ) of f(n). Hence this is a member of B which is not mapped to by any positive integer n. So f is not onto. So R is uncountable. 

15. What we just did was to show that B must be uncountable because any function from Z + to B that you propose to me as a bijection can’t be onto. We did this by taking the proposed bijection and constructing a number in B that is not mapped to by any integer. So no such bijection exists because any one that is proposed can be shown to not be onto. The technique used in this proof is called diagonalization (we went down the diagonal to assign our digits for the number we created). This technique can be used to prove that many other sets are uncountable. Once you understand this diagonalization argument from the previous proof, consider the following question: We know that the set of natural numbers is countable because we found a bijection between it and the positive integers. Hence we shouldn’t be able to prove that the set of natural numbers is uncountable (because it isn’t). We could tack on zeros to the front of any natural number leaving its value unchanged, just as we did at the end of real numbers. Why then can we not use this diagonalization technique to show that the set of natural numbers is uncountable?