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Then ax b, by a and u xy. Since ax y axy by a we have axy a1 or xy 1 the required value.
1.16. If M > 0, N > 0; and a > 0 but a 6 1, prove that loga
Let loga M x, loga N y.
M loga M loga N. N
Then ax M, ay N and so or loga M x y loga M loga N N
M ax y ax y a N
COUNTABILITY 1.17. Prove that the set of all rational numbers between 0 and 1 inclusive is countable.
Write all fractions with denominator 2, then 3; . . . considering equivalent fractions such as 1 ; 2 ; 3 ; . . . no 2 4 6 more than once. Then the 1-1 correspondence with the natural numbers can be accomplished as follows: Rational numbers Natural numbers 0 l 1 1 1 1 2 3 l l l 2 3 4 l l l 5 6 7
2 3 1 4 3 4 1 2 ... 5 5 l l 8 9 ...
Thus the set of all rational numbers between 0 and 1 inclusive is countable and has cardinal number Fo (see Page 5).
1.18. If A and B are two countable sets, prove that the set consisting of all elements from A or B (or both) is also countable.
Since A is countable, there is a 1-1 correspondence between elements of A and the natural numbers so that we can denote these elements by a1 ; a2 ; a3 ; . . . . Similarly, we can denote the elements of B by b1 ; b2 ; b3 ; . . . . Case 1: Suppose elements of A are all distinct from elements of B. Then the set consisting of elements from A or B is countable, since we can establish the following 1-1 correspondence.
A or B
NUMBERS
[CHAP. 1
Natural numbers
a1 l 1
b1 l 2
a2 l 3
b2 l 4
a3 l 5
b3 l 6
... ...
Case 2: If some elements of A and B are the same, we count them only once as in Problem 1.17. Then the set of elements belonging to A or B (or both) is countable. The set consisting of all elements which belong to A or B (or both) is often called the union of A and B, denoted by A [ B or A B. The set consisting of all elements which are contained in both A and B is called the intersection of A and B, denoted by A \ B or AB. If A and B are countable, so is A \ B. " The set consisting of all elements in A but not in B is written A B. If we let B be the set of elements " which are not in B, we can also write A B AB. If A and B are countable, so is A B.
1.19. Prove that the set of all positive rational numbers is countable.
Consider all rational numbers x > 1. With each such rational number we can associate one and only one rational number 1=x in 0; 1 , i.e., there is a one-to-one correspondence between all rational numbers > 1 and all rational numbers in 0; 1 . Since these last are countable by Problem 1.17, it follows that the set of all rational numbers > 1 is also countable. From Problem 1.18 it then follows that the set consisting of all positive rational numbers is countable, since this is composed of the two countable sets of rationals between 0 and 1 and those greater than or equal to 1. From this we can show that the set of all rational numbers is countable (see Problem 1.59).
1.20. Prove that the set of all real numbers in 0; 1 is non-countable.
Every real number in 0; 1 has a decimal expansion :a1 a2 a3 . . . where a1 ; a2 ; . . . are any of the digits 0; 1; 2; . . . ; 9. We assume that numbers whose decimal expansions terminate such as 0.7324 are written 0:73240000 . . . and that this is the same as 0:73239999 . . . . If all real numbers in 0; 1 are countable we can place them in 1-1 correspondence with the natural numbers as in the following list: 1 2 3 . . . We now form a number 0:b1 b2 b3 b4 . . . where b1 6 a11 ; b2 6 a22 ; b3 6 a33 ; b4 6 a44 ; . . . and where all b s beyond some position are not all 9 s. This number, which is in 0; 1 is di erent from all numbers in the above list and is thus not in the list, contradicting the assumption that all numbers in 0; 1 were included. Because of this contradiction it follows that the real numbers in 0; 1 cannot be placed in 1-1 correspondence with the natural numbers, i.e., the set of real numbers in 0; 1 is non-countable. $ $ $ 0:a11 a12 a13 a14 . . . 0:a21 a22 a23 a24 . . . 0:a31 a32 a33 a34 . . . . . .
LIMIT POINTS, BOUNDS, BOLZANO WEIERSTRASS THEOREM 1.21. (a) Prove that the in nite sets of numbers 1; 1 ; 1 ; 1 ; . . . is bounded. (b) Determine the least 2 3 4 upper bound (l.u.b.) and greatest lower bound (g.l.b.) of the set. (c) Prove that 0 is a limit point of the set. (d) Is the set a closed set (e) How does this set illustrate the Bolzano Weierstrass theorem
(a) Since all members of the set are less than 2 and greater than 1 (for example), the set is bounded; 2 is an upper bound, 1 is a lower bound. We can nd smaller upper bounds (e.g., 3) and larger lower bounds (e.g., 1). 2 2
CHAP. 1]
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