Page 21 - Schaum's Outline of Theory and Problems of Advanced Calculus

12 NUMBERS [CHAP. 1

A or B a 1 b 1 a 2 b 2 a 3 b 3 .. .
l l l l l l
Natural numbers 1 2 3 4 5 6 .. .
Case 2:Ifsome 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 þ 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.

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
B
which are not in B,wecan also write A B ¼ A B. 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 ð0; 1Þ, i.e., there is a one-to-one correspondence between all rational numbers > 1
and all rational numbers in ð0; 1Þ.Since these last are countable by Problem 1.17, it follows that the set of all
rational numbers > 1isalso 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 ½0; 1Š is non-countable.
Every real number in ½0; 1Š has a decimal expansion :a 1 a 2 a 3 ... where a 1 ; a 2 ; ... 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 ½0; 1Š are countable we can place them in 1-1 correspondence with the natural
numbers as in the following list:
1 $ 0:a 11 a 12 a 13 a 14 ...
2 $ 0:a 21 a 22 a 23 a 24 ...
3 $ 0:a 31 a 32 a 33 a 34 ...
. . .
. . .
We now form a number
0:b 1 b 2 b 3 b 4 .. .
where b 1 6¼ a 11 ; b 2 6¼ a 22 ; b 3 6¼ a 33 ; b 4 6¼ a 44 ; .. . and where all b’s beyond some position are not all 9’s.
This number, which is in ½0; 1Š is different from all numbers in the above list and is thus not in the list,
contradicting the assumption that all numbers in ½0; 1Š were included.
Because of this contradiction it follows that the real numbers in ½0; 1Š cannot be placed in 1-1 corre-
spondence with the natural numbers, i.e., the set of real numbers in ½0; 1Š is non-countable.

LIMIT POINTS, BOUNDS, BOLZANO–WEIERSTRASS THEOREM

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1.21. (a)Prove that the infinite sets of numbers 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, 1isa lower bound.
1
3
We can find smaller upper bounds (e.g., ) and larger lower bounds (e.g., ).
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