4 1. The Real Numbers

Example 1.1.3. If A is the collection of all intervals of the form [s,2] where

0 s 1, find A and A.

Solution: A number x is in the set

A =

s∈(0,1)

[s,2]

if and only if

(1.1.1) s ≤ x ≤ 2 for every positive s 1.

Clearly every x in the interval [1,2] satisfies this condition. We will show that no

points outside this interval satisfy (1.1.1).

Certainly an x 2 does not satisfy (1.1.1). If x 1, then s = x/2 + 1/2 (the

midpoint between x and 1) is a number less than 1 but greater than x, and so such

an x also fails to satisfy (1.1.1). This proves that

A = [1,2].

A number x is in the set

A =

s∈(0,1)

[s,2]

if and only if

(1.1.2) s ≤ x ≤ 2 for some positive s 1.

Every such x is in the interval (0,2]. Conversely, we will show that every x in this

interval satisfies (1.1.2). In fact, if x ∈ [1,2], then x satisfies (1.1.2) for every s 1.

If x ∈ (0,1), then x satisfies (1.1.2) for s = x/2. This proves that

A = (0,2].

If B ⊂ A, then the set of all elements of A which are not elements of B is called

the complement of B in A. This is denoted A \ B. Thus,

A \ B = {x ∈ A : x / ∈ B}.

Here, of course, the notation x / ∈ B is shorthand for the statement “x is not an

element of B”.

If all the sets in a given discussion are understood to be subsets of a given

universal set X, then we may use the notation

Bc

for X \ B and call it simply the

complement of B. This will often be the case in this text, with the universal set

being the set R of real numbers or, in later chapters, real n-dimensional space

Rn

for some n.

Example 1.1.4. If A is the interval [−2,2] and B is the interval [0,1], describe

A \ B and the complement

Bc

of B in R.

Solution: We have

A \ B = [−2,0) ∪ (1,2] = {x ∈ R : −2 ≤ x 0 or 1 x ≤ 2},

while

Bc

= (−∞,0) ∪ (1, ∞) = {x ∈ R : x 0 or 1 x}.