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 =
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 =
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
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
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
of B in R.
Solution: We have
A \ B = [−2,0) (1,2] = {x R : −2 x 0 or 1 x 2},
= (−∞,0) (1, ∞) = {x R : x 0 or 1 x}.
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