Real Analysis 175
As we mentioned in chapter 7, the quantifier symbol ∀ here is read as “for all” and the
symbol ∃ as “there exists,” and so this symbolization asserts that every epsilon has a delta
as desired. We can illustrate how these quantities relate in the following figure:
c
f (c)
±δ
|x − c| <δ=⇒|f (x) − f (c)| <
To explain it differently, a function f is continuous at a point c if, for any desired degree
of accuracy, there is a degree δ of closeness to c, such that any x that is that close to c
will have f (x) within the desired accuracy of f (c). In short, you can ensure that f (x)isas
close as you want to f (c) by insisting that x is sufficiently close to c.
15.2 Sums and products of continuous functions
Let us get a little practice with the definition by proving that the sum of continuous func-
tions is continuous. In the proof, we will use the triangle inequality, which asserts that
|u + v|≤|u| + |v| for any real numbers u and v. To prove this, observe that when u and
v have the same sign, we have equality, and when they have opposite signs, then we get
cancellation on the left but not on the right.
Theorem 132. The sum of two continuous functions is continuous. In other words, if f
and g are real-valued continuous functions on the real numbers, then so is the function
f + g, defined by addition
( f + g)(x) = f (x) + g(x).
Proof. Suppose that f and g are both continuous at a point c, and consider the function
f + g, whose value at c is f (c) + g(c). To see that this function is continuous at c,fixany
>0. Thus, also /2 > 0. Since f is continuous at c, there is δ
1
> 0 such that any x with
|x −c| <δ
1
has |f (x) − f (c)| </2. And similarly, since g is continuous at c, there is δ
2
> 0
with |x − c| <δ
2
implying that |g(x) − g(c)| </2. Let δ be the minimum of δ
1
and δ
2
.So
if x is within δ of c, then it is both within δ
1
of c and also within δ
2
of c. Consequently, we
know that |f (x) − f (c)| </2 and also that |g(x) − g(c)| </2.