1549901369-Elements_of_Real_Analysis__Denlinger_

250 Chapter 5 • Continuous Functions

Case 1 (a< b): Since I is an interval, [a, b] ~I. Let

A= {x E [a,b]: f(x) < w}.

Then A is nonempty, since a E A. Also, A is bounded above, since A~ [a, b].
Thus, by the completeness property of JR.,

3c =sup A.

To prove: f(c) = w.

y

v --------------

a c b x

I

Figure 5.7

By Theorem 2.3.5, since c = sup A, 3 sequence {an} of points of A such
that

(2)

Note that c E [a, b] since A ~ [a, b]. Since f : I -> JR. is continuous and
c E [a, b] ~ I, f must be continuous at c. Thus, by the sequential criterion for
continuity off at c applied to (2),

f(an) -> f(c). (3)

Now \:/n, an E A, so f(an) < w. Thus, since limits preserve inequalities

(Theorem 2.3.12), lim f(an)::; w. That is, by (3),

n-+oo

I f(c)::; w. I (4)

We shall now complete the proof by proving that f(c) 2:: w. First, note that
c =f. b, since f(c) ::; w < v = f(b). But, c E [a, b]. Thus, c < b.