Assume, f has no upper bound on [a,b], i.e., for every
n ? N there exists an bn ? [a, b] such that f(bn) > n. We construct
a sequence (bn), which has by Bolzano Weierstrass a convergent
subsequence (an) = (bkn
). Since f is continuous (an) ? a implies
f(an) ? f(a). But f(an) > kn = n, so there is no real number that
is approximated by f(an). Contradiction. The same argument
shows that also a lower bound necessarily exists.
Attachments:
math-camp.pdf
