We have already proved Rolle's Theorem without using boundedness or extreme values of continuous functions, but nonetheleess we can prove that continuous functions are bounded. As is always the case, it is easier here to intuit and reason about discontinuity than continuity, so we aim for the contrapositive, that unbounded functions are discontinuous.
Define a boundedness discontinuity on a function f
to be
an interval (a, c)
with the property that one of
(a, b]
and [b, c)
always has a bounded image under
f
, and the other always has an unbounded image. Without loss of
generality, suppose that it is the [b, c)
intervals that are
unbounded. Then if a function f
is defined on a larger interval
(a, d)
, we can derive a cut discontinuity at the endpoint
c
of any such boundedness discontinuity.
Construct a sequence l(n)
so that
f(l(n+1)) > f(l(n)) + 1
, then define S
to be the set
of all points in [c, d)
whose images have a distance of at least
1/3
from any f(l(n))
. If S
is incident
with l(n)
then we have a discontinuity. Otherwise pick some
decreasing sequence r(n)
so that f(r(n))
is within
distance 1/3
from the smallest possible corresponding
f(l(n'))
. If infinitely many of l(n)
have no
corresponding r(n)
then that subsequence will be incident with
r(n)
, despite having positive distance, giving a continuity.
Otherwise we can take the subsequence that do have corresponding
elements in r(n)
, and pick alternatingly between these two
sequences, l'(n) = l(2n)
and r'(n) = r(2n + 1)
. These
sequences will be incident, with non-incident images, once again giving a
discontinuity.