Boundedness

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 fto 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.