Question

state and prove boundedness theorem

Answer 1

Boundedness Theorem. Let a and b be real numbers with a<b, and let f be a continuous, real valued function on [a,b]. Then f is bounded above and below on [a,b].

Proof. Suppose not. Then for all natural numbers n we can find some xn∈[a,b] such that |f⁢(xn)|>n. The sequence (xn) is bounded, so by the Bolzano-Weierstrass theorem it has a convergent sub sequence, say (xni). As [a,b] is closed (xni) converges to a value in [a,b]. By the continuity of f we should have that f⁢(xni) converges, but by construction it diverges. This contradiction finishes the proof.