Definition: Let’s call a function “locally bounded at x” if there exists a neighborhood of x, N(x;ε_x) and a number M_x s.t. f(x) ≤ M_x in the intersection of N(x;ε_x) with the domain of f.

Definition: We say that f is bounded on a set S if there is a number M s.t. f(x) ≤ M for all x in S.

Theorem: If f is locally bounded at each point in I, a closed interval, then f is bounded on I.

Before we prove this, please note that this is FALSE for f locally bounded everywhere on an open interval. Example?

And note that this is false for an unbounded interval. Example?

Proof:

Suppose f is locally bounded on I a closed interval. Then consider the collection of open intervals:

C = { N(x;ε_x) | x in I } obtained by applying the definition of locally bounded. Since each x in I is the center of one of the intervals in C, then C is an “open cover” of I. But since I is compact, C contains a finite subcover. Let N₁, N₂, ... N_k be the finite subset of C . I.e. each N_i is one of the N(x;ε_x) from C. And let M_i be the bounds on f(x) in N_i. Finally let M = max{M_i|i=1,...,k}. Then since each x is in some N_i , f(x) ≤ M_i ≤ M for all x in I. QED

Comment: Note that this proof uses the finite subcover very strongly. Without finiteness we can’t choose an M as a max.