Prove that if f is uniformly continuous on a bounded set S then f is a bounded function on S

Proof:

Let S be a bounded set in R and let f be a uniformly continuous function on S. We need to show that f is bounded on S, i.e., there exists a positive number M such that |f(x)| ≤ M for all x ∈ S.

Since S is bounded, there exist real numbers a and b such that S ⊆ [a, b]. Let ε > 0 be given. Since f is uniformly continuous on S, there exists a δ > 0 such that for any x, y ∈ S with |x − y| < δ, we have |f(x) − f(y)| < ε.

Choose a positive integer N such that Nδ > b − a. Divide the interval [a, b] into N subintervals of length δ, and let xi = a + iδ for i = 0, 1, 2, ..., N. Then S is covered by the N subintervals [xi−1, xi] for i = 1, 2, ..., N.

By the uniform continuity of f on S, for each i = 1, 2, ..., N, there exists a positive integer Mi such that |f(x) − f(y)| < ε for all x, y ∈ [xi−1, xi] with |x − y| < 1/Mi. Let M = max{M1, M2, ..., MN}. Then for any x ∈ S, there exists an integer i such that xi−1 ≤ x ≤ xi, and hence |x − xi| < δ. Therefore, we have |f(x) − f(xi)| < ε by the choice of δ.

Since xi ∈ [a, b], we have |f(xi)| ≤ K for some positive number K by the boundedness of f on [a, b]. Hence, we have

|f(x)| ≤ |f(x) − f(xi)| + |f(xi)| < ε + K

for all x ∈ S. Since ε > 0 is arbitrary, we conclude that |f(x)| ≤ K for all x ∈ S, i.e., f is bounded on S