Elementary Classical Analysis, Marsden, 2nd ed, Chapter 4, Problem 28

Problem

Let $f:(0,1) \rightarrow \mathbb{R}$ be uniformly continuous. Must $f$ be bounded?

Answer

Since $f$ is a uniformly continuous, there exists a $\delta >0$ such that $|x-y|<\delta$  $\Rightarrow$  $|f(x)-f(y)|<1$. Suppose $f$ is not bounded. Then, we can choose a sequence $x_n \in (0,1)$ such that $f(x_{n+1}) > f(x_n) + 1$. Since $x_n$ is a sequence in the compact set $[0,1]$, there exists a convergent subsequence $x_{n_k}$. Since $x_{n_k}$ is a Cauchy sequence, there exists $K \in \mathbb{N}$ such that $|x_{n_k} - x_{n_l}|<\delta$ for all $k,l \geq K$. However, $|f(x_{n_K}) - f(x_{n_{K+1}})|>1$, which is contradiction.

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