Lemma
Suppose that \(E \subseteq \mathbb{R}\) and that \(f:E \rightarrow \mathbb{R}\) is uniformly continuous. If \(x_n \in E\) is Cauchy, then \(f(x_n)\) is Cauchy.

Proof

Let \(\epsilon > 0\) and choose \(\delta > 0\) such that for all \(x, a \in E\)

$$ \begin{align*} |x - a| < \delta \quad \text{ imply } |f(x) - f(a)| < \epsilon \quad \quad (1) \end{align*} $$

Since \(\{x_n\}\) is Cauchy, then choose \(N \in \mathbb{N}\) such that whenever \(n,m \geq N\),

$$ \begin{align*} |x_n - x_m| < \delta \end{align*} $$

But then this means by (1) that when \(n,m \geq N\),

$$ \begin{align*} |f(x_n) - f(x_m)| < \epsilon. \ \blacksquare \end{align*} $$

Notes: note here that there isn’t any condition on \(E\). \(E\) can be bounded, unbounded, open or closed.

References