I need to show that the following statement holds true:
Given $a, b \in \mathbb{R}$, $a < b$, $f: (a, b) \to \mathbb{R}$, $f$ continuous. Show that $f$ uniformly continuous $\Rightarrow$ $\forall$ cauchy sequences $(x_n)$ with $a < x_n < b \space (n \in \mathbb{N}): f(x_n)$ is a cauchy sequence.
I feel like I am very close to the proof but I just cannot see the last step. Can you please help me? My proof goes like this so far:
$f$ is uniformly continuous $\Leftrightarrow$ $\forall \epsilon > 0 \space \exists \delta > 0: |f(x) – f(x')| < \epsilon \space \forall x, x' \in (a, b)$ with $|x – x'| < \delta$.
Let $(x_n)_{n \in \mathbb{N}}$ be an arbitrary cauchy sequence with $a < x_n < b \Leftrightarrow \forall \epsilon > 0 \space \exists N \in \mathbb{N}: |x_n – x_m| < \epsilon \space \forall n, m \geq N$.
We need to show that $f(x_n)$ is cauchy. So let $\epsilon > 0$. Search for $N \in \mathbb{N}$ such that $|f(x_n) – f(x_m)| < \epsilon \space \forall n, m \geq N$.
Because of $f$ being uniformly continuous and $x_n, x_m \in (a, b) \space \forall n, m$ there exists $\delta > 0: |f(x_n) – f(x_m)| < \epsilon$ with $|x_n – x_m| < \delta$.
This is almost exactly the definition of $f(x_n)$ being cauchy but without the restriction of $|x_n – x_m| < \delta$. What am I missing to finish the proof?
Proof that if $f$ is uniformly continuous then for every Cauchy sequence $(x_n)$ with $a < x_n < b$ $f(x_n)$ is also cauchy.
cauchy-sequencescontinuityreal-analysisuniform-continuity
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Best Answer
This question is already very old but I still want to provide an answer since I learned that this problem is actually not hard at all.
Let $\epsilon > 0$ be given. Since $f: (a, b) \to \mathbb{R}$ is uniformly continuous there exists a $\delta_\epsilon > 0$ such that $|x-y| < \delta_\epsilon \Rightarrow |f(x) - f(y)| < \epsilon \ \forall x, y \in (a, b)$. Since $(x_n)$ is a Cauchy sequence there exists a $N_\epsilon \in \mathbb{N}$ such that $\forall n, m \geq N_\epsilon: |x_n - x_m| < \delta_\epsilon$. Because of uniform continuity this implies $|f(x_n) - f(x_m)| < \epsilon$ and therefore $(f(x_n))$ is also Cauchy. Note that $f(x_n)$ is actually well defined because $a < x_n < b \ \forall n \in \mathbb{N}$.