# Prove $\{g(x_n)\}_{n=1}^\infty$ converges

Let $g : (a, b) → R$ be uniformly continuous on $(a, b)$. Let $\{x_n\}_{n=1}^\infty$ be a sequence in $(a, b)$ converging to $a$. Prove that $\{g(x_n)\}_{n=1}^\infty$ converges.

The general idea here is to use the uniform continuity of g as well as the fact that since the $\{x_n\}$ converges it is Cauchy and to prove that $\{g(x_n)\}_{n=1}^\infty$ is Cauchy.

From uniform continuity we have $\forall\epsilon > 0, \exists\delta > 0$ such that $|x - y| < \delta$ with $x,y \in (a,b)$ then $|g(x) - g(y)| < \epsilon$.

And since $\{x_n\}_{n=1}^\infty$ is Cauchy we have $\forall\epsilon > 0, \exists N \in J$ such that $\forall n,m \ge N$ then $|x_m - x_n| < \epsilon$.

However, I'm not sure how to combine these two given properties in order to get the Cauchy property for $\{g(x_n)\}_{n=1}^\infty$.

Thanks for the help!

• One way to work you way through this kind of problem is to rename everything that's small and positive to $\varepsilon_1$, $\varepsilon_2$, etc. Then, you just have to find out which ones must be equal. – xavierm02 Dec 7 '14 at 23:19

Let $\varepsilon > 0$. Because $g$ is uniformly continuous there exists $\delta > 0$ s.t. $$|x-y| < \delta \implies |g(x)-g(y)| < \varepsilon, \; x,y, \in (a,b)$$ Because $(x_n)_{n=1}^\infty$ is Cauchy sequence, there exists $n_0 \in \Bbb{N}$ s.t. $|x_n - x_k| < \delta$ when $n \geq n_0$ and $k \geq n_0$. So, if $n, k \geq n_0$ holds, we have $$|g(x_n) - g(x_k)| < \varepsilon$$ so $(g(x_n))_{n=1}^\infty$ is a Cauchy sequence. Because $\Bbb{R}$ is complete, $(g(x_n))_{n=1}^\infty$ converges.
• How do you know $|g(x_n) - g(x_k)| < \epsilon$ though? – jlang Dec 7 '14 at 23:54
• We chose $n \geq n_0$ and $k \geq n_0$. Then $|x_n - x_k| < \delta$. The $\delta$ was chosen to be such that $|x-y| < \delta \implies |g(x)-g(y)| < \varepsilon$. So now $|g(x_k) - g(x_n)| < \varepsilon$. – desos Dec 8 '14 at 8:59