# Let $f$ be continuous on $X$ and $(x_n)$ be a Cauchy sequence on $X$. Show that $(f(x_n))$ doesnt have to be a Cauchy sequence…

Let $$f$$ be continuous on $$X$$ and $$(x_n)$$ be a Cauchy sequence on $$X$$. Show that $$(f(x_n))$$ doesnt have to be a Cauchy sequence. And show that $$(f(x_n))$$ always has to be a Cauchy sequence when $$f$$ is uniformly continuous.

My attempt:

We take the Cauchy sequence $$(x_n)=\frac{1}{n}$$ and choose $$f:\mathbb{R}^+\longrightarrow\mathbb{R}:x\mapsto\frac{1}{x}$$ which is a continuous function on given domain.

Since $$x_n \longrightarrow0\,\,\Longrightarrow (f(x_n))\longrightarrow+\infty$$

So eventhough $$(x_n)$$ is a Cauchy sequence, $$(f(x_n))$$ on the continuous $$f$$ isnt!

$$\exists f\in C^0(\mathbb{R}^+)$$ where the requirement does not hold!

Now we show that from the fact that $$f$$ is uniformly continuous we can always conclude:

Is $$(x_n)$$ a Cauchy sequence $$\Longrightarrow$$ $$(f(x_n))$$ is a Cauchy sequence

A function $$f$$ is uniformly continuous on $$X$$ when the Heine-Cantor theorem holds:

$$\forall \epsilon>0 \,\,\, \exists \delta>0 \,\,\,\forall x,y\in X:x\in \mathcal{U}_{\delta}(y)\Longrightarrow f(x)\in \mathcal{U}_{\epsilon}(f(y))$$

So for a $$N\in \mathbb{N}$$ we know that for all $$n>N:x_n\in \mathcal{U}_{\delta}(\xi)$$

Where $$\xi$$ is the point $$(x_n)$$ is converging to.

Following the Heine-Cantor theorem we know that if $$n>N:x_n\in \mathcal{U}_{\delta}(\xi) \Longrightarrow f(x_n)\in \mathcal{U}_{\epsilon}(f(\xi))$$

So when $$(x_n)$$ is a Cauchy sequence, $$(f(x_n))$$ has to be one aswell!

Would be great if someone could check my reasoning and give me advice for improving :)

• I'm convince by the first part. But not by the second one. Can you restate the definition of a Cauchy sequence? Also, I don't know why you mention Heine-Cantor theorem. Heine-Cantor theorem gives conditions implying that a map is uniform continuous. But here we're already supposing that it is uniform continuous. – mathcounterexamples.net May 30 '20 at 8:39
• The second part follows from definition of uniform continuity and definition of a Cauchy sequence. I don't understand why you are quoting a theorem for this. – Kavi Rama Murthy May 30 '20 at 8:51
• I thought that the cauchy sequence definition gives us, that a infinit amount of values lie in every epsilon region of our point of convergence. so there is this $\xi$ for which every $x_n$ with $N>n$ lies in any epsilon region around $\xi$. For any epsilon we might choose. And from the Heine Cantor theorem it follows that since the function is uniformly continuous our $(f(x_n))$ also has to lie in the close region of $f(\xi)$.. – CoffeeArabica May 30 '20 at 8:58
• Cauchy sequence does not need to converge. – N.Quy May 30 '20 at 9:05
• Ohh.. could you give me an example for that case? :) so I can ponder with that? – CoffeeArabica May 30 '20 at 9:13

Consider $$A = \{1, 1/2, 1/3, · · ·\}$$ and $$f(1/n) = \begin{cases}1, &if ~~n ~is~ odd\\−1,& if~~ n~ is ~even\end{cases}$$.
Then $$f$$ is continuous but not uniformly continuous. The sequence $$(x_n)= \frac{1}{n}$$ in $$A$$ is Cauchy but the sequence $$f(x_n) = (1, −1, 1, −1, · · ·)$$ is not Cauchy.
If $$f$$ is uniformly continuous on $$A$$, then given $$ε > 0$$ there is $$δ > 0$$ such that if $$x, y\in A$$ and $$|x−y| < δ$$, then $$|f(x) − f(y)| < ε$$. Let $$(x_n)$$ be a Cauchy sequence in $$A$$. Then for given $$δ > 0$$ there is $$M$$ such that if $$p, q > M$$, then $$|x_p − x_q| < δ$$, and thus $$|f(x_p) − f(x_q)| < ε$$, implying that $$(f(x_n))$$ is a Cauchy sequence.