How to prove that continuous function do not necessarily preserve cauchy sequences I am trying to construct a proof that continuous function do not
preserve Cauchy sequences
Every proof I can find is disprove by counter example, which is great
but these counter examples cannot be extend to how we can fix the
failure of Cauchy preservation through uniform continuity.



  Prove: Let $(x_n)$ be a Cauchy sequence on a metric space $(M,d)$, let
  $f: (M,d_M) \mapsto (N,d_N)$ be a continuous function, then $(f(x_n))$
  may not be Cauchy.


(Note: I worded the problem myself which maybe a little bit awkward, feel free to point out a good fix)



(Use definition + by contradiction)

Let $(x_n)$ be a Cauchy sequence on $(M,d_N)$. Then $(x_n)$ is Cauchy if $\forall \epsilon &gt; 0, \exists N \in \mathbb{N}$, such that $\forall n, m \geq N, d_M(x_n, x_m) &lt; \epsilon$

Let $f$ be a continuous function such that  $f: (M,d_M) \mapsto (N,d_N)$. Then given $x_o \in M, \forall \epsilon&gt;0, \exists \delta(\epsilon,x_o) &gt; 0 \text{ s. t. } \forall x \in M, d_M(x,x_o) &lt; \epsilon \implies d_N(f(x), f(x_o)) &lt; \delta $  

Suppose $(f(x_n))$ is Cauchy on $(N,d_N)$, then $\forall \epsilon &gt; 0, \exists N' \in \mathbb{N}$, such that $\forall n', m' \geq N', d_N(f(x_{n'}), f(x_{m'})) &lt; \epsilon$

At this point I thought I would be notice some inconsistencies between the definition thus revealing a problem but I am not confident with my conclusions...


Can we say that we reach a contradiction because $f$ cannot be
continuous since it isn't true for $\forall x$ since $(x_n)$ is only
Cauchy for $n \geq N$?
Can we say that since $M$ is not complete, therefore $(x_n)$ may fail
to convergence, thus $f(x_n)$ is not defined as $n \to \infty$?


What is the problem here exactly?
If $d$ is a complete metric on $X$ and $f:X\to X$ is continuous then any Cauchy sequence $(p_n)_{n\in N}$ has a limit point $p$,  and the continuity of $f$ requires that $(f(p_n))_{n\in N}$ converges to $f(p).$

So to find an exception we must use  an incomplete metric.

Example. Let $X$ be the positive reals with the usual metric $d(x,y)=|x-y|$. Let $f(x)=1/x.$ The image of the Cauchy sequence $(1/n)_{n\in N}$ is the sequence $(n)_{n \in N}.$

It is not necessary that the metric be unbounded. In the above example, replace the metric $d$ with the equivalent metric $e(x,y)=\min (1, |x-y|).$
The statement says that $(f(x_n))$ may not be Cauchy.

The keyword here is "may not", hence trying to prove by contradiction (in the manner you described) is not possible, as the negation of "may not" is still "may not".