In metric space, image of cauchy sequence under continuous map. Let $f:X \to Y$ be a function from a metric space $X$ to another metric space $Y$. For any Cauchy sequence {$x_n$} in $X$,

*

*If $f$ is continuous then {$f(x_n)$} is Cauchy sequence in $Y$


*If {$f(x_n)$} is Cauchy then {$f(x_n)$} is always convergent in $Y$


*If {$f(x_n)$} is Cauchy in $Y$ then $f$ is continuous


*{$x_n$} is always convergent in $X$
My Attempt:
If I take $f : (0,\infty) \to \Bbb R$ defined by $f(x) = \frac{1}{x}$ which is continuous and If I take {$x_n$} = {$\frac{1}{n}$} $\to 0$ as $n \to \infty$  but $f(x_n) =  f(\frac{1}{n}) = n \to \infty$ as $n \to \infty$
So option 1,2 discarded.
Also $x_n = \frac{1}{n} \to 0$ but $0 \notin X$ So option 4 discarded. What happen with option 1 when we replace continuous map to uniform continuous map.  Please provide me a detailed solution with counter examples. Thanks in advance.
 A: Your example discards option 1. and 4.
$f:(0, \infty) \rightarrow (0, \infty)$ defined by $f(x)=x$ discards option 2. as $f(\frac{1}{n})=\frac{1}{n}$ and $\{\frac{1}{n}\}$ is Cauchy but doesn't converge in $(0, \infty)$.
Option 3. is correct. Cauchy continuity implies continuity.
Let $c \in X$. If $f$ is not continuous at $c$ then $\exists~~\epsilon>0$ such that for all $\delta>0~~~\exists~~x$ for which
\begin{align}
d_X(c, x)<\delta ~~~\textit{but}~~~d_Y(f(c), f(x))>\epsilon
\end{align}
Now for each $n \in \mathbb{N}~~~\exists~~x_n$ such that
\begin{align}
d_X(c, x_n)<\frac{1}{n}~~~\textit{but}~~~d_Y(f(c), f(x_n))>\epsilon
\end{align}
Define a sequence $\{y_n\}$ by
\begin{align}
y_{2n-1}&=x_{2n-1}~~~\textit{and}\\
y_{2n}&=c
\end{align}
for $n=1, 2, 3, \cdots$. Then note that $y_n \longrightarrow c$, hence is Cauchy in $X$. But for any $n \in \mathbb{N}$
\begin{align}
0<\epsilon<d_Y(f(c), f(x_{2n-1}))=d_Y(f(y_{2n}), f(y_{2n-1})).
\end{align}
This shows that $\{f(y_n)\}$ cannot be Cauchy in $Y$, which is a contradiction! So $f$ must be continuous at an arbitrary $c$ and hence on $X$.
In 1. if you replace continuity with uniform continuity then $\{f(x_n)\}$ will be Cauchy. For any $\epsilon >0~~~\exists~~\delta>0$ such that $d_X(x_1, x_2)<\delta$ implies $d_Y(f(x_1), f(x_2))< \epsilon$. As $\{x_n\}$ is Cauchy for the $\delta>0~~~\exists~~N \in \mathbb{N}$ such that $d_X(x_m, x_n)<\delta$ for all $m, n >N$. Then from above $d_Y(f(x_m), f(x_n))<\epsilon$ for all $m, n >N$. Proving $\{f(x_n)\}$  Cauchy.
A: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $f:X\to Y$ be a function. Say that $f$ is Cauchy continuous provided that for every Cauchy sequence $(x_n)_{n\in\mathbb{N}}$ in $X$, the sequence $(f(x_n))_{n\in\mathbb{N}}$ is a Cauchy sequence in $Y$.
Lemma:

Every uniformly continuous function $f:X\to Y$ is Cauchy Continuous

Proof:

Exercise (if you would like hints, please ask but this should be a fairly straightforward epsilon-delta argument). $\square$

Claim:

There exists metric spaces $(X,d_X)$ and $(Y,d_Y)$ and functions $f:X\to Y$ which are Cauchy continuous but not uniformly continuous.

Proof:
Consider $X=Y=\mathbb{R}$ with the standard metric. Let $f:X\to Y$ be defined as $f(x)=x^2$. We claim that $f$ is Cauchy continuous but not uniformly continuous.
Let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $X$. Since $X$ is complete, then $(x_n)_{n\in\mathbb{N}}$ is convergent, say to $x_0\in X$. As $f$ is continuous, then the sequence $(f(x_n))_{n\in\mathbb{N}}$ converges to $f(x_0)=x_0^2\in Y$. Since every convergent sequence in a metric space is Cauchy, then $(f(x_n))_{n\in\mathbb{N}}$ is a Cauchy sequence in $Y$, hence $f$ is Cauchy continuous.
It is well known that $f(x)=x^2$ is not uniformly continuous on $X$ which can be seen here. $\square$
You can find more interesting facts about Cauchy continuity on wikipedia.
