Is functions of cauchy sequences is also Cauchy? Recently i saw in some book that if a sequence is Cauchy then function of that sequence is also Cauchy.I have confusion about this. Please help me.
 A: This is not generally true, even if the function is continuous.  For instance take the sequence $a_n=\frac{1}{n}$ and the function $f(x)=\frac{1}{x}$.  Applying the function to the sequence gives the new sequence $b_n=\frac{1}{a_n}=n$, which is not Cauchy.
A: But if the function $f$ is uniformily continous, then it is easy to see that $(f(x_n))_n$ is also Cauchy.
For $\varepsilon > 0$ choose $\delta > 0$ such that $|f(x) - f(y)| < \varepsilon$ for $|x-y|<\delta$ holds. Then choose $N \in \Bbb{N}$ with $|x_n - x_m| < \delta$ for $n,m \geq N$.
Hence,
$$
|f(x_n) - f(x_m)| < \varepsilon \,\forall n,m \geq N.
$$
A: As paw88789 says, this is not generally true. However, under some condition the claim does hold.
Let $(a_n)$ be a Cauchy sequence in a complete metric space $X$. Then $(a_n)$ has a limit $a$. If $f:X\to Y$ is continuous, where $Y$ is any metric space (not necessarily complete), then the sequence $(f(a_n))$ converges to $f(a)$, thus it is Cauchy.
A: Let $\{a_n\}$ be a Cauchy sequence in R. $  \,\,\,  $ Then $\{a_n\}$ has a limit $a$. 
If $f:R→R$  is continuous on $x=a$,  then the sequence $\,\,\{f(a_n)\}\,\,$ converges to $\,\,f(a)$, thus it is Cauchy.
