If {$a_j$}$_{j\in\mathbb{N}}$ $\subset$ E is a Cauchy sequence, is {$f$($a_j$)}$_{j\in\mathbb{N}}$ also a cauchy sequence? Assume $f$ : E -> $\mathbb{R}$ is a continuous function defined on some set E $\subset$ $\mathbb{R}$ and if {$a_j$}$_{j\in\mathbb{N}}$ $\subset$ E is a Cauchy sequence is it true that {$f(a_j)$}$_{j\in\mathbb{N}}$ is also a Cauchy sequence? 
 A: Not necessarily. 
A sufficient condition such that $\{f(x_j)\}$ is again Cauchy is that $f$ does not increase distances, as defined in this question: Function doesn't increase distance. 
This is because, for any continuous function $f$ such that $\{f(x_j)\}$ is no longer Cauchy, one has that $f$ must increase distances. In fact such an $f$, despite being continuous, would increase distances either just as rapidly or even more rapidly than the distances between terms in the Cauchy sequence decrease -- this would cause the sequence to no longer be Cauchy.
The standard example is the following: let $\{a_j\} = \frac{1}{j}$. Then let $f$ = $\frac{1}{x}$ which is continuous.
But $\{f(a_j)\}$ is the sequence $j$, which is clearly not Cauchy. Clearly the distances between terms for $f(a_j)$ do not decrease to zero -- instead they remain constant at $1$. Thus $f$ in some sense "exactly cancels out" the approach of the inter-term distances to zero which we had for $\{a_j\}$.
(To address how this relates to the comments on the question above -- the limit of the original sequence, $0$, is not in the domain of the function $f(x)=\frac{1}{x}$, hence there is no contradiction.)
