Let $X$ be a metric space and let $f : X → \mathbb{R}$ be a continuous function. Pick out the true statements.
(a) $f$ always maps Cauchy sequences into Cauchy sequences.
(b) If $X$ is compact, then $f$ always maps Cauchy sequences into Cauchy sequences.
(c) If $X = \mathbb{R^n}$, then $f$ always maps Cauchy sequences into Cauchy sequences.
(a) Let $\epsilon>0$ given, $\exists \delta>0$: $|f(x)-f(y)|<\epsilon$ such that $\forall x : d(x,y)<\delta$ ($\because$ $f$ is continuous). consider {$x_n$} be a cauchy sequence. For the above $\delta$ , $\exists N(\delta)$ : $d(x_n,x_m)<\delta$ $\forall m,n \ge N(\delta)$.
It is given in the answer key that (a) is false, proving some thing is true is easy compared to proving something is false. How to cook a counter example from the given data?
(b) Given that $X$ is compact $\implies$ $X$ is sequentially Compact($\because$ $X$ is a metric space). Suppose {$x_n$} be a cauchy sequence , $\exists$ a convergent subsequence converging in $X$.
How to proceed further?
(c) I hope (c) can be done with the help of(a).
Please help me. I am facing difficulty in cooking counter examples? in this region. How to learn making counter examples.
b) is correct because continuous function on compact set in metric space is uniformly continuous, this can be done by Lebesgue number of the sequentially compact set (sequentially compact = compact in metric space).
Or we can argue with the following fashion. Since $X$ is sequentially compact, given Cauchy sequence $(x_{n})$, it contains a convergent subsequence $(x_{n_{k}})$, say, $x_{n_{k}}\rightarrow x$, so $x_{n}\rightarrow x$ by Cauchy property. So $f(x_{n})\rightarrow f(x)$ by continuity.