$\qquad\mathit{(i)}\,$ We know that $\sin:\Bbb R\to\Bbb R$ is continuous. Show that, if $\,U=\Bbb R$, then $U$ is open, but $\sin U$ is not.
$\qquad\mathit{(ii)}\,$ We define a function $f:\Bbb R\to\Bbb R$ as follows. If $x\in\Bbb R$, set $\langle x\rangle=x-[x]$ and write $$\langle x\rangle=.x_1x_2x_3\ldots$$ as a decimal, choosing the terminating form in case of ambiguity. If $x_{2n+1}=0$ for all sufficiently large $n$, let $N$ be the least integer such that $x_{2n+1}=0$ for all $n\ge N$, and set $$f(x)=(-1)^N\sum_{j=1}^\infty x_{2N+2j}10^{N-j}.$$ We set $f(x)=0$ otherwise.
$\qquad$ Show that if $U$ is a non-empty open set, $f(U)=\Bbb R$ and so $f(U)$ is open. Show that $f$ is not continuous.
Note: consider $\Bbb R$ with the ordinary Euclidean metric.
I've done the first part, it's the second part where I'm badly stuck, I tried showing $f(U)=R$ but I seem to get reals that $f(x)$ can't get.
Any help will be appreciated.
