how to show that sin(x) is not continuous with this topology

Question

Consider the topology on $\Bbb R$ with basis $\{[a,b):a<b, \ a,b\in \Bbb R\}$. Could anyone give a hint to show that the function $f:\Bbb R \to\Bbb R$ given by $f(x)=\sin(x)$ is not continuous? Here, the domain and codomain equipped with the same (given) topology. Thanks!

Answer 1

I think you can look at the preimage by $\sin$ of the open $[1,2)$. You get $\pi /2 + 2 \pi \mathbb{Z}$ which is not open.

Answer 2

Consider the sequence {$ \pi +\frac{1}{n}$} which converges to $\pi$ in this topology but {$sin(\pi +\frac{1}{n})$} does not converge to $sin(\pi)=0$ w.r.t. this topology since there is a basic nbd of 0 which does not contain any negative.