Give an example of a continuous function $f : X \rightarrow Y$ such that the image $f(F)$ is not open in $Y$ for a open $F$ in X

Question

Give an example of a continuous function $f : X \rightarrow Y$ such that the image $f(F)$ is not open in $Y$ for a open $F$ in $X$

My attempts : I know that open map to open,,,here I'm confused how can I find the counter-example

thanks u

Answer 1

Consider the function $f(x)= \sin(x)$ and choose an open $U=(0, \pi)$.

Answer 2

From Will. M comment take $f : \mathbb{R} \rightarrow \mathbb{R}$ define by $f(x) = 1$, as $f(x) =\{1\}$ which is closed as singleton set is closed in $\mathbb{R}$