# 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

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

• Open need not map to open. The inverse image of an open is open. – user10354138 Nov 2 '18 at 17:20
• Try a singleton. – Will M. Nov 2 '18 at 17:20
• Consider $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=0$ for all $x\in\mathbb{R}$ where $\mathbb{R}$ is equipped with the usual topology. – user 170039 Nov 2 '18 at 17:21
• Or a constant function... did you try anything? – Will M. Nov 2 '18 at 17:21

## 2 Answers

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

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 $$\mathbb{R}$$

• More importantlyregarding the OP: $\{1\}$ is not open – Hagen von Eitzen Nov 2 '18 at 17:33
• @HagenvonEitzen..exactly u r right sir – jasmine Nov 2 '18 at 17:35