-1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Open need not map to open. The inverse image of an open is open. $\endgroup$ – user10354138 Nov 2 '18 at 17:20
  • 2
    $\begingroup$ Try a singleton. $\endgroup$ – Will M. Nov 2 '18 at 17:20
  • 1
    $\begingroup$ 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. $\endgroup$ – user 170039 Nov 2 '18 at 17:21
  • 1
    $\begingroup$ Or a constant function... did you try anything? $\endgroup$ – Will M. Nov 2 '18 at 17:21
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\mathbb{R}$

$\endgroup$
  • 4
    $\begingroup$ More importantlyregarding the OP: $\{1\}$ is not open $\endgroup$ – Hagen von Eitzen Nov 2 '18 at 17:33
  • $\begingroup$ @HagenvonEitzen..exactly u r right sir $\endgroup$ – jasmine Nov 2 '18 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.