I want to find a continuous function: $f:\textbf{R}^n \rightarrow \textbf{R}^m$ s.t. for some open subset $A$, $f(A)$ is not open, and for some closed $B$, $f(B)$ is not closed.

I am able to find some mappings that satisfy one condition, e.g. $f(x)=exp(-x)$ maps closed $[0,\infty)$ to not closed $(0,1]$, but cannot find an example which satisfies both conditions.


Consider the continuous map $f(x)=e^{-|x|}$. Then $f(\Bbb R)=(0,1]$. Note that $\Bbb R$ is both open and closed, but it's image $(0,1]$ is neither open nor closed.


Consider the continuous map :

$$f:\mathbb{R}\to\mathbb{R},x\mapsto\cases{1\quad\mathrm{if}\,x\le0\cr\frac 1{x+1}\quad\mathrm{otherwise}}$$

and the subsets $A=(-\infty,0)$, $B=[0,+\infty)$.

Then :

  • $A$ is open and $f(A)=\{1\}$ is not open

  • $B$ is closed and $f(B)=(0,1]$ is not closed


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.