Does anybody know an example of a continous map F: R --> R and a closed Subset B C R such that f(b) is not closed.
Let $\displaystyle f(x)=\frac{1}{1+x^2}$ and $\displaystyle B=[0,\infty).$ It is clear that B is a closed set, whereas $\displaystyle f(B)=(0,1]$ is not closed.
No, because by definition every continuous image of an open set is open, whence every continuous image of a closed set is closed.
Julien, you have the definition backward. The inverse image of an open set under a continuous function is open. Likewise, the inverse image of a closed set under a continuous function is also closed.
Last edited by Jay Gatsby; Dec 5th 2006 at 02:33 PM.