Open, closed and continuous

Question

I have some troubles to understanding something:

We were asked to find a function that is open and continuous but not closed and actually I found such a function, but our tutor gave us this example

$ e^x$ since this function is continuous as a mapping $$\exp: \mathbb{R} \rightarrow \mathbb{R}_{>0}$$ and the inverse function is continuous too, so this function is also open.

and the counterexpample was, that $\mathbb{R}$ is closed and is mapped to $\mathbb{R}_{>0}$ which is open.

then i thought, hey if the inverse function is continuous that means that the fiber of closed sets have to be closed, but $\ln^{-1}(\mathbb{R})$ is not closed at all. where am i wrong?

Answer 1

Your tutor is wrong. Every homeomorphism (such as $\exp:\mathbb R\to\mathbb R_{>0}$) is closed. The set $\mathbb R_{>0}$ is a closed set in the standard topology on $\mathbb R_{>0}$.

Answer 2

If you consider $\def\R{\mathbb{R}}\def\exp{\mathrm{exp}}\exp\colon\R\to\R$, then the example is correct: the map is not closed, because

$$\exp(\R)=\R_{>0}$$

which is not closed in $\R$. On the other hand, the image of an open set in $\R$ is open in $\R_{>0}$, so also open in $\R$. Hence the function is open but not closed.

If you consider $\exp\colon\R\to\R_{>0}$, then of course its image is both open and closed in the codomain, being the whole of it. As already observed, this is a homeomorphism, so it's obviously open and closed.

Answer 3

I think you can consider the following function instaed: $$\pi:\mathbb R^2\to\mathbb R\\\\ \pi\left((x,y)\right)= x$$