Does there exist a continuous bijection which is open but not closed?

Question

I think that there isn't, but I don't have any further justification than "I can't find one." If I take some function $f:X\to Y$, where $X$ has topology $\tau_x$, and $Y$ has topology $\tau_y$, then I need to somehow tie together the facts: $$\forall U\in\tau_y,\,f^{-1}(U)\in\tau_x$$ $$\forall U\in\tau_x,\,f(U)\in\tau_y$$ $$\exists U\in\tau_x \text{ such that} \,f(U^c)\neq V^c,\,\forall\, V\in\tau_y$$ $$\forall x,y\in X,\,f(x)=f(y)\implies x=y$$ $$\forall y\in Y,\, \exists x\in X\text{ such that } f(x)=y$$

and come up with a contradiction. Or, otherwise, find a function and some topological spaces which together satisfy all of these statements. I honestly believe that there is some problem here, but I can't seem to find what it is. I have a similar problem with trying to show that there is no continuous bijection which is closed but not open. I would really appreciate any hints to point me in the right direction!

Answer 1

HINT: Suppose that $f:X\to Y$ is an open bijection, and let $F$ be a closed set in $X$. Then $X\setminus F$ is open in $X$, so $f[X\setminus F]$ is open in $Y$. Now use the fact that $f$ is a bijection to say something about $f[F]$.

Answer 2

No, there does not. Let $f$ be open, and let $U$ be an arbitrary closed subset of $X$. Then,

$$\begin{align}f(U) &= f(X - (X - U)) \\ &= f(X) - f(X - U) \\ &= Y - f(X - U)\end{align}$$

with the last two equalities because $f$ is bijective. Then, $f(U)$ is the complement of an open set, and hence, is closed.

Swapping the words "open" for "closed" and vice versa will give you the other proof. Note that this proof does not require the continuity of $f$, only that it is bijective.