How to construct examples of maps that are open or closed or continuous but not the others?

Question

I came across this exercise, the main problem for me are the restrictions, i need to find examples for maps $f: \mathbb {R}^2 \rightarrow \mathbb {R}^2$ or subsets of them such that $f$ is only one or two of the three at the same time. (I.e $f$ is open but no closed nor continuous) But i don't know How to build such examples, as i also don't see the kind of patologies i should be looking for

Answer 1

This is all I currently know about all eight cases.

Open, closed, continuous: $f(x,y)=(x,y)$

Open, closed, not continuous: Very interesting A function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is open and closed, but not continuous.

Open, not closed, continuous: $f(x,y)=(e^x,y)$

Open, not closed, not continuous: Any strongly Darboux function.

Not open, closed, continuous: $f(x,y)=(0,0)$

Not open, closed, not continuous: $f(x,y)=(\lfloor x\rfloor,0)$

Not open, not closed, continuous: $f(x,y)=(x,0)$

Not open, not closed, not continuous: $f(x,y)=(x+\lfloor x\rfloor,0)$

If you simply want an open, closed function $f:A\to B$ that is not continuous, where $A,B\subseteq\mathbb{R}^2$, then $f(x,y)=(x-\lfloor x\rfloor,y):\mathbb{R}^2\to\{(x,y)\in\mathbb{R}^2:x\in[0,1)\}$ suffices.

Answer 2

A constant map is continuous, closed and not open.

Map the closed upper half of the plane to (1,1) and the open lower half to (0,0).
That map is not continuous, closed, not open.

Answer 3

If $f: A\to B$ is a continuous surjection, and $B$ is a subspace of $C,$ and $C$ is a subspace of $D,$ such that $B$ is not closed in $C$ and $B$ is not open in $D,$ then $f:A\to D$ is continuous and $f$ maps $A,$ which is an open-and-closed subset of $A$(!) onto $B,$ which is neither open nor closed in $D.$

For example $A=\Bbb R^2,\,B= (-\pi/2, \pi/2)\times \{0\},\, C=\Bbb R\times \{0\},$ and $D=\Bbb R^2.$ With $f(x,y)=<\arctan x,\,0>.$

( I am using round brackets $(...,...)$ in the definition of $B$ to denote an open real interval, and $<...,...>$ in the definition of $f$ to denote an ordered pair.)

Part of the idea is to start with $A,B,$ and $f,$ and look for $C$ such that $B$ is a dense proper subset of $C,$ and $C$ has empty interior in $D$. This is made easier by taking $1$-dimensional $B$ and $C$, and $D=\Bbb R^2$.