Some questions regarding continuity and open and closed sets.

Question

$f:X\to Y$ is a map between two topological spaces $X,Y$.

$1.$ $f $ is said to be open (closed) mapping iff $f$ maps open(resp.closed) sets in $X$ to open sets in $Y$.

$2.$ $f$ is said to be continuous if $f$ pulls back open(closed) sets to open(resp.closed) sets.

There are some obvious questions that would arise:

Is every continuous map open.Is every continuous map closed.Is every open map continuous or is every closed map continuous.The counterexamples to each of these false statements are as follows:

$f(x)=0$ on $\mathbb R$ is continuous but not open because $f(\mathbb R)=\{0\}$,not open in $\mathbb R$,although $\mathbb R$ is open in $\mathbb R$.

Next,$f(x)=\arctan(x),x\in \mathbb R$ which is continuous but not closed.

Next,$f:\mathbb R \to \mathbb R$ by,

$f(x)= 0,x\in \mathbb Q,f(x)=1 ,x\in \mathbb {R-Q}$ is closed but not continuous and not closed also.

Example of open and not continuous function is $f(x)=x$ if $x\neq 0,1$ and $f(0)=1,f(1)=0$.(I have doubt whether this one is correct)

Are the examples correct?Are there more interesting counterexamples to study?

Answer 1

The last example is not correct. For example, if you consider the image of the open set $[0, 1/2)$ under $f$, the result is $\{1\} \cup (0, 1/2)$, which is not open. There might be simpler examples, but you could try $$f : \Bbb{R} \to [-1, 1] : x \mapsto \begin{cases} \sin(1/x) & \text{if }x \neq 0 \\ 0 & \text{if } x = 0\end{cases}.$$ This is the usual example of a discontinuous Darboux function, which maps intervals to intervals. Note that the codomain is $[-1, 1]$; intervals with closed endpoints at $-1$ or $1$ will still be open.

Answer 2

Let $X = \mathbb{R}$ with the topology having as a basis all the open intervals $(a, b)$, where $a, b \in \mathbb{R}$ and $a < b$.

Let $Y = \mathbb{R}$ with the topology having as a basis all the "closed-open" intervals $[a, b)$, where $a, b \in \mathbb{R}$ and $a < b$.

Now let $f \colon X \to Y$ be the map $x \mapsto x$. Then $f$ is open (and closed) but not continuous.

On the other hand, the map $g \colon Y \to X$, $y \mapsto y$, is continuous but not open / closed.

Let $$ S^1 = \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \ x^2 + y^2 = 1 \ \right\} $$ as a subspace of $\mathbb{R}^2$, and let $Y = [0, 1)$. Now let $f \colon Y \to S^1$ be the mapping $$ t \mapsto \left( \, \cos 2 \pi t, \sin 2 \pi t \, \right). $$ Then $f$ is continuous but not open.