Continuous function of $R_0$-space into non-$R_0$

Question

Let $f:X\rightarrow Y$ be a continuous function of an $R_0$-space into a topological space (non-necessary $R_0$-space) (see Wiki to recall what an $R_0$-space is).

I wish to find out the conditions to ensure that if $X$ is $R_0$ then $f(X)$ also is.

For the momment, I have found that if $f$ is open and injective then $f(X)$ is $R_0$. Moreover, I know that if $f$ is not open or is not injective, then $f(X)$ may be no $R_0$:

• For if $f$ is not open, given two separated points $x,y\in X$ and two (open) neighbourhoods $V_x$ and $V_y$ such that $y\notin V_x$ and $x\notin V_y$, then $f(y)\notin f(V_x)$ and $f(x)\notin f(V_y)$, but $f(V_x)$ and $f(V_y)$ needn't to be neigbhbourhoods of $f(x)$ and $f(y)$ resp. An example illustrating that is the identity map of $\{0,1\}$ (discrete) onto $\{0,1\}$ (Sierpinski space).

• Similarly, if $f$ is open but not injective, despite that $f(V_x)$ and $f(V_y)$ are neighbourhoods of $f(x)$ and $f(y)$ and that $y\notin V_x$ and $x\notin V_y$, it is poosible that $f(x)\in f(V_y)$ and that $f(y)\in f(V_x)$ (for example if $f(x)=f(y)$).

However, I can't find an example illustrating the second case. My problem is that I can't find continuous open functions of an $R_0$-space into a non-$R_0$-space. My first try was to consider a map from the Sorgrenfey line onto the Sierpinski space (because it is disconnected) but that map was not open. Then I considered the identity from $[-1,1]$ with the usual subspace topology onto $[-1,1]$ endowed with the overlapping interval topology, but it was open neither.

Can you provide an example for the second case, please?

EDIT: The result is trivial if $f$ is constant. I'm interested when $f$ is non-constant, i.e. $f(X)$ has at least two points, so in the above, asuume that $f$ is non-constant.

Answer 1

It seems there was a simple example:

Let $X=\mathbb R$ with the usual topology and $Y=\{0,1\}$ be the Sierpinski space (with $\{1\}$ open). Consider $f:X\rightarrow Y$ defined as $f(0)=0$ and $1$ otherwise. Then:

• $f$ is continuous because $f^{-1}(\{1\})=(-\infty,0)\cup(0,+\infty)$ is open and $f^{-1}(\{0,1\})=\mathbb R$ is also open.

• $f$ is open because $f((a,b))=\{1\}$ if $0\notin (a,b)$ and $\{0,1\}$ if $0\in(a,b)$; in both cases they are open.

• The Sierpinski space is not symmetric because $0$ and $1$ are topologically distinguishable but $1$ belongs to each neighbourhood of $0$.