Continuity of a function with respect to various topologies (example) Let $f: \mathbb{R} \to \mathbb{R}$ be given by $$f(x)= \begin{cases}
-x-1, &\text{if }x≥0\\
1, &\text{if }x<0.
\end{cases}$$
Denote by


*

*$U$ the usual topology on $\mathbb{R}$.

*$H$ the half-open interval topology on $\mathbb{R}$.

*$C$ is the open half-line topology on $\mathbb{R}$. 


Is $f$: 


*

*$U$-$U$ continuous?

*$U$-$H$ continuous?

*$U$-$C$ continuous?

*$H$-$U$ continuous?

*$C$-$C$ continuous?

*$C$-$U$ continuous?

*$C$-$H$ continuous?

*$H$-$H$ continuous?


So, I kinda understand how to find things that are $U$-$U$ or whatever continuous, but the function is kinda throwing me off. So maybe some tips or suggestions as to where to start.  
 A: We should start by looking at what the inverse images of various basic open sets look like.  This will help us determine the continuity of the function. (And plotting the graph of the function is quite helpful.)
Let's start out with $U$, the usual topology.  The basic open sets in this topology are just the open intervals $(a,b)$ for $a<b$.  The inverse image of these sets break down into several cases.


*

*If $a < b \leq -1$, then $f^{-1} [ (a,b) ] = ( -b-1 , -a-1 )$;

*If $a < -1 < b \leq 1$, then $f^{-1} [ (a,b) ] = [ 0 , -a-1 )$;

*If $a < -1 < 1 < b$, then $f^{-1} [ (a,b) ] = [ 0 , -a-1 ) \cup ( -\infty , 0 )$;

*If $-1 \leq a < b \leq 1$, then $f^{-1} [ (a,b) ] = \varnothing$;

*If $-1 \leq a < 1 < b$, then $f^{-1} [ (a,b) ] = ( -\infty , 0 )$;

*If $1 \leq a < b$, then $f^{-1} [ (a,b) ] = \varnothing$.


As $[ 0 , 1 ) = [ 0 , -(-2)-1 ) = f^{-1} [ (-2,1) ]$ is not open with respect to the usual topology, then $f$ is not $U$-$U$ continuous.  A simple check shows that all of these sets are open in the half-open interval topology, and so $f$ is $H$-$U$ continuous. As $(-4,-2) = ( -3-1 , -1-1 ) = f^{-1} [ (1,3) ]$ is not open in the open half-line topology, then $f$ is not $C$-$U$ continuous
For the half-open interval topology you perform a similar analysis for the basic open sets of this topology, which are of the form $[ a , b )$ for $a < b$.  (If you know the connection between this topology and the usual topology, you should be able to see that the above actually answers the question of the possible $U$-$H$, $H$-$H$ and $C$-$H$ continuity of $f$ for two of these.)
Finally for the half open-line topology you consider the inverse images of all sets of the form $( a , + \infty )$.
A: Here is a breakdown of my attempt for each one of my questions:


*

*U-U Continuous:  $(-2,1)$ is open in the U topology. $f^{-1}((-2,1))=[0,1)$.  $[0,1)$ is not open in the U topology and therefore f is not U-U continuous.

*U-H Continuous:  $(-2,1)$ is also H-open, which implies that f is not U-H continuous.  

*U-C Continuous:  Yes, it is U-C continuous.  Why not sure...

*C-C Continuous:  The function $f^{-1}((a,+ \infty))= [0,a-1)$ where $a\lt-1$.  Which is not C-open and therefore not C-C continuous.

*H-U Continuous:  [-3,1) is open in the H topology.  $f^{-1}([-3,1)=[0,2]$.  [0,2] is not H-open and therefore it is not H-U continuos.

*C-U Continuos:  No, [0,1) is not C-open and therefore it is not C-U continuous or C-H continuous.

*C-H Continuous:  No, See #4.

*H-H Continuous:  No, similar to #5.
I am really not sure if this is correct and would really appreciate corrections or feedback!
