Find a function that is continuous in usual topology, discontinuous in lower limit topology I had found a function that was continuous in the lower limit topology but not the usual topology
Show the Heaviside step function is continuous in $(\mathbb{R}, \mathcal{T}_\text{lower limit})$
$$h(x):=\begin{cases} 1, & x \geq0\\ 0, & x <0 \end{cases}$$
$h(x)$ is not continuous in the usual topology because the preimage of $1$ is $[0, \infty)$ which is not open.

Now I need a function that is continuous in usual topology (meaning:
$f: \mathbb{R}_{usual} \to \mathbb{R}_{usual}$ is continuous)
But
discontinuous in lower limit topology $(f: \mathbb{R}_{lower limit} \to \mathbb{R}_{lower limit}$ is discontinuous).

I was thinking a function whose preimage is of the form $[a,b]$, is this line of thinking correct? What would be such a function?
 A: If you put the same topology on the domain and the codomain, a classical example of this is $f(x) = -x$, which is continuous as a map $f:(\mathbb{R}, \mathbb{R}_{\text{usual}}) \rightarrow (\mathbb{R}, \mathbb{R}_{\text{usual}})$ but not as a map $f: (\mathbb{R}, \mathbb{R}_{\text{lower}}) \rightarrow (\mathbb{R}, \mathbb{R}_{\text{lower}})$, the latter because $f^{-1}[[0,1)] = (-1,0]$ which is not open as $0$ is not an interior point. 
This also shows that $\mathbb{R}$ is not a topological group under addition in the lower limit topology. 
If you have the same topology on the codomain, then as $(\mathbb{R}, \mathbb{R}_{\text{usual}})$ is coarser than $(\mathbb{R}, \mathbb{R}_{\text{lower}})$, every function continuous on usual reals would be continuous in the lower limit topology. So it's essential we change the topology on the codomain as well.
A: (1). $h$ is discontinuous with the usual topology but not for the reason given. One equivalent def'n of continuity is that pre-images of closed sets are closed....  $ h^{-1}\{1\}=[0,\infty)$ while both $\{1\}$  and $[0,\infty)$ are closed. However, $(0,\infty)$ is open but $h^{-1}(0,\infty)=[0,\infty)$ is not open so $h$ is not continuous.
(2). Let $f(x)=x$ for $x\in [0,1)$ and $f(x)=0$ for $x\not \in [0,1).$ Then $f^{-1}(-\infty,1/2)=(-\infty,1/2)\cup [1,\infty)$ is not open in the usual topology, so $f$ is not continuous with respect to it. You may check that $f^{-1}[a,b)$ is always open in the Sorgenfrey (lower-limit) topology so $f$ is continuous with respect to it.
