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?

  • $\begingroup$ To be clear, when you say "discontinuous in the lower limit topology", you are putting the lower limit topology on both the domain and the codomain? $\endgroup$ Jun 5, 2016 at 4:11
  • $\begingroup$ @EricWofsey Yes sorry if that is not clear $\endgroup$
    – Fraïssé
    Jun 5, 2016 at 4:12
  • $\begingroup$ What do you mean by "a function whose preimage is of the form $[a,b]$"? $\endgroup$ Jun 5, 2016 at 4:13
  • $\begingroup$ Given some $V \in \mathcal{T}_{lower limit}$, I wish construct a function such that $f^{-1}(V) = [a,b]$ of some sort, so that the preimage is not an open set. I could be gravely wrong though, is $[a,b]$ not open in lower limit topology? $\endgroup$
    – Fraïssé
    Jun 5, 2016 at 4:15
  • $\begingroup$ How about the function $f(x)=-x$? $\endgroup$
    – bof
    Jun 5, 2016 at 4:24

2 Answers 2


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.


(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.


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