Let $f$ be a right continuous function then show that it is continuous from $\mathbb{R}_l \to \mathbb{R}$ The most simplest example of a right continuous function that could come to my mind is
$$f(x)= \begin{cases}x &x \le 2 \\  x+1 & x>2\\
\end{cases}$$
This function is discontinuous at the point $x=2$ but it is right continuous .The function is not continuous in $\mathbb{R}$ as if we consider the open set $(1,3)$ then the pre-image of this set is $(1,2] \cup (3,4) $ which is not open in $\mathbb{R}$ .
Is the set $(1,2] \cup (3,4)$ open in $\mathbb{R}_l$?I am not sure though . This is a question from Munkres and I need to prove this, so may be I am making some mistakes.
I think it will be of the form
$$(1,2] \cup (3,4) = \cup _{n \ge 2}[1- \frac{1}{n},2) \cup_{n \ge 2} [2,2-\frac{1}{n}) \cup_{n \ge 2} [3-\frac{1}{n},4-\frac{1}{n})$$
Here $\mathbb{R}_l$ denotes the lower limit topology. Is this way of writing it correct?
Edit 1:I was thinking of considering two cases :
Case 1: when $a \ne f(x_1-)$  and $ b \ne f(x_2-)$ then we consider the open interval $(a,b)$ .So the inverse $f^{-1}(a,b)=(f^{-1}(a),f^{-1}(b))$
Case 2: when $a = f(x_1-)$ then $f^{-1}(a,b)=[x_1,f^{-1}b)$
Case 3: when $b=f(x_2-)$ can also be dealt similarly .
Case4: when $a=f(x_1-)$ and $b=f(x_2-)$ (where $x_1 < x_2 $ then $f^{-1}(a,b)=[x_1,x_2)$.
This is how I am thinking about the general proof . Am I in the right path?I would encourage some hints rather than the complete answer.
 A: Your function $f$ is not right continuous, but left continuous (by which I mean that $\lim_{x \uparrow a} f(x) = f(a)$ for all $a \in \Bbb R$ (so taking limits from the left only; right continuous is about limits from the right). In your case the right limit towards $2$ is $3$ while $f(2)=2$, so $f$ is not right continuous.
Moreover $f^{-1}[(1,3)] = (1,2]$ which is not open in $\Bbb R_l$ (as $2$ is not an interior point). So $f: \Bbb R_l \to \Bbb R$ is not continuous. Which is to be expected if you want right continuous functions to work...
The modified function
$$g(x)=\begin{cases} x & x < 2\\ x+1 & x \ge 2\\ \end{cases}$$
is continuous from the right
and here we have $g^{-1}[(1,3)] = (1,2)$ e.g. which is open in $\Bbb R_l$ (as it's finer than the usual topology) and $g^{-1}[(1,4)] = (1,2) \cup [2,3) = (1,3)$, also open etc. and also $f^{-1}[(2,4)] = [2,3)$, open in $\Bbb R_l$ (also showing non-continuity wrt the standard topology on the domain).
So that looks better though it's by no means a complete proof that a right continuous $f$ is indeed continuous going from $\Bbb R_l$ to $\Bbb R$; for that a proper proof will be required.
