Are $\mathbb{R_k}$ and $\mathbb{R_L}$ really non-comparable topologies? First of all, we notice that: $$(a,b)=\bigcup_{a<x<b} [x,b).$$ 
Also, we notice that: $$(0,1)-\left\{\frac{1}{n} |\, n\in \mathbb{Z^+}\right\}=\left(\frac{1}{2},1\right)\cup \left(\frac{1}{3},\frac{1}{2}\right)\cup \left(\frac{1}{4},\frac{1}{3}\right)\cup \left(\frac{1}{5},\frac{1}{4}\right)\cup \ldots =$$
$$\bigcup_{a_i=1}^{\infty} \left\{ \left(\frac{1}{2a_i},\frac{1}{2a_i-1}\right)\bigcup\left(\frac{1}{2a_i+1},\frac{1}{2a_i}\right)\right\}$$
We also can express every basis element of $\mathbb{R_k}$ of the form $(a,b)-K$ in a similar way to $(0,1)$.
So every element of $\mathbb{R_k}$ can be expressed as a union of open intervals. But, we can express  any open interval as a union of intervals of the form $[x,b)$ as we have done in the beginning . So, we conclude that every element in $\mathbb{R_k}$ is expressible using elements of $\mathbb{R_L}$. So $\mathbb{R_L}$ is finer than $\mathbb{R_k}$.
My question is, what is the wrong with my presentation? I know that there is a proof for that both topologies are not comparable and it convinced me but I can't find where my way of thinking went wrong, any help is appreciated.
 A: It's true that $(0,1) \setminus K$ can be written as a union of open intervals, but it's not true that you can do it for all sets of the form $(a,b) \setminus K$. For example
$$(-1, 1) \setminus K = (-1, 0] \cup \bigcup_{n\geq 1} (\frac{1}{n+1}, \frac{1}{n})$$
Note that $(-1, 0]$ is not open in $\mathbb{R}_L$.
A: The issue lies in the assertion that any interval $(a,b) - K$ can be decomposed as a union of open intervals "in a way similar to $(0,1)$." Indeed, the procedure does not extend if $a<0<b$. Let's assume $b=1$, and $a<0$. Note that $(a,b)-K$ is not equal to $(a,0)\cup\big(\cup_{i=1}^\infty (\frac{1}{i+1},\frac{1}{i})\big)$, since no interval of the form $(\frac{1}{i+1},\frac{1}{i})$ contains zero.
However, your investigation shows that the lower limit topology on $\mathbb R$ is "almost" finer than the $K$-topology: all open sets of $\mathbb R_K$ not containing $0$ are also open in $\mathbb R_L$. This reflects the fact that the induced K-topology on $\mathbb R_K -\{0\}$ coincides with the usual subspace topology on $\mathbb R-\{0\}$. 
For posterity, here are a few words about how to show that the two topologies are incomparable (although I understand that you are already convinced). First check that $K=\{\frac{1}{n} : n \in \mathbb N  \}$ is closed in $\mathbb R_K$ but not in $\mathbb R_L$. To finish up, observe that $[0,1)$ is not open in $\mathbb R_K$. 
