Show that the Topologies of $\mathbb{R}_l$ and $\mathbb{R}_K$ are not comparable.

Here, $\mathbb{R}_l$ is the lower limit topology on $\mathbb{R}$ and $\mathbb{R}_K$ is the K-topology on $\mathbb{R}$. I understand the proof that these topologies are strictly finer than $\mathbb{R}$, but I am at a loss to begin how to show they aren't comparable. This is from Munkres book.

To show that they are not comparable, you just need to find an open set in each that is not open in the other. (As in Munkres, I will denote the set $\{ \frac{1}{n} : n \in \mathbb{Z}_+ \}$ by $K$.)

• The set $[2,3)$ is open in $\mathbb{R}_l$, but not in $\mathbb{R}_K$.
• $\mathbb{R} \setminus K$ is open in $\mathbb{R}_K$, but not in $\mathbb{R}_l$. (Every open set in $\mathbb{R}_l$ containing $0$ meets the set $K$.)

(1)
Let $[c, d)$ be a basis element for $\mathcal{T^{'}}$.
If $\mathcal{T^{''}}$ is finer than $\mathcal{T^{'}}$, there exists a basis element $(a,b)-K$ for $\mathcal{T^{''}}$ such that $c\in (a,b)-K\subset [c, d)$ by Lemma 13.3 on p.81.
We show $(a,c)\subset K$ doesn't hold by proof by contradiction.
Assume that $(a,c)\subset K$.
Let $a_1\in (a,c)$.
Then, $a_1=\frac{1}{n}$ for some $n\in\mathbb{Z}_{+}$.
If $\frac{1}{n+1}\notin (a,c)$, then $\frac{1}{n+1}\leq a$.
Let $a_2\in (a,a_1)\subset (a,c)$.
Then, $a_2=\frac{1}{m}$ for some $m\in\mathbb{Z}_{+}$.
Since $\frac{1}{m}=a_2<a_1=\frac{1}{n}$, $n<m$.
So, $n+1\leq m$.
So, $\frac{1}{m}\leq\frac{1}{n+1}$.
So, $a_2=\frac{1}{m}\leq a$.
This is a contradiction.
So, $\frac{1}{n+1}\in (a,c)$.
Let $a_3\in (\frac{1}{n+1},\frac{1}{n})\subset (a,c)$.
Then, $a_3=\frac{1}{k}$ for some $k\in\mathbb{Z}_{+}$.
$\frac{1}{n+1}<\frac{1}{k}<\frac{1}{n}$.
So, $n<k<n+1$.
This is a contradiction.
So, $(a,c)\subset K$ doesn't hold.
So, there exists $a_4\in (a,c)\subset (a,b)$ such that $a_4\notin K$.
$a_4\in (a,b)-K\subset [c,d)$.
But $a_4<c$ since $a_4\in (a,c)$.
This is a contradiction.
So, $\mathcal{T^{''}}$ is not finer than $\mathcal{T^{'}}$.

(2)
If $\mathcal{T^{'}}$ is finer than $\mathcal{T^{''}}$, there exists a basis element $[c,d)$ for $\mathcal{T^{'}}$ such that $0\in[c,d)\subset (-1,1)-K$ by Lemma 13.3 on p.81.
There exists $n\in\mathbb{Z}_{+}$ such that $0<\frac{1}{n}<d$ by the Archimedean ordering property of the real line on p.33.
So, $\frac{1}{n}\in (0,d)\subset [c,d)\subset (-1,1)-K$.
This is a contradiction.
So, $\mathcal{T^{'}}$ is not finer than $\mathcal{T^{''}}$.