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.
 A: 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 $K$.)

A: (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 $\mathcal{T^{''}}$.
