Consider the following topologies on $\mathbb{R}$ :
- $\mathcal{T}_1=$ the standard topology
- $\mathcal{T}_2=$ topology of $\mathbb{R}_K$
- $\mathcal{T}_3=$ the finite complement topology
- $\mathcal{T}_4=$ the upper limit topology, having all sets $(a,b]$ as basis
- $\mathcal{T}_5=$ the topology having all sets $(-\infty,a)$ as basis
$\mathbb{R}_K$ is the topology on $\mathbb{R}$ with basis elements $(a,b)$ and $(a,b)-K$ where $K=\left\{\frac{1}{n}:n\in \mathbb{N}\right\}$.
If you do not want to read all that is below, I am writing conclusion :
- $\mathcal{T}_3\subsetneq\mathcal{T}_1\subsetneq \mathcal{T}_2\subsetneq\mathcal{T}_4$
- $\mathcal{T}_5\subsetneq\mathcal{T}_1\subsetneq \mathcal{T}_2\subsetneq\mathcal{T}_4$
- $\mathcal{T}_3$ and $\mathcal{T}_5$ are not comparable.
Let $x\in (a,b)$ a basis element in standard topology.. This is also a basis element in $\mathbb{R}_K$ so we have $x\in (a,b)\subset (a,b)$.. So, $\mathcal{T}_2$ is finer than $\mathcal{T}_1$..
Let $0\in (a,b)-K$ a basis element in $\mathcal{T}_2$, for any basis element $U$ in $\mathcal{T}_1$ we have
$\epsilon>0$ such that $0\in (-\epsilon,\epsilon)\subset U$. There exists $N\in \mathbb{N}$ such that $\frac{1}{N}<\epsilon$.
So, we can not have $0\in U\in (a,b)-K$ as $\frac{1}{N}\in U$ but $0\notin \ (a,b)-K$.
So, $\mathcal{T}_2$ is finer than $\mathcal{T}_1$ but not the other way.
Let $x\in (a,b)$ a basis element in $\mathcal{T}_1$. We can not find $ \{x_1,\cdots,x_n\}$ such that
$x\in \mathbb{R}\setminus \{x_1,\cdots,x_n\}\subset (a,b)$ as $(a,b)$ is a bounded set where as
$ \mathbb{R}\setminus \{x_1,\cdots,x_n\}$ is unbounded.
Let $x\in U=\mathbb{R}\setminus \{c_1,c_2,\cdots,c_n\}$ a bais element in $\mathcal{T}_3$. We have $1\leq i\leq n$ least such that
$x<c_i$ then, $x\in (c_{i-1},c_i)\in \mathbb{R}\setminus \{c_1,c_2,\cdots,c_n\}$ if $i>1$
and $x\in (x-1,c_1)$ if $i=1$. In any case we have a basis element in $\mathcal{T}_1$ containing $x$ and contained
in $U$.
So, $\mathcal{T}_1$ is finer than $\mathcal{T}_3$ but not the other way.
Let $x\in (a,b)$ a basis element in $\mathcal{T}_1$ then we have $x\in (a,x]$ a basis element in
$\mathcal{T}_4$ such that $x\in (a,x]\subset (a,b)$.
For $x\in (a,x]$ a basis element in $\mathcal{T}_4$ we can have no $(c,d)$ such that $x\in (c,d)\subset (a,x]$
as if $d<x$ then $x\notin (c,d)$.
So, $\mathcal{T}_4$ is finer than $\mathcal{T}_1$ but not the other way.
Let $x\in (a,b)$ a basis element in $\mathcal{T}_1$ then there can be no open set $(-\infty,a)$ containing $x$
and contained in $(c,d)$ as $(c,d)$ is bounded where as $(-\infty,a)$ is unbounded.
Let $x\in (-\infty,a)$ a basis element in $\mathcal{T}_5$ then we have $x\in (x-1,a)$ a basis element
in $\mathcal{T}_1$ and $(x-1,a)\subset (-\infty,a)$..
So, $\mathcal{T}_1$ is finer than $\mathcal{T}_5$ but not the other way.
$\mathcal{T}_2$ is finer than $\mathcal{T}_3$ but not the other way clear.
Let $x\in (a,b)$ a basis element in $\mathcal{T}_2$ then we have $x\in (a,x]$ a basis element in
$\mathcal{T}_4$ such that $x\in (a,x]\subset (a,b)$.
Let $x\in (a,b)-K$ another basis element in $\mathcal{T}_2$. We can assume that $(a,b)\cap K\neq \emptyset$. We then have $\frac{1}{n+1}<x<\frac{1}{n}$ for some $n$ then we have $x\in (\frac{1}{n+1},\frac{1}{n})\subset (a,b)-K$.
Let $x\in (a,x]$ a basis element in $\mathcal{T}_4$. We know that $\mathcal{T}_1$ is not finer than $\mathcal{T}_4$.. So, if at all $\mathcal{T}_2$ is finer than $\mathcal{T}_2$ we must have $x\in (a,b)-K\subset (a,x]$ which would again mean that $x<b$ and $b<x$.. Contradiction.
So, $\mathcal{T}_4$ is finer than $\mathcal{T}_2$ but not the other way.
$\mathcal{T}_2$ is finer than $\mathcal{T}_5$ but not the other way. Clear. $\mathcal{T}_4$ is finer than $\mathcal{T}_3$ but not the other way. Clear.
$\mathcal{T}_3$ is not finer than $\mathcal{T}_5$ is clear.
Let $x\in \mathbb{R}\setminus \{x_1,x_2,\cdots,x_n\}$ (wlog we can assume that $x_1<x_2$) and that $x\in (x_1,x_2)$
then any open set containing $x$ of the form $(-\infty,a)$ contains $x_1$
as $x_1<x$. So, $\mathcal{T}_3$ is not finer than $\mathcal{T}_5$.
$\mathcal{T}_4$ is finer than $\mathcal{T}_5$ but not the other way. Clear.
Let me know if these justifications are correct…
Best Answer
In general you have the right ideas, and good on you for showing proper work! Some remarks:
You do not need to consider the topologies 1 and 4 separately, this will already fall out ot considering 1 and 2 and 2 and 4. If we have strict inclusion there, then a fortiori we will have it for 1 and 4, no need for a separate proof. Similary with 2 and 5, e.g.
What about 4 and 5? You just say "clear" but some argument seems in order.
You seem to assume that $x_1 < x_2$, because the open interval $(x_1,x_2)$ is non-empty (in 3 vs 5)? A bit sloppy.