Problem 7, section 13 in Munkres' Topology, 2nd edition: which topology contains which? Consider the following topologies on $\mathbb{R}$: 


*

*$ \mathscr{T}_1 = $ the standard topology.  

*$ \mathscr{T}_2 = $ the topology of $\mathbb{R}_K$ with the collection of all open intervals $(a,b)$ and all sets of the form $(a,b) \setminus \{1, 1/2, 1/3, \ldots\}$ as  basis.  

*$ \mathscr{T}_3 =$  the finite complement topology.    

*$ \mathscr{T}_4 = $ the upper limit topology, having all sets $(a,b]$ as basis.

*$ \mathscr{T}_5 = $ the topology having all sets $(-\infty, a) = \{x | x< a \}$ as basis.
Then which of these topologies contains which (not to mention trivial self-inclusions of course)? 
I think $\mathscr{T}_1$ contains $\mathscr{T}_3$ and $\mathscr{T}_5$, whereas $\mathscr{T}_1$ is itself contained in $\mathscr{T}_2$ and $\mathscr{T}_4$. 
I also think that $\mathscr{T}_2$ and $\mathscr{T}_4$ are incomparable. 
Am I right? 
What is the relation between $\mathscr{T}_3$ and $\mathscr{T}_5$? 
Which other inclusions are possible? 
 A: You have correctly placed $\mathscr{T}_1$ with respect to the other four topologies. $\mathscr{T}_2$ actually is contained in $\mathscr{T}_4$. You already know that each open interval $(a,b)$ is in $\mathscr{T}_4$, so consider a set $U=(a,b)\setminus K$, where $K=\left\{\frac1n:n\in\Bbb Z^+\right\}$. Let $x\in U$. If $(a,x)\cap K=\varnothing$, then $(a,x]$ is a $\mathscr{T}_4$-nbhd of $x$ contained in $U$. If $(a,x)\cap K\ne\varnothing$, there is a smallest $n\in\Bbb Z^+$ such that $\frac1n\in(a,x)\cap K$, and in that case $\left(\frac1n,x\right]$ is a $\mathscr{T}_4$-nbhd of $x$ contained in $U$. Thus, every $x\in U$ has a $\mathscr{T}_4$-nbhd contained in $U$, and there $U\in\mathscr{T}_4$.
$\mathscr{T}_3$ is the smallest possible $T_1$ topology on $\Bbb R$, so $\mathscr{T}_3$ is contained in every $T_1$ topology on $\Bbb R$, and in particular in $\mathscr{T}_1,\mathscr{T}_2$, and $\mathscr{T}_4$. Moreover, every topology on $\Bbb R$ that contains $\mathscr{T}_3$ is a $T_1$ topology. Thus, $\mathscr{T}_3\nsubseteq\mathscr{T}_5$. But the only member of $\mathscr{T}_5$ with a finite complement is $\Bbb R$, so $\mathscr{T}_5\nsubseteq\mathscr{T}_3$: the two topologies are incomparable.
Clearly $\mathscr{T}_5\subseteq\mathscr{T}_1$, and you already know that $\mathscr{T}_1\subseteq\mathscr{T}_2$ and $\mathscr{T}_1\subseteq\mathscr{T}_4$, so $\mathscr{T}_5\subseteq\mathscr{T}_2$ and $\mathscr{T}_5\subseteq\mathscr{T}_4$
A: For reference:
\begin{equation*}
    \begin{matrix}
 & \tau_1       & \tau_2 & \tau_3 & \tau_4 & \tau_5\\ 
\tau_1 & =       & \supset & \subset & \supset & \subset\\ 
\tau_2 & \subset & = & \subset & \supset & \subset\\ 
\tau_3 & \supset & \supset & = & \supset & X\\ 
\tau_4 & \subset & \subset & \subset & = & \subset\\ 
\tau_5 & \supset & \supset & X & \supset & =
\end{matrix}
  \end{equation*}
The j-th collumn is contained($\subset$) / contains($\supset$) / is equal($=$) /is incomparable($X$) with the i-th row where all set inclusions are proper.
