Prove that there exists a unique topology $T$ for which $ P $ is its subbasis and that topological space $ ( T, \mathbb{N ^{*}})$ is metrisable . Let  $ \infty = \left\{ \mathbb{N} \right\}$ and $ \mathbb{N ^{*}} = \mathbb{N} \cup \left\{ \infty \right\}$. Let $$ P = \left\{ \left\{1\right\} \right\} \cup 
\left\{ \left\{n, n+1 \right\} \mid n \in \mathbb{N} \right\} \cup
\left\{ \left\{ k \in \mathbb{N} \colon k \geq 2n \right\} \cup \left\{ \infty \right\} \mid n \in \mathbb{N} \right\} $$
Prove that there exists a unique topology $T$ for which  $ P $ is its subbasis and that topological space $ ( T, \mathbb{N ^{*}})$ is metrisable . 
Proving that there exists a unique topology should be simple enough by showing that $P$ covers $ \mathbb{N ^{*}}$ . But as for the metrisability I am thinking maybe it can be metrisable with discrete metric, but I can't prove that $ \left\{ \left\{ \mathbb{N} \right\} \right\}$ is open ( all other sets $ \left\{ n \right\} $ are open)
 A: This topology is basically the one point compactification of $\mathbb N$, which is homeomorphic to $X=\{0\}\cup\{\frac{1}{n}\}_{n\in \mathbb N}$ in the usual metric topology via the map 
\begin{align}& n\mapsto\frac{1}{n},  \quad n\in \mathbb N \\ &\infty\mapsto0 \end{align}
It is easy to prove that every element of $P$ is open in the metric topology. We consider the reverse direction. Let $U$ be an open set (in the metric topology) and $x\in U$. We want to find an open set $V$ (in the topology generated by $P$) such that $x\in V\subset U$. 
If $x\neq0$, then $x=\frac{1}{n}$ for some $n$. we can let $V=\{\frac{1}{n}\}$. If $n=1$, then $V$, being an element of $P$, is open. If $n\neq1$, notice $$V=\left\{\frac{1}{n-1},\frac{1}{n}\right\}\bigcap\left\{\frac{1}{n},\frac{1}{n+1}\right\},$$ which is the intersection of two elements of $P$, hence open. 
If $x=0$, there exists $n\in \mathbb N$ such that $0\in[0,\frac{1}{n}]\cap X\subset U$ ($\because$every open set in the metric topology is a union of open balls). Without loss of generality, we can assume $n$ is even. Then we can simply let $V=[0,\frac{1}{n}]\cap X$. Again, $V$ being an elements of $P$, is open. 
