clean up deprecated proof of topology

spaces/S000044/README.md

@@ -15,19 +15,3 @@ and {S194}.
 
 Defined as counterexample #52 ("Nested Interval Topology")
 in {{doi:10.1007/978-1-4612-6290-9}}.
-
-<!-- [[Proof of Topology]]
-1. Note that $X \in N$ by definition and $U_1=(0,1-\frac{1}{1})=(0,0)= \emptyset$ thus $\emptyset \in N$.
-
-2. Let $I$ be any indexing set of $n \in \mathbb{N}$ where $\bigcup\limits_{n \in I} U_n$ where $U_n \in N$. Thus
-\begin{align*}
-\bigcup\limits_{n \in I} U_n &= U_{max(n) \in I}, or X
-\end{align*}
-Which are both open.
-
-3. Let $I$ be any finite indexing set of $n \in \mathbb{N}$. Thus
-\begin{align*}
-\bigcap\limits_{n \in I} U_n &= U_{min(I)}
-\end{align*}
-which is an open set so any finite union of open sets is open.
- -->