I am reading the section on compactness from Munkres their is a certain part I don't understand.
Consider the following subspace of $\Bbb R$:
$X = \{0\} \cup \{\frac{1}{n} : n \in \mathbb{Z^{+}} \}$
Given an open covering $A$ of $X$, there is an element $U$ of $A$ containing $0$. The set contains all but finitely many of the points $\frac1n$; (Why is that ?) Is it because $0$ is a convergent point for the sequence $\frac1n$ ?
On
Yes. Assume we are using the standard topology on $\mathbb R$. As $A$ is an open covering, we know that $U$ is a neighbourhood of $0$, so there exist $a,b > 0$ such that $(-a, b) \subseteq U$. Let $N = \lceil 1/b \rceil - 1$, a nonnegative finite integer. Then $(-a, b)$ can only fail to cover at most $N$ points, namely: $$ 1, \frac{1}{2}, \ldots, \frac{1}{N} $$
The basis open sets of $\mathbb{R}$ under the standard topology are the open intervals of $\mathbb{R}$. So if $U$ is an open set with $0\in U$, by definition of a basis for a topology, there exists a basis open set (an open interval) of the form $(a,b)$ with $0\in (a,b)\subseteq U$. So since $0\in (a,b)$, we have $a<0<b$. By the Archimedian property of $\mathbb{R}$ there exists $n\in\mathbb{Z}_{>0}$ s.t. for every $m\geq n$, $0 < \frac{1}{m} < b$, so only finitely many of the $\frac{1}{k}$ for $k\in\mathbb{Z}_{>0}$ can lie outside of $(a,b)$.