I know that $$A=\{ \frac{1}{n}: n \,\epsilon \, \mathbb N \}\, \subseteq\, \mathbb R $$ is not compact
However, I am confused why $$A\, \cup\, \{0\}$$ is compact.
My attempt at understanding:
Let $B=A \cup \{0\}$ and suppose $U_n$ is an open cover for $B$, then there exists a $U_{\lambda_0}$ such that $0\, \epsilon\, U_{\lambda_0} $. Since $U_{\lambda_0} $ is open there exists a $\delta\, \gt 0$ such that $B(0,\delta)\subset U_{\lambda_0}$