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 $B(0,\delta)\subset U_{\lambda_0}$
There exists some $n_0$ such that $1/n_0 < \delta$. Hence $\{\frac1n:n \ge n_0\} \subset U_{\lambda_0}$. That leaves only finitely many other points that need to be covered.
The set $A$ is definitely bounded. The only limit point it is missing is the point $0$, so $A \cup 0$ is both closed and bounded. By the Heine-Borel theorem, that implies that $A \cup 0$ is compact.
Sanity check: Since $\lim_{n\rightarrow\infty} \frac{1}{n}=0$ an open cover for $0$ covers all but finitely many terms in $B$. Therefore, for any cover of $B$, pick the open set that covers $0$. This set will cover all but finitely many elements of $B$. Then pick the covers of those finitely many elements. The collection is finite, hence for any cover of $B$ you can find a finite subcover. Hence $B$ is compact.
The following fact is useful
A subset of R is compact iff it is closed and bounded.
