How to show $A$ is compact in $\Bbb{R}$ with standard topology? Let $A = \{0\} \cup \{ \frac{1}{n}| n \in \Bbb{N}\} \subseteq \Bbb{R}$. I am struggling with showing whether $A$ is compact or not. I do not even know how to start.
Is $u=\{0\} \cup \left\{\left(\frac{1}{n+1}, \frac{1}{n-1}\right)| n \in \Bbb{N}\right\} \cap \{(1/2, 2)\}$ an open cover of A with no finite subcover? If so, what is the next step?
 A: Hint : Compacts in $\mathbb{R}$ are precisely the closed and bounded sets. $A$ is clearly bounded. What about $L=\lim_{n}\frac{1}{n}$ ? Does $L\in A$ ?
A: This is a nice exercice for practicing with definitions.
So $X$ is compact if EVERY open cover has a finite sub cover. In order tho prove that $A$ is compact, you then have to start with a generic open cover of $A$ (you cannot choose the cover because the proof has to work for every cover). Say $U=\{U_i\}_{i\in I}$ is an open cover. Then there is at least one of the $U_i$, say $U_{i_0}$ which contains $0$. Since $U$ is open for the standard topology, and contains $0$, it must also contain an interval $(-\epsilon,\epsilon)$, which contains all but finitely many points of $A$. (Precisely, it contains all points $1/n$ for wich $n>1/\epsilon$). For any of those finitely many points $1/n$ left outside $U_{i_0}$ there is an $U_{i_n}$ containing it. So the family formed by $U_0$ and those finitely many $U_{i_n}$ is a finite cover of $A$.
A: The given set is $A=\{0\} \cup \big\{ \frac1n : n \in \mathbb{N} \big\}$
To show that $A$ is compact set in $\mathbb{R}$, we need to show that any open cover of $A$ has a finite subcover.
Consider any open cover of $A= \bigcup_{i \in I} U_i$. Then since $0 \in A$, then for some $i_0 \in I, 0 \in U_{i_0}$. since $U_{i_0}$ is an open set. Hence there exists $\epsilon>0$ such that $(-\epsilon,\epsilon ) \subset U_{i_0}$.
Archimedean Property

For any $\epsilon >0$, there exists $n \in \mathbb{N}$ such that $\frac1n < \epsilon$

Using the Archimedean Property, choose an $n_0$ such that $\frac{1}{n_0} < \epsilon$. Thus for $n \ge n_0$ we have $\frac1n \le \frac{1}{n_0} < \epsilon$. Thus all but finite number of elements of $A$ are contained in $(-\epsilon,\epsilon)$. Thus they are all contained in $U_{i_0}$. Now choose a finite number of open sets from $U_i$ to create a finite subcover of $A$
A: If you can use the Heine-Borel theorem is easy to see that the set of acumulation points $A' =\{0\} \subseteq A$ therefore $A$ is closed and observe that $\forall x \in A$ we have that $ |x - 1| \leq 1$ hence $A$ is bounded. So $A$ is a closed and bounded set, by Heine-Borel theorem we can conclude that $A$ is compact.
