Let $A = \{ \frac{1}{n} : n \in \Bbb{N} \}$, show that $\overline{A} = A \cup \{0\}$ 
Let $A = \{ \frac{1}{n} : n \in \Bbb{N} \}$, show that $\overline{A} = A \cup \{0\}.$

We have $A \subseteq \overline{A}$ by the definition of closures. To show that $\{0\} \subset \overline{A}$ we need to show that for every open set $U$ containing $0, U \cap A \not= \emptyset$. The elements of $A$ converge to $0$, since $\forall \epsilon > 0 \  \exists N \in \Bbb{N}, \ s.t \ \forall n > N \ \frac{1}{n} \in B_{\epsilon}(0).$ So that any open set $U$ containing $0$, we have an $\epsilon$ s.t $0 \in B_{\epsilon}(0) \subseteq U $. And since $B_{\epsilon}(0)$ belongs to $A$, we have that $A \cap U \not= \emptyset \ \forall \  U.$
This establishes that $A \cup \{0\} \subseteq \overline{A}$. 
I'm not too sure how to show this direction $\overline{A} \subseteq A \cup \{0\}$ .
 A: It suffices to show that $\overline{A}-A=\{0\}$.
To show that $\overline A-A \subseteq \{0\}$ Suppose there exists some $a \in \overline{A}-A$ so that $a \neq 0$.
Suppose $a<0$. Then $\mathcal{B}(a,\frac{a}{2})$ shows a contradiction.
Supposes $a\geq 1$, and you will find a contradiction for the same reason.
Finally, suppose that $a \in (0,1)$. Then there exists $n \in \mathbb{N}$ so that $\frac{1}{n+1}<a<\frac{1}{n}$. Well, define an open ball $\frac{1}{2n^2+2n}$ around $a$ to show a contradiction.
There are slicker ways to reach this conclusion, but this is the most "Straightforward." (( for example, $\{\frac{1}{n}\}$ is a convergent sequence, so you can say something about any of its convergent subsequences )
A: Here is another way to do it which I find easier:
Since $\overline{A}$ is defined as the set $A$ and all its limit points, by definition $\overline{A}=A\cup L$, where $L$ is the set of limit points of $A$.  
Now you only need to show that $L=\{0\}$, i.e. that $A$ only has only $0$ as its limit point. The result then follows immediately.
A: Let $ a_n $ be a convergent sequence with its terms in $ \bar{A} $. If it has finitely many distinct elements, then it has to become constant after a certain index, and therefore its limit is in $ \bar{A} $. Otherwise, the sequence $ a_k $ contains arbitrarily small elements of $ A $, so there is a subsequence $ b_k $ converging to $ 0 $. Then, we have $ \lim_{n\to \infty} a_n = \lim_{n\to \infty} b_n = 0 $ as a sequence and its subsequences cannot converge to different limits. As every sequence with its terms in $ \bar{A} $ has its limit in $ \bar{A} $, the result follows.
