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 $\overline{A} \subseteq A \cup \{0\}$ .

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.
