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

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.

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.

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 )

• Hence the assumption that $a \in \overline{A}-A$. – Andres Mejia May 28 '16 at 22:34