If $E = \bigcup_{n = 1} ^ \infty \left(\frac{1}{n+1}, \frac{1}{n}\right) $. Show $\bar E = [0,1]$ is the closure of $E$ If  $E = \bigcup_{n = 1} ^ \infty \left(\frac{1}{n+1}, \frac{1}{n}\right) $.
Show $\bar E = [0,1]$ is the closure  of $E$
Attempt: Since $ (\frac{1}{n+1}, \frac{1}{n}) $ is a subset of [0,1], so $E \subset [0,1]$. Suppose 
 $a\in [0,1]$, where $a\in E$. Choose $\delta > 0$ such that $a > 0$, so $\frac {1}{n+1}< a < \frac{1}{n}$. So there is an open ball having $a$ such that $(\frac{1}{n+1}, \frac{1}{n}) \cap E = \emptyset $.
I don't know how to continue. Please can someone please help? How can I show $[0,1]$ is the closure?
Thank you in advance.
 A: First, lets notice that
$$E=(0,1]\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$$
By double contention:
$$E\subseteq (0,1]\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$$
be $x\in (0,1)\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$, let be $n$ such as $\frac{1}{n+1}<x<\frac{1}{n}$ thus $x\in E$.
$$(0,1]\backslash \left\{ \frac{1}{n} \right\}_{n\geq1} \subseteq E$$
be $x\in E$, then, $\exists n$ such as $\frac{1}{n+1}<x<\frac{1}{n}\Rightarrow x\in(0,1]\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$.
And now its easy to see the closure.
Be $a\in [0,1]$, if a is in $E$ we're done, if not, let be $\epsilon>0$, then $B_\epsilon \cap E\neq \emptyset$.
If $a\notin [0,1]$ let $\epsilon <|a-[0,1]|$, then $B_\epsilon\cap E = \emptyset$. $a$.
A: Notice first that
$$
\left(\frac{1}{n+1},\frac1n\right)\subset [0,1]  \quad \forall n\ge 1,
$$
we have
$$
E=\bigcup_{n\ge 1}\left(\frac{1}{n+1},\frac1n\right)\subset [0,1],
$$
and therefore
$$
\overline{E}\subset \overline{[0,1]}=[0,1].
$$
Since
$$
\left(\frac{1}{n+1},\frac1n\right)\subset E  \quad \forall n\ge 1,
$$
we have
$$
\left[\frac{1}{n+1},\frac1n\right]=\overline{\left(\frac{1}{n+1},\frac1n\right)}\subset \overline{E} \quad \forall n\ge 1.
$$
It follows that
$$
(0,1)\subset (0,1]=\bigcup_{n\ge1}\left[\frac{1}{n+1},\frac1n\right]
\subset \overline{E}\subset [0,1],
$$
i.e. $\overline{E}$ is a closed set containing $(0,1)$. But $[0,1]$ is the smallest closed set containing $(0,1)$, thus $[0,1]\subset \overline{E}$. This shows that $\overline{E}=[0,1]$.
