$E=\{x|x=\sum_{x=1}^{\infty}\frac{a_k}{10^k}, a_k\in\{3,7\}\} $, find E' $$E= \left\lbrace x : x=\sum_{x=1}^{\infty}\frac{a_k}{10^k}, a_k \in \left\lbrace 3,7\right\rbrace \right\rbrace $$
Find $E'$ where $E'$ is the set of all accumulation points of $E$.
I think $E'=\emptyset$ because there must be another rational number between any element in E
According to the comments, I realized there must be accumulation points, any other hints that I can earn?
 A: Claim:$\;E'=E$.

Here's an outline of a proof (you'll need to fill in a few details) . . .

Let $x\in E$.

Then we can write 
$$x=\sum_{k=1}^{\infty}\frac{a_k}{10^k}$$
where $a_k \in \{3,7\}$ for all $k$.

For each positive integer $n$, choose $y_n\in E$ such that
$$y_n=\sum_{k=1}^{\infty}\frac{b_k}{10^k}$$
where $b_k=a_k$ for $1\le k\le n$, but $b_m\ne a_m$ for some $m > n$.

Then it's easily seen that the sequence $y_1,y_2,y_3,...$ is an infinite sequence of elements of $E\,{\setminus}\{x\}$ converging to $x$, hence $x\in E'$.

It follows that $E\subseteq E'$.

For the reverse inclusion, let $z\in E'$.

We can't have $z\le 0$, since the least element of $E$ is ${\large{\frac{1}{3}}}$, and we can't have $z\ge 1$, since the greatest element of $E$ is ${\large{\frac{7}{9}}}$.

Thus we must have $z\in (0,1)$.

Suppose $z\notin E$.

Our goal is to derive a contradiction.

Let $z$ have the decimal expansion 
$$z=\sum_{k=1}^{\infty}\frac{g_k}{10^k}$$
where $g_k \in \{0,...,9\}$ for all $k$, and no tail of the expansion has all digits equal to $9$.

Let $n$ be the least positive integer such that $g_n\notin\{3,7\}$. Such an $n$ must exist else we would have $z\in E$.

Define elements $p,q,r,s\in E$ as follows . . .


*

*$p={\displaystyle{\sum_{k=1}^{\infty}\frac{a_k}{10^k}}}$, where $a_k=g_k$ if $k < n$, and $a_k=3$ if $k\ge n$.$\\[4pt]$

*$q={\displaystyle{\sum_{k=1}^{\infty}\frac{b_k}{10^k}}}$, where $b_k=g_k$ if $k < n$, $b_n=3$, and $b_k=7$ if $k > n$.$\\[4pt]$

*$r={\displaystyle{\sum_{k=1}^{\infty}\frac{c_k}{10^k}}}$, where $c_k=g_k$ if $k < n$, $c_n=7$, and $c_k=3$ if $k\ge n$.$\\[4pt]$

*$s={\displaystyle{\sum_{k=1}^{\infty}\frac{d_k}{10^k}}}$, where $d_k=g_k$ if $k < n$, and $d_k=7$ if $k\ge n$.


It's clear that $p < q < r < s$.


*

*If $z < p$, then $p$ is the closest element of $E$ to $z$.$\\[4pt]$

*We can't have $p < z < q$, else $d_n=3$.$\\[4pt]$

*If $q < z < r$, then one of $q,r$ is the closest element of $E$ to $z$.$\\[4pt]$

*We can't have $r < z < s$, else $d_n=7$.$\\[4pt]$

*If $z > s$, then $s$ is the closest element of $E$ to $z$.$\\[4pt]$
In all valid cases, there a closest element of $E$ to $z$, which contradicts $z\in E'$.

It follows that $E'\subseteq E$.

Thus we have both inclusions, hence $E'=E$, as claimed.
