How to show given set Compact and Perfect?

Question

Let $E$={$x\in [0,1]$| every decimal digit is 4 and 7}.show that it is closed and perfect?

My attempt:
$t\in E'$ so $\forall \epsilon >0 ,\exists y\in E $ , $ d(y,t)<\epsilon$
Claim :$t\in E$

On contary assume that $t\notin E $ that is it has digits other that 4 and 7.
I wanted to show this is contradication .
But how to argue using metric that i don't know.

To show it is perfect.We have to show that no point is isolated, as already we have closedness .
On contrary say $m\in E$ is isolated point so there exist $r>0 $ such that $(m-r,m+r)$ contain no point of E. That there is no point other than m in it which contain 4 and 7 .It is intutionaly I know not possible because Rational are dense.But i don't know how to write above in a rigorous way.Any Help will be appreciated.
Any help will be appreciated

Answer 1

Your approach is valid but has led to a wall. This can be overcome but I'll offer an alternate approach. Let $x\in E$ and show that it is a limit point. This can be done by writing $x=0.x_1x_2x_3...$ where $\{x_i\}_{i=1}^\infty\subset\{4,7\}$ is a sequence. Now construct a sequence $\{y_i\}_{i=1}^\infty\subset E$ that gets arbitrarily close to $x$ by modifying the decimal positions of $y_i$.

Answer 2

$E$ is closed

It is equivalent to prove that $E$ complement in $[0,1]$ is open. So take $x \in [0,1] \setminus E$.

Let's name $x \equiv 0,x_1 x_2 \dots x_n \dots$ the digits of $x$.

By hypothesis, one of the digits of $x$ is not in $\{4, 7\}$. Let's say the $n$-th digit and name $\overline{x_n} \equiv 0,x_1 x_2 \dots x_n 0 \dots$. Then all the reals of the open interval $(\overline{x_n} - 10^{-n-1}, \overline{x_n}+ 10^{-n-1})$ have at least a digit not in $\{4, 7\}$: the $n$-th one is equal to $x_n$ for $y \in [\overline{x_n}, \overline{x_n}+ 10^{-n-1})$ and to $x_n-1 \notin \{4,7\}$ for $y \in (\overline{x_n}-10^{-n-1}, \overline{x_n})$.

This proves that $E$ is closed.

$E$ has no isolated point

Take $x \equiv 0,x_1 x_2 \dots x_n \dots \in E$, which means that all the $x_i$'s are in $\{4,7\}$. Then the sequence of reals $(X_m)$ where all the digits of $X_m$ are equal the one of $x$ except the $m$-th one, which is equal to $4$ if $x_m=7$ or to $7$ if $x_m=4$ is a sequence of distinct elements of $E$ which converges to $x$. Hence $E$ has no isolated point.