We say a set is perfect if it is closed and every point is limit point.
The question is from baby Rudin$(2.18)$. I understood the construction given in solution manual by J. D. Taylor. I have constructed two examples and wanted to verify if they are correct.
$1.$ Before giving first example, I introduce one terminology.
Let $S\subset \mathbb{R}$. Let $x\in S$ but is not a limit point of
$S$. So we can find $\epsilon>0$ such that $B(x,\epsilon) \cap S=x$.
Then we choose $n_1\in \mathbb{N}$ big enough so that
$x+\frac{1}{n_1}\in B(x,\epsilon)$. Then $\forall k>n_1, k\in
\mathbb{N}$, we have $x+\frac{1}{k}\in B(x,\epsilon)$. These elements
$\{x+\frac{1}{k}:k\geq n_1\}$, we will call neighbouring elements of
x and this process of 'choosing appropriate $\epsilon, n_1$ for
such $x$ and finding it's neighbouring elements', we will call
"setting up neighbourhoood for x with respect to S".
Now let $S_1=\sqrt{2}\cup \{\sqrt{2}+\frac{1}{n}:n\in \mathbb{N}\}$. Only $\sqrt2$ is limit point of $S_1$. So we 'set up neighbourhood with respect to $S_1$' for all other elements and let denote that entire collection(i.e. $S_1$ with all those neighbouring elements) by $S_2$. Now, all numbers present in $S_1$ are limit points of $S_2$. But we have whole new bunch of numbers in $S_2$ which are not limit points of $S_2$. So we do same with $S_2$ what we did for $S_1$(but here we set up neighbourhood with respect to $S_2$)and call it $S_3$. Let
$S=\cup_{n=1}^{\infty}S_n$. All numbers in $S$ are irrationals as they all are of the form $\sqrt2 +r$, for some rational $r$. Each number will be limit point because if that number is in $S_i$, in the next step, i.e. in $S_{i+1}$, we set up neighbourhood for it. But I am bit skeptical about the closeness of $S$. Actually inspecting my construction closely, I think that each $S_i$ is countable, and thus $S$ is countable but we know that perfect set is uncountable, so I think my this example won't work. But I would like to know your thoughts on this. Like can I somehow make this perfect doing some small changes. And if $S$ is not closed, I tried to find it's limit points which are not in $S$, but I am not seeing it straightaway.
$2$. The idea behind this construction is using previous exercise. Let $E$ be the set of all $x\in[0,1]$, whose decimal expansion contains only the digits $4$ and $7$. I know this set is perfect. The only problem here is that $E$ contains rationals too. Suppose we translate $E$ by some irrational $y$ in such a way that $y+E$ contains irrationals only, then we are done I guess. We can find such $y$, one such is $y\in(0,1)$ having only $0$ and $1$ in it's decimal expansion. Am I correct?
Thanks.
Best Answer
Let's tweak the $4,7$ example a bit. Consider decimal expansions of the form
$$.\text{_}0\text{_}00\text{_}000\text{_}0000\text{_}00000\text{_} \dots$$
where each $\text{_}$ is either a $4$ or $7.$ Denote the set of all numbers having such a decimal expansion by $E.$ Then $E$ is perfect.
Let $x\in E.$ If $x$ were rational, then its decimal expansion would be repeating after some point. But that can't happen because of the arbitrarily large stretches of $0$'s that occur in the expansion of $x.$
Therefore $E$ is a perfect set with no rationals.