# Baby Rudin Problem Chapter 2, Problems 17(c) and (d)

Let $E$ be the set of all $x \in [0,1]$ whose decimal expansion contains only the digits $4$ and $7$. Then I've managed to show that (a) $E$ is not countable, and (b) $E$ is not dense in $[0,1]$.

How to determine if $E$ is closed (and hence compact)?

And, if closed, then how to determine if $E$ is perfect?

My effort:

Suppose $y \in [0,1] - E$, and let $n$ be the smallest natural number such that the $n$th digit after the decimal point $d_n$ in the decimal expansion of $y$ is neither $4$ nor $7$.

Let us take $\epsilon$ such that $$0 < \epsilon < \frac{\min (|d_n-4|, |d_n-7|)}{10^{n+1}}.$$ Then is it true that this $\epsilon$-neighborhood of $y$ contains no point of $E$? If so, then how to show this fact rigorously? If this holds, then of course no point $y$ not in $E$ can be a limit point of $E$, implying that $E$ is closed.

To see if $E$ is perfect, let $x$ be an arbitrary element of $E$, and let the decimal expansion of $x$ be as follows: $$x \colon = \sum_{i=1}^\infty \frac{d_i}{10^i},$$ where each $d_i$ is either $4$ or $7$. Let $\epsilon > 0$ be arbitrary. Let's choose $n$ such that $$\frac{4}{10^{n}} < \epsilon.$$ Now we can find an element $y$ of $E$ whose decimal expansion is given by $$y \colon = \sum_{i=1}^\infty \frac{d^\prime_i}{10^i},$$ where $$d^\prime_i \colon= \begin{cases} d_i \mbox{ if } i \neq n; \\ 4 \mbox{ if } i =n \mbox{ and } d_n = 7; \\ 7 \mbox{ if } i=n \mbox{ and } d_n = 4.\end{cases}$$ Then evidently, we have $$d(x,y) = \frac{3}{10^n} < \epsilon,$$ and moreover $y \in E$ and $y \neq x$. So $x$ is a limit point of $E$. Thus every point of $E$ is a limit point.

Now is $E$ closed?

## 1 Answer

Yes $E$ is closed.

Take $y \in [0,1] - E$, then for any $x\in E$, suppose $x$ and $y$ has the first different decimal at the position of $\dfrac{1}{10^m}$, then we can show $|x-y| \geq \dfrac{1}{3\cdot 10^{m+1}}$. Because the largest compensation coming from all the later decimanls is $\sum_{k=m+1}^{+\infty}\dfrac{6}{10^{k}} = \dfrac{2}{3\cdot 10^m}$

Now let $n$ be the smallest natural number such that the nth digit after the decimal point in the decimal expansion of $y$ is neither $4$ nor $7$, as you defined, then we see $m \leq n$, then $$|x-y| \geq \dfrac{1}{3\cdot 10^{m+1}}\geq \dfrac{1}{3\cdot 10^{n+1}},\forall x\in E$$

• thank you so much, but I guess you might wanna edit your answer for possible (typing or other) mistakes. Or, you might like to add some more detail. As it is, it's not fully comprehensible to me. – Saaqib Mahmood Nov 4 '14 at 12:41
• @SaaqibMahmuud which part is not fully comprehensible to you? – Petite Etincelle Nov 4 '14 at 13:04