How to prove that the set of irrationals between $0$ and $1$ whose decimal expansion contains only $4$'s and $7$'s is closed?

Question

I'm following Baby Rudin and completed exercise 2.17, which states that the set of points, $E$, between $0$ and $1$ whose decimal expansion contains only $4$'s and $7$'s is perfect. Exercise 2.18 asks you to find a perfect set containing only irrationals.

I am familiar with the Cantor set and know that you can use something similar to it to prove this, but all proofs I've seen use the concept of measure, which I haven't learned yet and would like to avoid using.

I figured that the set of irrationals in $E$, $E_1 = E \setminus \mathbb{Q}$, is also perfect, which someone else claimed, too, in this answer.

However, the only part I can't prove is that $E_1$ is closed. Intuitively, it does not make sense to me: for any rational $p \in E$, and any $\epsilon > 0$, can't we find an irrational in $E_1$ whose decimal expansion agrees with enough digits of $p$ that it is epsilon-close to $p$, making a rational (which is not a member of $E_1$) a limit point of $E_1$, meaning $E_1$ is not closed and therefore not perfect?

Does this mean my proof is wrong and $E_1$ is actually not perfect, or am I missing something?

Answer 1

You cannot prove it since it is false. Consider the number $a=0.4444\dots\in E$. For each $n\in\Bbb N$, let $a_n\in(0,1)$ be the number whose decimal expansion after the dot consists of $n$ $4$'s, one $7$, two $4$'s, one $7$, three $4$'s and so on; for instance,$$a_3=0.44474744744474\ldots$$Each $a_n$ is irrational, but the sequence converges to $a$, which is a rational number (it is equal to $\frac49$).

Answer 2

Here is how to build such a perfect set: take a union $X$ of disjoint open intervals covering $\mathbb{Q}$, having irrational bounds, so that the distance between two such intervals is positive. Take $P = {\mathbb R}\setminus X$. It is a perfect set containing only irrational points.

More precisely, let $(q_1, q_2, ...)$ be an enumeration of ${\mathbb Q}$ Let us construct a sequence of non empty open intervals $I_k = (\alpha_k, \beta_k)$,where $\alpha_i, \beta_i\in {\mathbb R}\setminus{\mathbb Q}$ by the following method: let $q$ be the first rational in the enumeration that does not belong to $A_k=\bigcup_{i<k} I_i$ (we start with $A_1=\emptyset$) We choose $I_k$ such that $q\in I_k$ and $A_k\cap I_k = \emptyset$. We can choose $\alpha_k<\beta_k$ irrational so that none of them is equal to one of the previously chosen $\alpha_i$ or $\beta_i$.

Let $A_\infty=\bigcup_i I_i$. Clearly, $A_\infty$ is an open set (as a union of open sets) that contains ${\mathbb Q}$. Let $P = {\mathbb R}\setminus A_\infty$. It is a closed set containing only irrational numbers.

Suppose that $x\in P$ is an isolated point. It means that there are non empty intervals $(a,x)$ and $(x, b)$ contained in $A_\infty$. But as $A_\infty$ is a disjoint union of non empty open intervals it means that $(a, x)$ must be included in one of the intervals $I_k$ and similarly for $(x, b)\subset I_\ell$. This is a contradiction because the upper bound of $I_k$ would be equal to the lower bound of $I_\ell$.

Finally, $P$ is a perfect set of irrational numbers.

Conversely, let us prove that every perfect set of irrational numbers has this structure, provided we also allow the bounds $-\infty$ and $\infty$ for the intervals $I_k$.

For this, let $P$ be a perfect set of irrational numbers and $A={\mathbb R}\setminus P$, an open subset of ${\mathbb R}$. We can decompose $A$ as the disjoint union of its connected components $A=\bigcup_k I_k$. If $A$ is not empty, the number of connected components is finite or countable, each $I_k=(\alpha_k, \beta_k)$ is an open interval which bounds belong to $P\cup\{-\infty, \infty\}$ (because $A$ is open). When $\ell\not = k$, one has $\beta_k\not = \alpha_l$ otherwise $\beta_k$ would be an isolated point in $P$, contradicting the perfection of $P$. QED