Proving that a certain set is dense using a useful characterization of the irrational numbers. (Based on Chapter 4 Exercise 25 (b) Baby Rudin)

Question

The following is exercise 25 (b) Chapter 4 in Rudin's Principle's of Mathematical Analysis:

Let $\alpha$ be an irrational real number. Let $C_1$ be the set of all integers, let $C_2$ be the set of all $n\alpha$ with $n \in C_1$. Show that $C_1$ and $C_2$ are closed subsets of $\mathbb{R}$ whose sum $C_1+C_2$ is not closed, by showing that $C_1+C_2$ is a countable dense subset of $\mathbb{R}$.

I am writing to check if the answer presented below is correct. Furthermore, at the end of the proof I give a characterization of the irrationals which makes this problem much simpler and I wanted to share this characterization with the members here.

Answer 1

Hints for people that are working on this problem:

(There are three solutions that I am aware of and below are hints that yield each of these solutions.)

1. Construct a subsequence of $C_1+C_2$ that converges to zero, then, using an arbitrarily small element of this convergent subsequence, partition $\mathbb{R}$ into a union of arbitrarily small neighborhoods and argue that every element of $\mathbb{R}$ is a limit point of $C_1+C_2$

2. A clever application of the pigeonhole principle will also solve this problem. Take the fractional part of any member of $C_1+C_2$, call it $\lambda$ and then look at the fractional part of $k\lambda$ for $k=1,2,...,m$ while observing that each of these fractional parts is a member of $C_1+C_2$. The pigeonhole principle implies that there are two members of this set of multiples that are within $\frac{1}{m-1}$ of one another. Their difference is also a member of $C_1+C_2$ so you can use this arbitrarily small element to partition $\mathbb{R}$ and proceed as in hint 1) above.

3. Use the following characterization of the irrationals: $\alpha$ is irrational if and only if $\forall m \neq 0 \in \mathbb{Z}$ $\exists!$ $ n \in \mathbb{Z}$ such that $0<m\alpha-n<1$. You can then argue that the boundedness of a subset of $C_1+C_2$ yields a convergent subsequence. Showing that the series is Cauchy will give you an arbitrarily small element of $C_1+C_2$ by taking the difference of arbitrarily close members of the convergent subsequence. Proceed from here as in 1).

Proof: Let $C=C_1+C_2$. Notice that there exists an injection from $\mathbb{Z} \times \mathbb{Z}$ to $C$ because otherwise $\alpha$ would be rational. Further notice that each member of $C$ has a unique decomposition into $a+b\alpha$ for $a,b \in \mathbb{Z}$ because if such decomposition were not unique then $\alpha$ must be rational, which is a contradiction. This decomposition constitutes an injection from $C$ to $\mathbb{Z} \times \mathbb{Z}$. Thus by the Cantor-Schroder-Bernstein Theorem, $C$ is countable.

To see that $C_1$ and $C_2$ are both closed observe that $C_1^c$ and $C^c_2$ are both open.

We conclude by showing that $C$ is dense in $\mathbb{R}$ by using a useful characterization of the irrationals. If $\alpha$ is irrational then $\forall m \neq 0 \in \mathbb{Z}$ $\exists!$ $ n \in \mathbb{Z}$ such that $0<m\alpha+n<1$. Now this is a subset of $C$ that is a bounded sequence in $\mathbb{R}$. The Bolzano-Weierstrass Theorem tells us that this bounded sequence contains a convergent subsequence which must be Cauchy. Thus we make take any $\epsilon >0$ and find two members of our convergent subsequence that are within that $\epsilon$ of one another. Now the positive difference of these two elements is less than $\epsilon$ and is also a member of $C$. We know this difference is non zero because every element of $C$ is distinct. Furthermore all integer multiples of this element are also in $C$. Therefore every element of $\mathbb{R}$ is within an arbitrary $\epsilon$ of some member of $C$ and is thus a limit point of $C$. Hence $C$ is dense in $\mathbb{R}$.

I wanted to write to check if this is correct and most of all share the characterization of the irrationals that makes this proof easier; that is: $\alpha$ is irrational if and only if $\forall m \neq 0 \in \mathbb{Z}$ $\exists!$ $ n \in \mathbb{Z}$ such that $0<m\alpha-n<1$. (In the above proof I changed the sign of $n$ to make the connection between the definition of the irrationals and the set $C$ more natural.)