Prob. 25(b), Chap. 4 in Baby Rudin: The set $C_1 + C_2$ need not be closed in $\mathbb{R}$ even for closed sets $C_1$ and $C_2$ Here is Prob. 25, Chap. 4 in Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

If $A \subset \mathbb{R}^k$ and $B \subset \mathbb{R}^k$, define $A + B$ to be the set of all sums $x+y$ with $x \in A$, $y \in B$. 
(a) If $K$ is compact and $C$ is closed in $\mathbb{R}^k$, prove that $K+C$ is closed. [This I think I've managed to prove.]
Hint: . . . 
(b) 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 $\mathbb{R}^1$ whose sum $C_1 + C_2$ is not closed, by by showing that $C_1 + C_2$ is a countable dense subset of $\mathbb{R}^1$. 

My effort: 

Let $p$ be a real number that is not an integer, and let $\delta$ be any real number such that $$0 < \delta < \min \left\{ \ p - \lfloor p \rfloor, \ \lceil p \rceil - p \ \right\}. $$ 
  Then the $\delta$-neighborhood of $p$ in $\mathbb{R}^1$ does not intersect the set of integers at all, showing that every point $p$ of $\mathbb{R}^1 - C_1$ is an interior point and hence that $C_1$ is closed in $\mathbb{R}^1$. Am I right?
For any two distinct points $x$ and $y$ of $C_2$, we note that 
  $$\vert x-y \vert \geq \vert \alpha \vert > 0.$$ 
  So if $p$ is any real number that is not in $C_2$ and if $\delta$ is any real number such that 
  $$0 < \delta < \min \left\{ \ \left\vert p - \alpha \lfloor \frac{p}{\alpha} \rfloor \right\vert, \ \left\vert \alpha \lceil \frac{p}{\alpha} \rceil - p \right\vert \ \right\},$$ 
  then the $\delta$-neighborhood of $p$ in $\mathbb{$}^1$ --- which equals the segment $(p-\delta, p+\delta) --- does not intersect the set $C_2$ at all, for is $x \in \mathbb{R}$^1$ and $$ p-\delta < x < p+\delta, $$
  then we must have 
  $$ \alpha \lfloor \frac{p}{\alpha} \rfloor < x < \alpha \lceil \frac{p}{\alpha} \rceil,$$ 
  and $\lfloor \frac{p}{\alpha} \rfloor$ and $\lceil \frac{p}{\alpha} \rceil$ are two successive integers, which implies that $x$ lies strictly in between two successive elements of $C_2$. Thus every point $p$ of $\mathbb{R}^1- C_2$ is an interior point, from which it follows that $C_2$ is closed. Am I right?
Moreover, we also note that if $n$ is non-zero, then $n \alpah$ is irrational, by Prob. 1, Chap. 1 in Baby Rudin, 3rd edition. So the sets $C_1$ and $C_2$ intersect in $0$ only. Am I right? 
Now the set $C_1 + C_2$ is given by 
  $$C_1 + C_2 = \left\{ \ m + n\alpha \ \colon \ m \mbox{ and } n \mbox{ are integers } \ \right\}.$$ Am I right?
Now the map $m + n \alpha \ \mapsto \ (m, n)$ is an injective mapping of $C_1 + C_2$ into the Cartesian product $C_1 \times C_1$, and this Cartesian product is of course countable. So $C_1 + C_2$ is countable. Am I right? 

Now how to show that $C_1 + C_2$ is dense in $\mathbb{R}^1$? 

Once we have shown this then we know that $C_1 + C_2$ cannot be closed, because if $F$ is a closed set in a metric space $X$ and if $F$ is dense in $X$, then we have 
  $$X = \overline{F} = F,$$
  but $C_1 + C_2$, being a countable subset of $\mathbb{R}^1$, is of course a proper subset of $\mathbb{R}^1$, since $\mathbb{R}^1$ is uncountable. Am I right? 

So, if what I've established so far is correct, then my only question is how to (rigorously) show that $C_1 + C_2$ is dense in $\mathbb{R}^1$? 
 A: Hint : Take $a_n=n\alpha-[n\alpha]$. Since $\alpha$ is irrational, $a_n$'s are all distinct. This shows that there are infinitely many such numbers in $(0,1)$. Can you choose two such that they are within $\epsilon$- distance from each other? What about the integer multiple of the difference?
Expansion: There must be a half open interval of the form $\left(\frac{k-1}{N-1},\frac{k}{N-1}\right)$ for $k=1,2,\ldots,N-1$ containing two of the numbers $a_1,a_2,\ldots,a_N$ for there are $N-1$ such intervals and all these numbers are distinct. The inequalities will tell you that for some $i,j$ we have $0 \lt (i\alpha-[i\alpha])-(j\alpha-[j\alpha]) \lt \frac{1}{N-1}$ which implies that $a=(i-j)\alpha+([j\alpha]-[i\alpha]) \in \left(0,\frac{1}{N-1}\right)$ and $a$ is an element in $C_1+C_2$
Now go ahead show that there is such a point in every interval of the form $\left(\frac{k}{n}, \frac{k+1}{n}\right)$ for any positive integer $n$ and any integer $k$. 
A: Here is an example: Let $C_1=\{2n+2^{-2n}: n\in \mathbb N\}$ and $C_2=\{1-2n-2^{1-2n} :n\in \mathbb N\}.$ Then $1\not \in C_1+C_2$ but $1-2^{-2n}=(2n+2^{-2n})+(1-2n-2^{1-2n})\in C_1+C_2$ for all $n\in \mathbb N.$
