Exercise 4.25

(a): Following the hint, take $z\notin K+C$ and put $F=\{z\}-C$. $K$ and $F$ are disjoint since if there were a $y\in F\cap K$ then there would be $x\in C$ such that $y=z-x$, but then $z=y+x$ contradicts $z\notin K+C$. $F$ is closed since it is the inverse image of $C$ by the continuous map $x\mapsto z-x$. Hence by Exercise 21 there is a $\delta >0$ such that $|x-y|>\delta$ if $x\in K$ and $y\in F$. If $x+w\in K+C$, then $|(x+w)-z|=|x-(z-w)|>\delta$, so that the open neighborhood of $z$ of radius $\delta$ is disjoint from $K+C$. Hence the complement of $K+C$ is open.

(b): Each element of $C_1+C_2$ can be associated with a unique pair of integers, since if $n_1+n_2\alpha =m_1+m_2\alpha$ for $n_2\ne m_2$, then this equation can be rewritten to express $\alpha$ as a rational number. Since the collection of such integer pairs is countable, $C_1+C_2$ is countable.

Consider the fractional parts of the integer multiples of $\alpha$, the set of $\mathit{p\alpha }-[\mathit{p\alpha }]$ for $p$ an integer, which are all in $C_1+C_2$. These are all distinct by the argument above. Let $n$ be any positive integer. By the pigeonhole principle, there are at least two such fractional parts in one of the subintervals $(0,1∕n),(1∕n,2∕n),\dots ,((n-1)∕n,1)$ of $(0,1)$. That is, there are integers $p,q$ such that $0<(p-q)\alpha -([\mathit{p\alpha }]-[\mathit{q\alpha }])<1∕n$, so that $C_1+C_2\cap (0,1∕n)$ is nonempty.

Let $x$ be any real number and let $𝜖>0$. Let $n$ be a large enough positive integer such that $1∕n<𝜖$, and let $y\in C_1+C_2\cap (0,1∕n)$. Then some multiple of $y$ lies in $[x,x+1∕n)$, so that some element of $C_1+C_2$ is within $𝜖$ of $x$. Hence the closure of $C_1+C_2$ is $R$, and since it is a proper subset of $R$, it is not closed.