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 $\mathbb{R}^1$?

  • 1
    $\begingroup$ This problem is intimately related to irrational rotation on the circle. For $\alpha \in [0,1) \setminus \mathbb Q$, define the map $T_\alpha : [0,1) \to [0,1)$ by $T_\alpha x = x + \alpha \, (\text{mod } 1)$. It is well known that for any $x \in [0,1)$ (including $x = 0$) the orbit $\{ T^n x\}_{n \in \mathbb Z}$ is dense in $[0,1)$ (see, e.g., en.wikipedia.org/wiki/Equidistribution_theorem). $\endgroup$ – A Blumenthal Apr 7 '17 at 14:16
  • $\begingroup$ I do not see a definition of $C_1 $ in your effort. $\endgroup$ – DanielWainfleet Apr 7 '17 at 21:03
  • $\begingroup$ @user254665 the set $C_1$ is the set of all the integers. It has been defined in Rudin's statement of the problem. Can you please read through my original post from the beginning? $\endgroup$ – Saaqib Mahmood Apr 8 '17 at 10:56
  • $\begingroup$ You have correctly shown that $C_1$ and $C_2 $ are closed. An easier way to do it is that if $x\ne 0$ then $S=\{nx: N\in \mathbb Z\}$ is closed because its complement $\mathbb R$ \ $S= \cup_{n\in \mathbb Z}(nx,(n+1)x)$ .... (because if $mx\in (nx,n+1)x) $ then $m$ is between $n$ and $n+1$,)..., so its complement is a union of open intervals.... With $x=1$ we have $S=C_1$. ... I felt like giving a simpler example (in my A) of two closed sets whose sum is not closed. $\endgroup$ – DanielWainfleet Apr 8 '17 at 12:15

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 $k$.

  • $\begingroup$ yes the $a_n$ are all distinct and all lie in $(0, 1)$, and thus there are infinitely many $a_n$ in $(0, 1)$. But I'm not sure if we can find any two of these that are within $\varepsilon$ distance from each other. And, all the $a_n$, except one, are irrational, and so all their integer multiples (except one) would also be irrational. Am I right? Can you please expand upon your answer? $\endgroup$ – Saaqib Mahmood Apr 7 '17 at 14:40

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 $n\in \mathbb N.$


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