Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(a) $\{m+nπ:m,n\in \mathbb{Z}\}$

(b) $\{m+nπ:m,n \textrm{ are positive integers}\}$

I know the interior of a and b are both empty and the answer said the closure is $\mathbb{R}$,i got troubled in how to find the closure.

share|cite|improve this question
Please make the question contained in the body of the post, not in the title. – Asaf Karagila Feb 12 '13 at 1:34
@Asaf Done...but good pointer for the question asker. Also, descriptive titles are good; but most details should be left for the body of the question itself, i.e. aim for one-liner titles. – amWhy Feb 12 '13 at 1:37
up vote 1 down vote accepted

I’ve completed (a) and given a hint for (b).

(a) This answer contains a proof that if $\alpha$ is any irrational number, $\{n\alpha-\lfloor n\alpha\rfloor:n\in\Bbb Z\}$ is dense in $[0,1]$, where $x-\lfloor x\rfloor$ is the fractional part of $x$. Let $D=\{n\alpha-\lfloor n\alpha\rfloor:n\in\Bbb Z\}$. For any $m\in\Bbb Z$ the set $m+D=\{m+d:d\in D\}$ is dense in $[m,m+1]$, so $\bigcup_{m\in\Bbb Z}(m+D)$ is dense in $\Bbb R$, and clearly $\bigcup_{m\in\Bbb Z}(m+D)\subseteq\{m+n\alpha:m,n\in\Bbb Z\}$.

(b) The set $S=\{m+n\pi:m,n\in\Bbb Z^+\}$ is very different. Clearly $S$ is an unbounded set of positive real numbers. Fix $s\in S$, and suppose that $m$ and $n$ are positive integers such that $m+n\pi\le s$. Then $$m\le\lfloor s\rfloor\quad\text{and}\quad n\le\left\lfloor\frac{s}{\pi}\right\rfloor\;,$$ so $\{t\in S:t\le s\}$ is finite. (In fact it has at most $\lfloor s\rfloor\cdot\left\lfloor\frac{s}{\pi}\right\rfloor$ elements.) Use this to show that $S$ is a closed, discrete set in $\Bbb R$: it has empty interior and no limit points and is its own closure.

share|cite|improve this answer
Thank you for your link. I didn't know the pigeonhole principle can be used like this. – Jebei Feb 12 '13 at 10:22
@frame99: You’re welcome. That’s one of my favorite applications of it, simply because it is a little unexpected. – Brian M. Scott Feb 12 '13 at 10:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.