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(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.

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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
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1 Answer 1

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.

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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
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