How to find interior, closure, limit points, and isolated points of subsets of $\mathbb{R}$? (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.
 A: 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.
