Writing interval as unions of sets Given that $a,b \in \mathbb{R}$
Can the interval $(a,b)$ be expressed as $(a,b) = \mathop{\bigcup_{n \in \mathbb{Z}}}(a,\frac{2b}{\pi}\tan^{-1}(n))$. 
I thought, as n gets larger, one eventually gets a bunch of sets in union with $(a,b)$ therefore giving us $(a,b)$... Sure, I could get away with saying an arbitrary union of $(a,b)$ gives us the desired set... 
 A: This is false in general.
For example, if $a = -2$ and $b = -1$, we have that $$(-2, 0) = \left(-2, \frac{-2}{\pi}\tan^{-1}{(0)}\right)\subseteq\bigcup_{n\in\mathbb Z}\left(-2, \frac{-2}{\pi}\tan^{-1}(n)\right),$$ which makes the equality impossible.
A: 
See Dennis's answer for a counterexample

This proof only works for $b\geq 0$
Let $S_1=(a,b)={x\in\mathbb{R}\vert a<x<b}$
Let $S_2=\mathop{\bigcup_{n\in\mathbb{Z}}(a,\frac{2b}{\pi}\tan^{-1}(n))}$
Then, to prove that $S_1=S_2$, we must prove that $\forall{a}\in{S_1},{a}\in{S_2}$ and vice versa.
Let $x\in{S_1}$. Then, $x\in{S_2}\iff \exists n\in\mathbb{Z}[x\in(a,\frac{2b}{\pi}\tan^{-1}(n))]$. $\forall n\in\mathbb{Z}\forall y\in\mathbb{R}[y\in(a,\frac{2b}{\pi}\tan^{-1}(n))\iff a<y<\frac{2b}{\pi}\tan^{-1}(n)]$.
$\forall x\in S_1[\because x\in(a,b)\therefore a<x]$
Thus, to prove that $S_1 \subseteq S_2$, we just need to prove that $\forall x\in S_1[\exists n\in\mathbb{Z}[x<\frac{2b}{\pi}\tan^{-1}(n)]]$.
$\lim_{n\to\infty}\tan^{-1}n=\frac{\pi}{2}$
We can say that $\frac{2b}{\pi}$ is a constant for our purposes; thus, $\lim_{x\to\infty}\frac{2b}{\pi}=\frac{2b}{\pi}$.
$\lim_{n\to\infty}\frac{2b}{\pi}\tan^{-1}(n)=\lim_{n\to\infty}\frac{2b}{\pi}\lim_{n\to\infty}\cdot\tan^{-1}(n)=\frac{2b}{\pi}\cdot\frac{\pi}{2}=b$
$\therefore\forall x\in S_1[\exists n\in\mathbb{Z}[x<\frac{2b}{\pi}\tan^{-1}(n)]]$
$\therefore S_1\subseteq S_2$
That's the first part of the proof. The second part is $S_2\subseteq S_1$.
We can prove this by showing that $\forall x\in S_2[x\in S_1]$. We can make this simpler using modus tollens (sorry Kenny for not using intuitionistic logic :P) and show that $\forall x\notin S_1[x\notin S_2]$
There are two cases for which $x\notin S_1$.
Case 1: $x<a$
The lower bound of $S_2$ is the minimum of the lower bounds of all of the sets whose union is $S_2$; since the lower bound of all such sets is $a$ then the lower bound of $S_2$ is also $a$.
$\therefore\forall x<a[x\notin S_2]$
Case 2: $x>b$
The upper bound of $S_2$ is the maximum of the upper bounds of all of the sets whose union is $S_2$. The codomain of $\tan^{-1}$ is $[-\frac\pi 2,\frac\pi 2]$ therefore $\forall n\in\mathbb{Z}[\tan^{-1}(n)\le\frac\pi 2]\therefore\forall n\in\mathbb{Z}[\frac{2b}{\pi}\tan^{-1}(n)\le{b}]$
$\therefore\forall x>b[x>b\ge max({\frac{2b}{\pi}\tan^{-1}(n)\vert n\in\mathbb{Z}})]$
$\therefore \forall x\notin S_1[x\notin S_2]\therefore \forall x\in S_2[x\in S_1]\text{ by modus tollens}$
$\therefore S_2\subseteq S_1$
$\therefore S_1=S_2$
Note that this proof needs to be slightly modified for negative $a$ and/or $b$ but the idea remains the same.
