Indexed Sets Problem, union and intersection For $_r$ is an element of $\mathbb R_+$, define $B_r = [-1-r,r)$. Determine $\bigcup_{r\in\mathbb{R}_+} B_r$ is an element of $\mathbb R_+$ $B_r$ and  $\bigcap_{r\in\mathbb{R}_+} B_r $ is an element of $\mathbb R_+$ $B_r$. 
$\bigcup_{r\in\mathbb{R}_+} B_r = \mathbb{R}$
For $\bigcap_{r\in\mathbb{R}_+} B_r $ = [-1,0]

Notice that $B_x\subseteq B_y$ whenever $x\leq y$, so the family of sets are what we call "nested", meaning that if there was a "smallest" set in our family, the intersection over the entire family would simply be that smallest set.
In our case, however, there is no "smallest" but it can still give us some intuition as to what is going on.  If we were indexing over all non-negative (zero included) sets, then the smallest would have been $B_0 = [-1,0)$.
Since zero is not one of the values over which we are indexing, notice then that $B_\epsilon = [-1-\epsilon, \epsilon)$ for all $\epsilon>0$ and in particular then that $0\in [-1-\epsilon,\epsilon)$.

Steps to complete:


*

*Prove that $[-1,0]\subseteq B_r$ for all $r\in \Bbb R^+$

*Prove that any $x\in \left(\Bbb R\setminus [-1,0]\right)$ has at least one $r$ for which $x\notin B_r$

*Use the definition of intersection over a family to conclude that $\bigcap\limits_{r\in \Bbb R^+} B_r = [-1,0]$
