Verifying That A Collection Is A Partition of $\mathbb R^3$ Can someone possibly explain this to me, I'm having difficulties visualizing it:
 For each $r\in\mathbb{R}$, let $A_r=\{(x, y, z)\in\mathbb{R}^3 : x+y+z=r\}$. How can I tell if this is a partition of $\mathbb{R}^3$?
EDIT: Okay, so I pretty much understand this problem, but what if it were like:For each $r\in\mathbb{R}$, let $A_r=\{(x, y, z)\in\mathbb{R}^3 : x^2+y^2+z^2=r^2\}$. Is it correct to think that it would not be a partition because $A_r$ and $A_{-r}$ would have a value in common?
 A: Recall the definition of a partition $\mathcal A\subseteq P(\mathbb R^3)$:


*

*For every $x\in\mathbb R^3$ there exists $A\in\mathcal A$ such that $x\in A$.

*For every $A,B\in\mathcal A$ if $A\neq B$ then $A\cap B=\varnothing$.

*$\varnothing\notin\mathcal A$.


If so, we first need to make sure that every vector is somewhere in the partition, which is trivial: if $(x,y,z)\in\mathbb R^3$ then let $r=x+y+z$ and thus $(x,y,z)\in A_r$.
Now we need to see that those are all disjoint, which is again simple. If $(x,y,z)$ in $A_r$ and in $A_s$ then $s=x+y+z=r$ so the intersection cannot be nonempty for $r\neq s$.
Lastly, we need to see that for every $r\in\mathbb R$ we have that $A_r\neq\varnothing$ and indeed $(r,0,0)\in A_r$.

As for the edited question: indeed in order to show that a certain collection is not a partition we only need to contradict one of the three properties above. Just as you write, $(1,0,0)\in A_1\cap A_{-1}$ proving that the partition is not pairwise disjoint.
A: Hint:  Can any point be in more than one $A_r$?  Are all points in some $A_r$?
