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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?

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What properties do you need to verify? Which ones are you having trouble with? – Qiaochu Yuan Feb 15 '12 at 23:56
It's just trying to determine if A is a partition of R3 or not. Like for any r there are infinitely many triples of real numbers that can sum to it - does that make it not a partition? – roboguy12 Feb 15 '12 at 23:58
That's not the question I'm asking. There are some properties that a partition satisfies. What are those properties? Which ones are you having trouble showing hold in this case? – Qiaochu Yuan Feb 16 '12 at 0:09
In your second example, $A_r$ and $A_{-r}$ would not simply have a value in common; rather they would both be the same set. – Michael Hardy Feb 16 '12 at 1:20
up vote 3 down vote accepted

Recall the definition of a partition $\mathcal A\subseteq P(\mathbb R^3)$:

  1. For every $x\in\mathbb R^3$ there exists $A\in\mathcal A$ such that $x\in A$.
  2. For every $A,B\in\mathcal A$ if $A\neq B$ then $A\cap B=\varnothing$.
  3. $\varnothing\notin\mathcal A$.

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.

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Ok, the logic makes sense (for every problem like this I just have to test those three conditions). But how would I draw this partition? I just can't visualize dividing up R3 like this. – roboguy12 Feb 16 '12 at 0:04
@roboguy12: There is no guarantee that you can visualize a partition. For example, let me partition $R^3$ into ten sets, being the mod 10 sum of the first 10 decimal digits after the point of each coordinate. I can't imagine that one. In this case, the sets are planes perpendicular to (1,1,1). – Ross Millikan Feb 16 '12 at 0:08
Not everything you can draw or imagine. Try to understand which vectors would be mapped to $1$; try to understand how the same idea would work on $\mathbb R^2$ and use that to understand how it would look like in $\mathbb R^3$. – Asaf Karagila Feb 16 '12 at 0:08
Okay, that makes sense. I'm just new to this is all..does the logic in my edit make sense? – roboguy12 Feb 16 '12 at 0:11
@roboguy12: Yes, indeed it makes sense. – Asaf Karagila Feb 16 '12 at 0:12

Hint: Can any point be in more than one $A_r$? Are all points in some $A_r$?

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