Given the following:

For each $r \in \mathbb{R}$, let $A_r = \lbrace(x,y) \in \mathbb{R} \times\mathbb{R} : x - y = r\rbrace$

Prove: $\lbrace A_r : r \in \mathbb{R}\rbrace$ is a partition of $\mathbb{R} \times \mathbb{R}$.

Here's my attempted proof:

Let $(x,y) = (1,0), (2,1), ... \in A_1$

Let $(x,y) = (2,0), (3,1), ... \in A_2$

$A_1 \cup A_2 \cup\text{ ... } A_n = A$ where $n = \infty$ (guess).

Note that $A_1 \cap A_2 = \emptyset$

Is this proof complete?

  • $\begingroup$ IMHO to show the $A_r$ family is a partition of ${\Bbb R}^2$ you need to show it fits partition properties, that is 1) sets are disjoint: $A_r \cap A_s = \emptyset$ for all $r\ne s$ and 2) their union is your space: $\bigcup\limits_{r\in \Bbb R}A_r = {\Bbb R}^2$. You shown only a little piece of that just for $A_1$ and $A_2$. $\endgroup$ – CiaPan Jun 9 '15 at 14:54
  • $\begingroup$ $\mathbb{R}^{2}$ is the same as $\mathbb{R}\times \mathbb{R}$? $\endgroup$ – Kevin Meredith Jun 9 '15 at 14:55
  • $\begingroup$ BTW, there is no such $n$ in natural or in real numbers, that $n=\infty$. $\endgroup$ – CiaPan Jun 9 '15 at 14:56
  • 1
    $\begingroup$ Yes, for sets the natural 'power' is a Cartesian product of a set with itself: $X^2 = X\times X, \ \ X^3 = X\times X\times X$. $\endgroup$ – CiaPan Jun 9 '15 at 14:57
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    $\begingroup$ No, the proof is not complete at all. I would have given it zero points if I were to grade it. $\endgroup$ – Asaf Karagila Jun 9 '15 at 14:59

You need to show that $\mathbb{R} \times \mathbb{R} = \cup_{r \in \mathbb{R}} A_r$ and that $A_r \cap A_s = \varnothing$ whenever $r \neq s$. For the first of these, it should be fairly clear that given a pair $(x,y)$, it belongs in some $A_r$ (which one?). For the second part, show that if we assume that $(x,y) \in A_r \cap A_s$ when $r \neq s$ then we arrive at a contradiction.

  • $\begingroup$ Could you please explain what this notation means: $\cup_{r \in \mathbb{R}} A_r$? $\endgroup$ – Kevin Meredith Jun 21 '15 at 18:11
  • $\begingroup$ It's the union of the sets $A_r$, for each $r \in \mathbb{R}$. $\endgroup$ – lokodiz Jun 21 '15 at 19:47

Let $(x,y)\in\mathbb{R}^2$, then $(x,y)\in A_{x-y}$. So $\mathbb{R}^2\subset\bigcup_{r\in\mathbb{R}} A_r$. Also any two $A_s, A_r$ for $s\neq r$ are disjunct (let $(x,y)\in A_s$, then $x-y=s$, so $(x,y)\notin A_r$, and vice versa. So $A_r\cap A_s=\emptyset$ for $r\neq s$). So the set of all $A_r$ is a partition of $\mathbb{R}^2$.


From wikipedia definition of the partition:

A partition of a set X is a set of nonempty subsets of X such that every element x in X is in exactly one of these subsets

Trivially, none of the sets are empty. That means, for partition $P = \{A_r:r\in \mathbb{R} \}$ you'll need to verify these two conditions are met:

  1. The joint set of all the $A_r$ is $\mathbb{R}\times \mathbb{R}$
  2. For any $r,s$ such that $r \neq s$, $A_r \cap A_s = \emptyset$

For a given $(x,y)$ from $\mathbb{R}\times \mathbb{R}$, select $r = x - y$. Now, $r$ is of course real and therefore $A_r$ is a set from the partition $P$. That verifies condition 1.

Now, given $r,s$ such that $r \neq s$, we assume (for contradiction) that there is a tuple $(x,y)$, that is present both in $A_r$ and $A_s$. That would mean, that $x - y = r$ and $x - y = s$. Hence, $r$ must be equal to $s$, which is in contradiction with the assumption. Therefore, the sets $A_r$ and $A_s$ are disjoint. That verifies condition 2 and concludes the proof.


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