$A_r=\{(x, y) \in \mathbb{R} \times \mathbb{R} : x+y=r\}$

Question

I have figured out how to prove $A_r$ is nonempty but I am stuck on the last two parts of proving a partition of Real numbers. For part 2 I have

"Let r, s $\in \mathbb{R}$ with x+y=r and x+y=s, so there exists $A_r$ and $A_s$."

Answer 1

I'm assuming you are trying to show that $\{A_r\}_{r \in \Bbb{R}}$ forms a partition of $\Bbb{R}^2$.

Clearly for each $r \in \Bbb{R}$, the point $(r,0) \in A_r$, hence $A_r \neq \emptyset$.

To prove: $\Bbb{R}^2=\bigcup_{r \in \Bbb{R}}A_r$.

Let $(a,b) \in \Bbb{R}^2$, then consider the real number $r_0=a+b$. Corresponding to this $r_0$ we have the set $A_{r_0}$. Now $(a,b) \in A_{r_0} \subseteq \bigcup_{r \in \Bbb{R}}A_r$. Thus $\Bbb{R}^2 \subseteq \bigcup_{r \in \Bbb{R}}A_r$. The other containment is obvious.

To prove: $A_r \cap A_s = \emptyset \iff r \neq s$.

Assume $r \neq s$ and suppose $(c,d) \in A_r \cap A_s$. Then $(c,d) \in A_r$, which means $c+d=r$. Also $(c,d) \in A_s$, which means $c+d=s$. But this contradicts the fact that $r \neq s$. So the intersection is empty. The other implication is straightforward.

Answer 2

Consider the line in the Cartesian plane joining $(1,0)$ and $(0,1)$. This set of points of this line is your $A_1$.

Now consider ALL the lines in the plane that are parallel to this $A_1$. As parallel lines do no intersect they are disjoint.

And also given any point in the plane not in the line $A_1$ there is a (unique) line passing through that and parallel to $A_1$. So every point is covered by these lines.

Any such line is not parallel to the $x$-axis and so it will intersect it at some point $(r,0)$, and it is the line $A_r$ in your problem. Thus you can see that these sets provide a partition.