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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 $A_s$."

Anurag A
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NeedHelp
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    What exactly is the result you are trying to show? – Robert Thingum Nov 12 '18 at 04:10
  • $y=-x+r$ and $y=-x+s$ each is a straight line with slope $-1$, with $y$ intercepts $r$ and $s$ respectively. If $r\not=s$ these two lines are disjoint (and these two lines are the same as the sets $A_r$ and $A_s$). – Mirko Nov 12 '18 at 04:13
  • I hope that it is clear that each $A_r$ is nonempty. Basic work with linear equations shows that if $r\neq s,$ then $A_r$ and $A_s$ are disjoint. Is every pair $(x,y)$ a member of some $A_r?$ If so, you have a collection of nonempty, disjoint sets whose union is all of $\mathbb{R} \times \mathbb{R},$ which I believe is what you are trying to show (not that you have a partition of $\mathbb{R}$). – Chris Leary Nov 12 '18 at 04:17
  • I am attempting to first show 1) $A_r =A_s$ or the intersection of $A_s$ and $A_r$ is empty. 2) The U $A_r$ = $\mathbb{R} \times \mathbb{R}$ – NeedHelp Nov 12 '18 at 04:52

2 Answers2

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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.

Anurag A
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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.

P Vanchinathan
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