0

$\Bbb{R}^2=\Bbb{R}\times\Bbb{R}$, if I then try and imagine the intersection of $\Bbb{R}^2$ and $\Bbb{R}$, I see a plane built by fixing two real number lines perpendicular to each other and try to imagine how $\Bbb{R}$ intersects with this. Does it intersect with both number lines or just one? In other words, if I define

$$\Bbb{R}^2\cap\Bbb{R},$$

do I expect to find all points $\{(0,x),(x,0)|x\in\Bbb{R}\}$ or is that over counting since I'm including both "axis" of the $\Bbb{R^2}$ plane?

Charlie
  • 685
  • 7
    Actually the intersection is empty. If you have anything in the intersection, this must be at the same time a couple $(a,b) \in \Bbb R^2$ and a real number $x \in \Bbb R$. But a couple is not a real number. – Crostul Apr 02 '20 at 22:33
  • 7
    By pure set theory the intersection is empty. That being said, $\mathbb{R^2}$ contains infinitely many subsets which are naturally isomorphic to $\mathbb{R}$, for example ${(x,0):x\in\mathbb{R}}$. – Mark Apr 02 '20 at 22:34
  • Ah I see that makes sense, I was trying to think a little bit too much in terms of baby geometry. Thanks for the answers. – Charlie Apr 02 '20 at 22:35

1 Answers1

0

In the interest of this post having an answer, the intersection of $\Bbb R$ and $\Bbb R^2$ is empty.

Charlie
  • 685