Why an open triangle is a 2-dimensional Borel set in $R^2$? For example, $\{(x,y)\in R^2: 0<y<x, 0 < x < 1  \}$. I googled and found some proofs. Basically their logic is, since every rectangle is a Borel set, we can bisect rectangles to fill the triangle. 
However, I am not convinced. To fill the triangle, for each point on the subspace $\{(x,y) \in R^2: x=y, 0< x < 1 \}$, there must be one rectangle. Therefore, the number of rectangles is at least uncountable, because the numbers on $\{x\in R: 0< x < 1\}$ are uncountable.  
 A: It looks like you're working from the bare-bones definition of Borel, starting from rectangles - pretty restrictive, but you've got to start somewhere. So here's a covering with rectangles:
For any point $(x,y)$ in the triangle, we have $y < x$. Because the rationals are dense, there exists a rational $q$ with $y < q < x$. Let $R_q$ be the open rectangle with vertices $(q,q)$, $(1,q)$, $(1,0)$, and $(q,0)$. This rectangle covers $(x,y)$. Now, for every point in the triangle there is such a $q$, so the union of all these rectangles covers everything - but even though there are uncountably many points, there are only countably many $R_q$. So $T$ is the union of open rectangles $R_q$ for $q \in \mathbb{Q}$, which makes $T$ Borel.
I'm assuming here you meant to say that the triangle ends at $y = 0$ - if not, and you meant for the "triangle" to extend into an infinite vertical strip below the horizontal axis, then just stretch all the rectangles vertically. I'm also assuming you meant to say $0 < x < 1$, not $0 \leq x \leq 1$, in which case the triangle is not open. It would still be Borel, though, just a little trickier to show.
