1

I know that the Cartesian product $[0,1]\times[0,1]$ represents a unit square in the first quadrant. Is it possible to write the triangular region with vertices $(0,0)$, $(0,1)$ and $(1,1)$, interior included, as a Cartesian product o two sets? I think one can have $A=\{x\in\mathbb{R}\, |\,0\leq x\leq 2\}$ and $B=\{y\in\mathbb{R}\, |\, x\leq y\leq 2, 0\leq x\leq 2\}$. Then one can consider $A\times B$.

However, consider the ordered pair $(1,0.5)$, which is not in the region in question. Is this ordered pair in $A\times B$? Clearly $x=1\in A$, and $y=0.5\in B$ if one picks $x=0$ (which is in $A$), so that $y=0.5$ satisfies $0\leq 0.5\leq 2$. Or does the $x$ that I choose to determine the $y$ have to be $x=1$? So that for $x=1$, the $y$ values in $(1,y)$ can only come from $1\leq y\leq 2$? If so, then it works. Thanks!

Asaf Karagila
  • 393,674
user210190
  • 147

2 Answers2

1

The problem with $B$ as you have stated it in your question is that it is not well-defined: it depends on the value of $x$ you choose. I think the error you are making is that for an arbitrary element of the cartesian product $(x,y) \in A \times B$, the value $y$ cannot depend on the value $x$.

You can't have such a triangle as the product of two subsets of $\mathbb{R}$. Suppose you could write it as $A \times B$ for some $A,B \subset \mathbb{R}$. Then we would need $(0,0) \in A \times B$ (so that $0 \in A$ and $0 \in B$) and $(1,1) \in A \times B$ (so that $1 \in A$ and $1 \in B$). But then the point $(1,0)$ is in $A \times B$ (since $1 \in A$ and $0 \in B$), which is not in the triangle you described.

Geometrically, you can think of the cartesian product $A \times B$ as putting a (vertical) copy of $B$ at every $x$-coordinate in $A$. So it will always look rectangular (perhaps with some holes).

jl00
  • 806
1

No, you can't, I'm afraid. The $x$ in the definition of $B$ is not the same as the $x$ in the definition of $A$. Indeed, the definition of $B$ is a little ambiguous regarding $x$. Is the condition $x \le y \le 2$ for all $0 \le x \le 2$? Or is it $x \le y \le 2$ for some $0 \le x \le 2$? In the former case, we would have $B = \{2\}$ (and $A \times B$ would be a line segment) or in the latter case, we would have $B = [0, 2]$, in which case $A \times B$ is a square.

In general, $A \times B$ must be a disjoint union of "rectangles", where we use the term in a very loose sense: we include points and open/closed/semi-open rectangles/line segments. It cannot be the triangle you described. Why? If $(0, 0) \in A \times B$, then $0 \in A$ and $0 \in B$. Similarly if $(1, 1) \in A \times B$, then $1 \in A$ and $1 \in B$. This implies that $(1, 0) \in A \times B$, which is not inside the triangle.

Theo Bendit
  • 50,900