# Is it possible to write the set $T =${$(x, x, y) : x,y \in \mathbb{N}$} or $S =${$(x, x^2, y) : x,y \in \mathbb{N}$} as Cartesian product of 3 sets? [closed]

Is there some kind of uniqueness condition that might be applicable?

$$\mathbb{N}$$ denotes the set of Natural Numbers.

• No, because in both cases the second component depends on the first. – Brian M. Scott Sep 27 '20 at 18:07
• What is $N$? The natural numbers? – Eric Wofsey Sep 27 '20 at 20:15
• @EricWofsey yes – Just another person Sep 28 '20 at 7:21

Firstly, note that $$T = T' \times \mathbb{N}$$ where $$T' = \{(x,x) : x \in \mathbb{N}\}$$.
Now, suppose that $$T' = T_1 \times T_2$$. Since $$(1,1)\in T'$$ and $$(2,2)\in T'$$, we see that $$(1,2)\subset T_1$$ and $$(1,2) \subset T_2$$ and so $$(1,2)\times(1,2) \subset T_1\times T_2.$$
But set $$(1,2)\times (1,2) = \{(1,1), (1,2), (2,1), (2,2)\}$$ contains elements $$(1,2)$$ and $$(2,1)$$ which are not in $$T'$$ – here is contradiction.
Generally, for any non-constant function $$f: A \to B$$, set $$F = \{(x,f(x)) : x\in A\}$$ could not be represented in a such form: consider any two points $$x_1, x_2 \in A$$ such that $$f(x_1) \neq f(x_2)$$ and do the same reasoning as above.