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

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

  • 3
    $\begingroup$ No, because in both cases the second component depends on the first. $\endgroup$ – Brian M. Scott Sep 27 '20 at 18:07
  • $\begingroup$ What is $N$? The natural numbers? $\endgroup$ – Eric Wofsey Sep 27 '20 at 20:15
  • $\begingroup$ @EricWofsey yes $\endgroup$ – 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.


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