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Is the set $\theta=\{\big((x,y),(3y,2x,x+y)\big):x,y ∈ \mathbb{R}\}$ a function? If so, what is its domain, codomain, and range?

This is probably a dumb question. I understand what a function is, but the three elements in the ordered pair got me confused.

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    @Andrew: The usual set-theoretic formalization of "function" identifies a function with its graph.2012-08-02

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Yes it is, presumably one from $\mathbb{R}^2$ to $\mathbb{R}^3$, although the domain and codomain could potentially be smaller. You have an ordered pair in which the first element is itself an ordered pair (of real numbers), and the second is an ordered triple (of real numbers).

I'm used to codomain and range meaning the same thing. If you meant image for one of them, I can't think of a better description than $\{(3y,2x,x+y):x,y\in\mathbb{R}\}$.

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    @JayeshBadwaik: Yes, exactly.2012-08-02
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A general definition of function follows that of relation, a subset of the cartesian product of two sets $A\times B$, with the further prescription that each element of $A$ is in relation with exactly one element of $B$.

In such view, your set $\theta$ seems a subset of $\mathbb{R}^2\times \mathbb{R}^3$.

The domain is $\mathbb{R}^2$ and the range or image is a subset of $\mathbb{R}^3$. When $\mathbb{R}^3$ is viewed as a linear space and given the explicit linear and homogeneous form of the given formulas, this subset is a subspace.