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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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Where did this problem come from? – J. M. Aug 2 '12 at 13:19
Book of Proof by Richard Hammack. Chapter 12.1, question #12. – laser295 Aug 2 '12 at 13:20
Yes, it is a function. Domain is $\mathbb{R^{2}}$ and co-domain is $V\times W$ where $V$ is a vector space over $R$ of dimension $2$ and $W$ is a vector space over $R$ of dimension 3. – Jayesh Badwaik Aug 2 '12 at 13:22
In this case, the notation $(3y,2x,x+y)$ refers to an ordered triple of numbers. As Jayesh Badwaik, I would consider it an element of $\mathbb R^3$ – Ross Millikan Aug 2 '12 at 13:28
@Andrew: The usual set-theoretic formalization of "function" identifies a function with its graph. – Henning Makholm Aug 2 '12 at 13:47
up vote 1 down vote accepted

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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I understand now, thank you. – laser295 Aug 2 '12 at 13:28
@Mathh Pressland: I do not believe the co-domain is $\mathbb{R}^{3}$, rather I believe the co-domain is a vector space $W$ where an element of $W$ is $w$ of the form $(u,v)$ where $u \in \mathbb{R}^{2}$ and $v \in \mathbb{R}^{3}$. This space might be isomorphic to $\mathbb{R}^{3}$ though. – Jayesh Badwaik Aug 2 '12 at 13:44
@JayeshBadwaik: No -- the elements of the set that represents the function are of the form $(u,v)$ meaning that $f(u)=v$. Each value of the function is a $v$ that lives in $\mathbb R^3$. – Henning Makholm Aug 2 '12 at 13:46
The codomain is not the same as the range. A function always has a domain and a codomain, but the codomain can contain more points than the range of the function. It is impossible to actually say what the codomain of a function is just by looking at it's "graph" set - you can only with certainly say what the range is, and that the codomain must contain the range. – Thomas Andrews Aug 2 '12 at 13:47
@HenningMakholm: Okay, thanks. I was not aware of this notation. So you are saying the notation is equivalent to the below? \begin{equation} \theta(x,y) = (3y,2x,x+y) \end{equation} – Jayesh Badwaik Aug 2 '12 at 13:48

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.

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