Range of a vector function

Question

May someone help me. Suppose that there exist two real functions $f_1$ and $f_2$ defined on $A\subset\Bbb{R}^n$. The images of them are denoted by $f_1(A)$ and $f_2(A)$, respectively. Let us define a vector function by $f=(f_1,f_2): A \to \Bbb{R}^2$. Am I right or not if I write $f(A)=f_1(A) \times f_2(A)$?. Is there a notion of the range for a vector function? If yes, the set $f(A)$ can be said the range of $f$ or not?

Answer 1

No, consider $f_{1}(x)=x$ and $f_{2}(x)=x^{2}$ for $x\in A:=[-1,1]$, then $f_{1}(A)=[-1,1]$ and $f_{2}(A)=[0,1]$ but the point $(-1/2,1/2)\notin f(A)$, if so, then $(-1/2,1/2)=(x,x^{2})$, and clearly no solution. Note that $(-1/2,1/2)\in[-1,1]\times[0,1]$.

Answer 2

Yes, of course there's a notion of the range of a vector function — just like for any function, it's the set of all its outputs. In other words: if $f:X\to Y$ is a function, then its range is the set $\{y\in Y\mid \exists x\in X \text{ such that }f(x)=y\}$. In particular, $X$ can be $\mathbb{R}^m$ and $Y$ can be $\mathbb{R}^n$ — the concept of the range is still the same as always.

And the range, or rather image, of a subset of the domain is also defined in the same way as for any function: if $f:X\to Y$ is a function and $A\subseteq X$ is a subset of the domain, then its image is the set $f(A)=\{y\in Y\mid \exists x\in A \text{ such that }f(x)=y\}$.

However, in the situation that you're describing, it is NOT true that "$f(A)=f_1(A) \times f_2(A)$". The reason is that the cartesian product will create all possible pairings of elements of $f_1(A)$ and $f_2(A)$, while the actual image consists only of the pairs that appear simultaneously as $f(x)=(f_1(x),f_2(x))$ for some element $x$.

Here's a quick counterexample. Let $A=\mathbb{R}$, $f_1:\mathbb{R}\to\mathbb{R}$ is defined via $f_1(x)=x$, and $f_2:\mathbb{R}\to\mathbb{R}$ is defined via $f_2(x)=x$.

• Then $f_1(A)=f_1(\mathbb{R})=\mathbb{R}$ and $f_2(A)=f_2(\mathbb{R})=\mathbb{R}$, so $f_1(A)\times f_2(A)=\mathbb{R}^2$.
• But $f(x)=(f_1(x),f_2(x))=(x,x)$, so $f(A)=\{(x,x)\mid x\in\mathbb{R}\}$ is the line $y=x$, not the entire plane $\mathbb{R}^2$.

Answer 3

I believe the way you have defined $f$ is as $f : A \to \mathbb{R}^2$, $f(a) = (f_1(a), f_2(a))$. $f$ does have a range -- the range of $f$, $f(A)$, is defined as

$$f(A) = \{ (x, y) | (x, y) = f(a) \text{ for some } a \in A \} {,}$$

the same way it is defined for other functions. Additionally, we have that

$$f_1(A) = \{ x | x = f_1(a) \text{ for some } a \in A \}$$ $$f_2(A) = \{ y | y = f_2(a) \text{ for some } a \in A \} {,}$$

so that

$$f_1(A) \times f_2(A) = \{ (x, y) | x \in f_1(A) \text{ and } y \in f_2(A) \}.$$

But these sets are not equal. Let $(x, y) \in f(A)$. Then there is an $a \in A$ such that $(x, y) = f(a)$, and hence $(x, y) = (f_1(a), f_2(a))$; thus $(x, y)$ is also in $f_1(A) \times f_2(A)$, and, since it was arbitrary, we have $f(A) \subseteq f_1(A) \times f_2(A)$.

However, if we were to attempt to show the reverse inclusion, we would run into the issue of showing that an arbitrary element of $f_1(A) \times f_2(A)$ belongs to $f(A)$. Consider, for example, the functions $f_1 : \mathbb{R} \to \mathbb{R}$, $f_1(x) = \sin{x}$; $f_2 : \mathbb{R} \to \mathbb{R}$, $f_2(x) = \cos{x}$; and $f : \mathbb{R} \to \mathbb{R}^2$, $f(x) = (f_1(x), f_2(x))$. Immediately we see that $(0, 0) \in f_1(\mathbb{R}) \times f_2(\mathbb{R})$, but $(0, 0) \notin f(\mathbb{R})$.