I understand that a function $f: \mathbb{R}\to \mathbb{R}$ cannot provide more than one value per input without failing to be a function, but what about $g: \mathbb{R} \to X$ or $h: X \to X$ where $X$ is the set of subsets of $\mathbb{R}$. Can $\sqrt{x}$ be considered a function from $\mathbb{R} \to X$ in this circumstance without restricting its range to only positive numbers since a subset of $\mathbb{R}$ is still a single member of $X$? I'm not sure if I'm missing something or not, but for example when the vertical line test is talked about regarding a relation being a function or not it seems like it's specifically about a function from $\mathbb{R} \to \mathbb{R}$. What am I not understanding?
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1Nothing. The vertical line test is for functions from subsets of $\Bbb R$ into $\Bbb R$. – José Carlos Santos May 01 '22 at 16:52
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1It's not specifically from $\mathbb{R}$ to $\mathbb{R}$, but you are correct that each member of the domain can only map to one member of the codomain (or range). Functions are "left total" and "right unique". – Joe May 01 '22 at 16:54
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Interesting question. The issue is trickier than may appear at first glance. Your example of the relation $x=y^2$ can be converted to a function $\Bbb R^{\ge0}$ to $\Bbb R^{\ge0}$ just by erasing the lower leg of the parabola. That is, you’ve restricted the target space (codomain) from all of $\Bbb R$ to the nonnegative reals.
But look at the example of the relation $y^2-x-2xy+x^2=0$. Call up Desmos.com and see that it’s a parabola with diagonal axis, so a parabola that’s rotated from its usual opening-up position by $45^\circ$ counterclockwise. In fact if you use Quadratic Formula to solve for $y$, you get $y=x\pm\sqrt x$.
Now, you see what the function is that you want to convert this relation to, it’s just $y=x+\sqrt x$, where as usual the radical means “nonnegative root only”. But you can’t get this from the original relation by refusing to allow certain function-values of your codomain.
Lubin
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