So, I encountered a function defined by
$$f(x):=\frac{1}{\sqrt{x-|x|}}$$
Clearly, since $x$ can't be greater than $|x|$ for any real $x$, the function has its domain $\mathcal D(f)$ equal to the empty set $\emptyset$.
I wonder what is its range. Is it the empty set $\emptyset$ or do we say that it is not defined or does not exist?
In general, let $A$ be the empty set and $B$ be any arbitrary set. We can define a function $f$ from $A$ to $B$. Clearly, $f=\emptyset$, when we take functions as a special case of a relation.
I think by definition, the range is the set of second elements of the ordered pairs in the function. So, $\mathcal R(f)=\emptyset$. Is it correct or am I misunderstanding things?
Best Answer
The image of a function is the set of all $y$ in the codomain such that there exists an $x$ in the domain so that $f(x)=y$. If there are no $x$ in the domain, then the image is empty. Your reasoning in the last paragraph is correct for the range of $f$.