# Confusion about the definition of a multifunction/multi-valued function

I had this definition of a multi-function (for the case of complex-valued functions)

A multi-function on a open subset $$U$$ is $$f:U\to \mathcal{P}(\mathbb{C}).$$

Or at least how I interpret this definition is that $$f(z)\subset \mathbb{C}$$ for each $$z\in U.$$

However, I am now thinking wouldn't this definition imply that, informally speaking, single-valued functions $$\subset$$ multifunctions?

Since if I take any single-valued function, say $$f$$. Then for each $$z\in U$$, $$f(z)$$ will just be a singleton subset of $$\mathbb{C}$$ and therefore it makes $$f$$ a multifunction, by definition? Or are we saying that, for this $$f$$, $$f(z)$$ is an element of $$\mathbb{C}$$ and thus not a subset of $$\mathbb{C}$$?

So I guess what I trying to ask is do we, by convention, include single-valued functions as a subset of multi-functions?

Many thanks!

• Every $f:\Bbb{C} \to \Bbb{C}$ induces a map $\tilde{f} : \Bbb{C} \to \mathcal{P}(\Bbb{C})$ simply as $\tilde{f}(z):= \{f(z)\}$; as per your terminology, this $\tilde{f}$ is a multi-function, but of course, strictly speaking $f\neq \tilde{f}$ (because they have different target spaces so of course they're not the same map). Conversely, every $\tilde{f}:\Bbb{C} \to \mathcal{P}(\Bbb{C})$ such that for every $z\in \Bbb{C}$, $|\tilde{f}(z)| = 1$ (i.e a set with a single element) induces a map $f:\Bbb{C} \to \Bbb{C}$. Jul 29, 2020 at 22:50