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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!

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    $\begingroup$ 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 $f:\Bbb{C} \to \Bbb{C}$. $\endgroup$
    – peek-a-boo
    Jul 29, 2020 at 22:50

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Yes. You are exactly right; single-valued functions can be viewed as the special case of multi valued functions which send each input to a singleton set.

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