Consider the single-valued functions $f: \mathbb{R}^n \to \mathbb{R}$, and the set valued function $F: \mathbb{R}^n \to 2^{\Bbb{R}}$.

Is $f$ a set-valued function as well?

My only trouble in understanding the difference is because the image of $F$ consists of sets, whereas the image of $f$ consists of an element in the reals, i.e., a number $6$ which is not a set $\{6\}$.

Can someone make clear to me the distinction between $f$ and $F$? Can we think of the set of $f$ as a sub-class of set-valued functions? What is the distinction between $f$ and a function whose output consists of singleton sets?

  • $\begingroup$ Technically no, they are not the same, but often people are sloppy and do not distinguish between a function such as $x \mapsto Ax$ and the set-valued function $x \mapsto \{Ax\}$. $\endgroup$ – littleO Apr 25 '17 at 21:24
  • $\begingroup$ So, for example, two different functions $f(x)=Ax$ and $F(x)=\{Ax\}$ might often be given the same name (just $f$), even though they are different functions. This is a bit sloppy and creates a danger of ambiguity, but, we can hope it is clear from context which of the two functions is being referred to. (In this example $A$ is a matrix, by the way.) $\endgroup$ – littleO Apr 25 '17 at 21:34

I would say yes, with the caveat that an identification has to be made.

Often in mathematics there is a bit of a discrepancy between how we technically define things and how we think of them. For example, what is $\mathbb{R}^3$? Is it $$(\mathbb{R} \times \mathbb{R}) \times \mathbb{R}$$ or $$\mathbb{R} \times (\mathbb{R} \times \mathbb{R})?$$ Technically these would be different, since the first has elements like $((a,b),c)$ and the second has elements $(a,(b,c))$. However, this distinction is not generally useful, so we identify them - that is, we forget about this difference between them. If you want to get technical, you can talk about isomorphisms, category theory, natural transformations and such. But it's generally considered fine to sweep this small technical difference under the rug.

Similarly, technically speaking a function $f: \mathbb{R} \rightarrow 2^{\mathbb{R}}$ cannot be a function $f: \mathbb{R} \rightarrow \mathbb{R}$ because no subset of $\mathbb{R}$ is also an element of $\mathbb{R}$. However, we can pretend that $\{2\} = 2$ (or in more usual jargon, identify $\{2\}$ with $2$). This is called an abuse of notation, and is generally considered fine as long as it is clear to the reader what the author intends. Math exposition can be very difficult, so most authors will take occasional shortcuts if it makes things easier to understand. Whether or not different kinds of abuse of notation are acceptable depends on the subject matter and the intended audience (and, if you're taking a class, what is expected by the grader.)

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    $\begingroup$ What is actually done when we "identify" one object with another? "Identify" sounds a little bit fast and loose for me. Are we constructing a map that relates the two sets? $\endgroup$ – Shamisen Expert Apr 25 '17 at 21:44
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    $\begingroup$ I was always confused by the term "identify", because my reaction was that you can't simply "identify" two different objects -- it didn't seem to make sense. We can't just turn off the part of our brain that knows these two objects are not the same. But, now I think when people say this it simply means that we will refer to the two different objects by the same name, and hope it is clear from context which object we are referring to. $\endgroup$ – littleO Apr 25 '17 at 21:49
  • $\begingroup$ Yes, it is a bit fast and loose. You're correct - one way to do this formally would be to construct a map $T$ from the set of functions $\mathbb{R} \rightarrow \mathbb{R}$ to the set of functions $\mathbb{R} \rightarrow 2^\mathbb{R}$ by setting $Tf$ to be the function $(Tf)(x) = \{f(x)\}$. $\endgroup$ – Jair Taylor Apr 25 '17 at 21:49
  • $\begingroup$ @littleO: Yes, maybe that's a better way to think of identification. $\endgroup$ – Jair Taylor Apr 25 '17 at 21:51

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