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I am confused with the following expression of a function:

$$K : \mathcal{Q} → \mathbb{R}^m_+$$

Here, it says that $K$ operates on $\mathcal{Q}$ and returns a vector of size $m$. But shouldn't a function return a single unique real number as per the definition of the funciton? (Bijection, Surjection $ Injection)

The above expression was taken from 3rd page of this journal

GENIVI-LEARNER
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1 Answers1

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A mapping (a generalization of a function, sometimes these terms are used as the same ones) should return a single unique element of the codomain. It can be a number or a vector or a function etc.

Defenition: $f$ is called a mapping if $f=(A,B,\Gamma_f)$, where $A$ and $B$ are sets ($A$ is called domain and $B$ is called codomain) and $\Gamma_f$ is a graph, i.e. $\Gamma_f \subset A\times B$ such that for every $a\in A$ there is a unique $b\in B: (a,b)\in\Gamma_f$.

The nature of $A$ and $B$ can be different.

In your case the codomain of $f$ is $\mathbb{R}_+^m$, whose elements are $m$-dimensional vectors with non-negative coordinates.

Maxim Nikitin
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  • you mentioned "single unique element of the codomain". How can vector in the codomain is single element? Isnt it multiple elements? Also you mentioned "nature of $A$ and $B$ can be different. Can you please elaborate what you mean by that? – GENIVI-LEARNER May 17 '20 at 19:01
  • @GENIVI-LEARNER, we consider vectors as ordered sets of $m$ elements. So, $\mathbb{R}$ is 'a set of vectors'. Each vector is it's single element. About the nature of sets: for instance, let $A$ be a set of people, $B$ is a set of people's names and let $f:A\rightarrow B$ matches one's name to it's holder. Then $f$ is a function. Of course, this is just a simple example. In fact, domains and codomains can consist of complex objects like sets, groups, modules, morphisms, topological spaces and so on. – Maxim Nikitin May 17 '20 at 19:12
  • Allright makes sense. What is the nature of $Q$ then? Is it also a vector in the given expression? – GENIVI-LEARNER May 17 '20 at 19:55
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    @GENIVI-LEARNER, as I understood from the journal, $Q$ is a convex, bounded area in $\mathbb{R}$. So, yes, every $q\in Q$ is a vector and $f$ matches a vector to a vector. – Maxim Nikitin May 17 '20 at 22:41
  • Ok, So just to clarify a little more. $f$ is mapping $\mathbb{R}$ to $\mathbb{R}^m$, right? $$f : \mathbb{R} → \mathbb{R}^m$$ – GENIVI-LEARNER May 18 '20 at 12:26
  • Oh, sorry, I mean bounded area in $\mathbb{R}^m$. So $f:\mathbb{R}^m\rightarrow\mathbb{R}^m$. – Maxim Nikitin May 18 '20 at 15:03