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I am familiar with the set-theoretical definition of functions.

A function $f : A \rightarrow B$ from a set $A$ to a set $B$ is a subset $f \subseteq A \times B$ of the Cartesian product $A \times B$ such that the following conditions hold:

  1. $\forall a \in A : \exists b \in B : (a,b) \in f$
  2. $\forall a \in A : \forall b, b' \in B : (a,b), (a,b') \in f \rightarrow b=b'$

Is there a purely set-theoretical definition of the function evaluation $f(a)$ for every $a \in A$? I get as far as defining $f(a) = \{ b \in B : (a,b) \in f \}$ but that defines rather the set of values of a multi-valued function.

shuhalo
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1 Answers1

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$f(a)$ is the unique $b \in B$ such that $(a,b) \in f$. In other words $$ f(a) = \bigcup \{ b \in B \mid (a,b) \in f \}. $$

Stefan Mesken
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    Depending on how familiar you are with unions, this may require a short proof which depends on the axiom of extensionality. – Stefan Mesken Jan 29 '18 at 17:48
  • Your answer is inconsistent. You define $f(a)$ as a set (namely as the set of values of $f$ at $a$) but before you speak of it as an element of $B$. Do we have $f(a) \in B$ or $f(a) \subset B$? – shuhalo Jan 29 '18 at 18:33
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    @shuhalo It's not inconsistent. In both cases $f(a)$ is an element of $B$. – Stefan Mesken Jan 29 '18 at 18:37
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    @shuhalo - Stefan Mesken's answer is correct, on the proviso that $b$ is a set, which is actually the case in a set-theoretic context, where everything is a set. – Taroccoesbrocco Jan 29 '18 at 19:18
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    @shuhalo I think you might be misparsing "$\bigcup$." "$\bigcup X$" denotes the union of the elements of $X$ - e.g. $\bigcup{a, b, c}=a\cup b\cup c.$ In case $X$ has only one element, then $\bigcup X$ is exactly that element - that is, $$\bigcup{x}=x$$ for every set $x$. So "$\bigcup{b\in B: (a, b)\in f}$" is in fact the unique $b$ such that $(a, b)\in f$. Of course, there is an implicit assumption here that $\bigcup X$ makes sense for all sets $X$; this ultimately rests on the set-theoretic context, in which indeed all objects are sets. – Noah Schweber Jan 29 '18 at 19:22
  • @NoahSchweber: Indeed, I was misparsing the union sign. – shuhalo Jan 29 '18 at 21:18