Choice function

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A choice function is a mathematical function $f$ whose domain $X$ is a collection of nonempty sets such that for every $S$ in $X$ , $f (S)$ is an element of $S$ . In other words $f$ chooses exactly one element from each set in $X$ .

Ernst Zermelo introduced choice functions along with the axiom of choice (AC) in a 1904 paper that gave a proof of the well-ordering theorem,^[1] which states that every set can be well-ordered. AC states that every set of nonempty sets has a choice function. A weaker form of AC, the axiom of countable choice (AC_ω) states that every countable set of nonempty sets has a choice function. However, in the absence of either AC or AC_ω, some sets can still be shown to have a choice function.

If $X$ is a finite set of nonempty sets, then one can construct a choice function for $X$ by picking one element from each member of $X .$ This requires only finitely many choices, so neither AC or AC_ω is needed.

If every member of $X$ is a well-ordered nonempty set, then it is possible to pick the least element of each member of $X .$ In this case infinitely many choices may be required, but there is a rule for making the choices, so again neither AC or AC_ω is needed. The distinction between "well-ordered" and "well-orderable" is important here: if the members of $X$ were merely well-orderable, it would be necessary to choose a well-ordering of each member, and this might require infinitely many arbitrary choices, and therefore AC (or AC_ω, if $X$ were countably infinite).

If every member of $X$ is a nonempty set, and the union $\bigcup X$ is well-orderable, then it is possible to choose a well-ordering for this union, and this induces a well-ordering on every member of $X$ , so a choice function will exist as in the previous example. In this case it was possible to well-order every member of $X$ by making just one choice, so neither AC nor AC_ω was needed. (This example shows that the well-ordering theorem implies AC. The converse is also true, but less trivial.)