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Mon, 25 Apr 2022
Unordered pairs and the axiom of binary choice
A few days ago I demanded a way to construct an unordered pair !![x, y]]!! with the property that $$[x, y] = [y, x]$$ for all !!x!! and !!y!!, and also formulas !!P_1!! and !!P_2!! that would extract the components again, not necessarily in any particular order (since there isn't one) but so that $$[P_1([x, y]), P_2([x, y])] = [x, y]$$ for all !!x!! and !!y!!. Several readers pointed out that such formulas surely do not exist, as their existence would prove the axiom of binary choice. This is a sort of baby brother of the infamous Axiom of Choice. The full Axiom of Choice (“AC”) can be stated this way:
(From this it also follows that $$\{ \mathcal C(i) \mid i\in \mathcal I \},$$ the collection of elements selected by !!\mathcal C!!, is itself a set.) This is all much more subtle than it may appear, and was the subject of a major investigation between 1914 and 1963. The standard axioms of set theory, called ZF, are consistent with the truth of Axiom of Choice, but also consistent with its falsity. Usually we work in models of set theory in which we assume not just ZF but also AC; then the system is called ZFC. Models of ZF where AC does not hold are very weird. (Models where AC does hold are weird also, but much less so.) One can ask about all sorts of weaker versions of AC. For example, what if !!\mathcal I!! and !!\mathscr F!! are required to be countable? The resulting “axiom of countable choice”, is strictly weaker than AC: ZFC obviously includes countable choice as a restricted case, but there are models of ZF that satisfy countable choice but not AC in its fullest generality. Rather than restricting the size of !!\mathscr F!! itself, we can consider what happens when we restrict the size of its elements. For example, what if !!\mathscr F!! is a collection of finite sets? Then we get the “axiom of finite choice”. This is also known to be independent of ZF. What if we restrict the elements of !!\mathscr F!! yet further, so that !!\mathscr F!! is a family of sets, each of which has exactly two elements? Then we have the “axiom of binary choice”. Finite choice obviously implies binary choice. But the converse implication does not hold. This is not at all obvious. Binary choice is known to imply the corresponding version of binary choice in which each member of !!\mathscr F!! has exactly four elements, and is known not to imply the version in which each member has exactly three elements. (Further details in this Math SE post.) But anyway, readers pointed out that, if there were a first-order formula !!P_1!! with the properties I requested, it would prove binary choice. We could understand each set !!\mathscr F_i!! as an unordered pair !![a_i, b_i]!! and then $$ \{ \langle i, P_1([a_i, b_i])\rangle \mid i\in \mathcal I \},$$ which is the function !!i\mapsto P_1(\mathscr F_i)!!, would be a choice function for !!\mathscr F!!. But this would constitute a proof of binary choice in ZF, which we know is impossible. Thanks to Carl Witty, Daniel Wagner, Simon Tatham, and Gareth McCaughan for pointing this out. [Other articles in category /math] permanent link |