[Math] Are there uncountably many essentially inequivalent versions of Mathematics

axiomsgm.general-mathematicslo.logicset-theory

Hi everyone,

Disclaimer 1: logic and set theory are a long way from my field, so apologies in advance if I demonstrate extreme ignorance or stupidity, and please correct me if (when?) I write stupid things. But hopefully my basic meaning should be fairly clear to everyone even if I get some details wrong.

Disclaimer 2: I admit this question might be slightly subjective. But I feel it's not too subjective, and is fairly natural and interesting to most mathematicians, out of mere curiosity.

Framework: Throughout, let's assume that standard ZF set theory is consistent, and take it as our basic mathematical foundation. (I don't necessarily think this is best, but I prefer to pin down the discussion).

We all know that Mathematics comes in several distinct flavours: e.g. you can believe or disbelieve the Continuum Hypothesis, and both points of view are (equally?) valid; they are really just matters of opinion. Thus there are at least 2 different versions. Of course we have infinitely many different versions: each number $m=1,2,3,\ldots$ gives a different flavour of Mathematics, given by the axiom $2^{\aleph_0} = \aleph_m$.

Subquestion Does the value of $m$ really matter very much? $2^{\aleph_0} = \aleph_1$ seems a particularly special case; but I find it hard to believe there'd be very much meaningful distinction (in terms of theorems anyone would want to consider) between the axioms $2^{\aleph_0} = \aleph_{103}$ or $2^{\aleph_0} = \aleph_{275}$, for example.

If desired, we could regard these different versions of Mathematics as essentially equivalent (in a rough sense): the axioms all look very similar, given by a single parametrisation. We could also throw in versions with $2^{\aleph_\alpha}$, etc.

Alternatively, we could remove these difficulties completely by not even considering cardinals beyond $\aleph_2$ or $\aleph_3$, say; (or any $\aleph_m$ with finite $m$).

It would be really amusing if we could do the following, for then we would have (at least) $2^{\aleph_0}$ different flavours of Mathematics! (Although I suppose there might be technical difficulties with nonconstructive infinite 0,1 strings…!) We'd have an explicit injective function $f$ from $[0,1]$ into the class of all possible versions of Mathematics!

Main question

Can we find (or prove the existence of) an infinite sequence of axioms $A_1, A_2, A_3, \ldots$, for which every sequence of true/false assignments is consistent? (e.g. the infinite string 1011001110… would mean that $A_j$ is true for $j=1,3,4,7,8,9,\ldots$ and false for $j=2,5,6,10,\ldots$; we want every string to be consistent).

If so, can it be done with $A_1, A_2, \ldots$ all being essentially different kinds of axioms? [maybe it's stupidly optimistic to hope for this]. Can it be done without ever considering $\aleph_k$ for $k>3$, say (or 4, or any fixed finite number)?

If not, what's a reasonable known lower bound $K$ on the number of $A_1, \ldots, A_K$ which are known to exist, so that we have at least $2^K$ essentially different versions of Mathematics?

Best Answer

Your question is essentially asking about the structure of the Lindenbaum–Tarski algebra of ZF. Jason gave a concrete example showing that one can embed the free Boolean algebra on countably many generators inside the Lindenbaum–Tarski algebra of ZF. In fact, it can be shown that the Lindenbaum–Tarski algebra of ZF is a countable atomless Boolean algebra. (There is nothing very special about ZF here, one only needs that the theory is consistent, recursively axiomatizable, and that it encodes a sufficient amount of arithmetic.) Since there is only one countable atomless Boolean algebra up to isomorphism, this completely determines the structure of the Lindenbaum–Tarski algebra of ZF.

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