Are there uncountably many essentially inequivalent versions of Mathematics?

Question

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?

Answer 1

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.

Answer 2

Main Question:

(1) Yes, let $A_j: 2^{\aleph_j} \neq \aleph_{j+1}$ (i.e., GCH does not hold at $\aleph_j$). We can do this by simultaneously forcing (via a countable product of posets adding Cohen subsets) $2^{\aleph_j} = \aleph_{j+1+s_j}$ where $s_j$ represents the truth value at $j$.

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Edited Additions: You can also let $A_j$ be the statement "$\aleph_{j}^{L}$ is a cardinal (in $V$)" (i.e., the $j^{th}$ uncountable cardinal of the constructible universe is a cardinal in the actual universe). In this case, you could simultaneously force over $L$ (via a countable product of posets from $L$ collapsing cardinals) to add a surjection from $\aleph_{j-1}^L$ to $\aleph_{j}^L$ exactly when $s_j = 0$ so that the cardinal $\aleph_{j}^L$ becomes an ordinary ordinal of size $|\aleph_{j-1}^L|$ in the forcing extension. In the case that all of the $s_j$'s are $0$, the first $\aleph_0$ many cardinals of $L$ all become countable ordinals from the perspective of the forcing extension whereas if they're all $1$'s, then we have done trivial forcing and so the forcing extension is $L$.

Now after showing the desired relative consistency results as above, you can note (for your If so part) that you are only considering countable ordinals here from the perspective of most universes. For example, if a certain type of Real exists in your universe, mainly $0^{\sharp}$, then the true $\aleph_1$ will be inaccessible in $L$ and more so all of the $\aleph_j^L$'s for $j \in \mathbb{N}$ will be very puny countable ordinals in the said universe. Of course, this is probably cheating, but I thought I'd mention it anyway.

Also to your subquestion, $2^{\aleph_0} = \aleph_1$ and $2^{\aleph_0} = \aleph_2$ are very meaningful distinctions. But also under ZFC, $2^{\aleph_0}$ needs to be quite large in order to extend the Lebesgue measure to a countably additive measure on the full powerset of $\mathbb{R}$.

François already gave a very nice general answer to your main question so I think I'll leave my answer at that.