# Are there uncountably many essentially inequivalent versions of Mathematics?

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?

-
Zen, logic is also very remote from my field, so maybe I am missing something. But how is your main question not answered by Goedel's incompleteness theorems? –  Alex B. Feb 17 '11 at 10:20
Supposing ZFC is consistent and given a finite set of axioms $I_0, I_1, \ldots, I_n$ that are all independent of ZFC, the axioms of ZFC union $\{I_0, I_1, \ldots, I_n\}$ (or the negations of any of the $I_j$) will still be computable because you're only adding a finite set. Consequently, by Goedel's incompleteness theorem, you will have a statement that's independent of the collection, $I_{n+1}$, and can add it (or its negation) to the list. Therefore there will actually be infinitely many computable extensions of ZFC. –  Jason Feb 17 '11 at 10:56
I was trying to think of explicit natural examples though. I'll try to think about it more later. –  Jason Feb 17 '11 at 10:58
Regarding the title question (but perhaps not the main question): I was under the impression that mathematics as practiced is encoded by finite (and perhaps bounded) chunks of information encoding the rules and conventions we follow. In that case, you can only have countably many inequivalent versions of mathematics that can be distinguished in a manner that can be communicated in finite time. –  S. Carnahan Feb 17 '11 at 15:38
@Scott: Fixing an enumeration of formulas in a countable language, there are indeed only countably many computably enumerable theories. However, you can also consider continuum many consistent extensions of a computable consistent theory (even though you won't be able to effectively prove all of its theorems). Also, let me remark separately that for my previous comment, each $I_j$ should be independent of ZFC union $\{I_0, I_1, \ldots, I_{j-1}\}$ and not just of ZFC. –  Jason Feb 17 '11 at 23:06

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.

-
Thanks very much; I'm still thinking about it. I like this answer and also Jason's answer, and can't yet decide which answer to choose, as in the meta discussion: tea.mathoverflow.net/discussion/178/dilemma-in-choosing-answers/… –  Zen Harper Feb 18 '11 at 5:03

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 $j$.

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 $\mathbb{R}$.

If you have any follow-up questions, feel free to ask them here. The general idea with collapsing cardinals is that a model of set theory can think that sets are larger than they really are in the "true" universe. For example, if we have a countable ZFC model $M$ for which $m \in M$ implies $m \subseteq M$ (i.e., transitive), then its $\aleph_1$, denoted $\aleph_1^M$, is countable since $\aleph_1^M \subseteq M$ and $M$ is countable. But if we were to add a bijection to this model from $\aleph_0$ into $\aleph_1^M$ through forcing, then the extension would realize that $\aleph_1^M$ is countable. –  Jason Feb 20 '11 at 7:22
In the extension, $\aleph_1^M$ is no longer a cardinal because there is a bijection from the smaller $\aleph_0$ into it. So the forcing extension $N$ thinks that $\aleph_1^M < \aleph_1^N$ where $\aleph_1^N$ will be the least uncountable ordinal in $N$. The constructible universe $L$ is an inner model (transitive and contains all ordinals of $V$) and is the minimal inner model in the sense that it is contained in any other inner model of ZF. Assuming certain large cardinal hypotheses, it turns out to be quite small as mentioned above. –  Jason Feb 20 '11 at 8:57