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Removed redundant language
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Noah Schweber
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Also, it occurs to me that we don't even need all of $ZFC_2$ to decide CH. Look at second-order analysis - that is, the natural second-order version of the first-order theory which is commonly called, annoyingly enough, "second-order arithmetic" (so I guess its second-order counterpart should be called "second-order analysis"). This will be enough to decide CH, since the arguments above will all go through.

Also, it occurs to me that we don't even need all of $ZFC_2$ to decide CH. Look at second-order analysis - that is, the natural second-order version of the first-order theory which is commonly called, annoyingly enough, "second-order arithmetic" (so I guess its second-order counterpart should be called "second-order analysis"). This will be enough to decide CH, since the arguments above will all go through.

Also, it occurs to me that we don't even need all of $ZFC_2$ to decide CH. Look at the natural second-order version of the first-order theory which is commonly called, annoyingly enough, "second-order arithmetic" (so I guess its second-order counterpart should be called "second-order analysis"). This will be enough to decide CH, since the arguments above will all go through.

Forgot to actually answer the question
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Noah Schweber
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I just realized that I didn't answer your actual question.

As Andreas says above, your statement is not correct: both $ZFC+CH$ and $ZFC+\neg CH$ are recursively axiomatizable, and consistent (assuming $ZFC$ is), and one of them is a subtheory of $ZFC_2$ (although we can't tell which). You could try to add some effectiveness criterion to your statement - something along the lines of, "There is no recursively axiomatizable consistent theory $T$ which decides CH and that is provably a subtheory of $ZFC_2$" - but it's unclear to me how to do this in a way that results in a non-trivial, but also not false, statement. The moral is that second-order logic is really nasty. For instance, it wouldn't even make sense to ask for a derivation of CH (or $\neg CH$) from $ZFC_2$, since there's no meaningful proof system for second-order logic. To understand how ridiculously awful this is, there are proof systems for some infinitary logics that are very useful in model theory and proof theory - Lopez-Escobar developed one that Barwise used (altered? my history is a little vague on this point), but I don't know a good reference - and logics that can express concepts like "is uncountable" or can quantify over automorphisms of certain kinds of structures are even compact. Basically, second-order logic is totally unusable (although, as always, there are exceptions).

Also, it occurs to me that we don't even need all of $ZFC_2$ to decide CH. Look at second-order analysis - that is, the natural second-order version of the first-order theory which is commonly called, annoyingly enough, "second-order arithmetic" (so I guess its second-order counterpart should be called "second-order analysis"). This will be enough to decide CH, since the arguments above will all go through.


I just realized that I didn't answer your actual question.

As Andreas says above, your statement is not correct: both $ZFC+CH$ and $ZFC+\neg CH$ are recursively axiomatizable, and consistent (assuming $ZFC$ is), and one of them is a subtheory of $ZFC_2$ (although we can't tell which). You could try to add some effectiveness criterion to your statement - something along the lines of, "There is no recursively axiomatizable consistent theory $T$ which decides CH and that is provably a subtheory of $ZFC_2$" - but it's unclear to me how to do this in a way that results in a non-trivial, but also not false, statement. The moral is that second-order logic is really nasty. For instance, it wouldn't even make sense to ask for a derivation of CH (or $\neg CH$) from $ZFC_2$, since there's no meaningful proof system for second-order logic. To understand how ridiculously awful this is, there are proof systems for some infinitary logics that are very useful in model theory and proof theory - Lopez-Escobar developed one that Barwise used (altered? my history is a little vague on this point), but I don't know a good reference - and logics that can express concepts like "is uncountable" or can quantify over automorphisms of certain kinds of structures are even compact. Basically, second-order logic is totally unusable (although, as always, there are exceptions).

Also, it occurs to me that we don't even need all of $ZFC_2$ to decide CH. Look at second-order analysis - that is, the natural second-order version of the first-order theory which is commonly called, annoyingly enough, "second-order arithmetic" (so I guess its second-order counterpart should be called "second-order analysis"). This will be enough to decide CH, since the arguments above will all go through.

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Noah Schweber
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Think of it this way: Let $V$ be a model of $ZFC_2$. Then I claim CH holds in $V$ if and only if $CH$ is actually true (note that in order for second-order logic to make sense, we have to make a commitment to an underlying "real" universe of sets). The proof of this is as follows. First, $\omega^V$ has order type $\omega$: clearly it has a subset of order type $\omega$, and by the second-order version of the powerset axiom, $P^V(\omega^V)=P(\omega^V)$, so if $\omega^V$ had the wrong order type $V$ would "see" the error. A fortiriori, we can deduce that $\omega^V$ is countable.

By similar reasoning, $P^V(P^V(\omega^V))=P(P(\omega^V))$. Now CH is false if and only if $P(P(\omega^V))$ contains three infinite sets $X, Y, Z$ no two of which have the same cardinalities (left-to-right is trivial; right-to-left follows from the countability of $\omega^V$).

Suppose $CH$ is false; let $X, Y, Z$ be as above. Since $P(P(\omega^V)=P^V(P^V(\omega^V))$, we have $X, Y, Z\in V$; by the axiom of extensionality, $V$ sees that the cardinalities of $X$, $Y$, and $Z$ are different, and by the second-order powerset axiom $V$ sees that $X$, $Y$, and $Z$ are infinite. So $CH\implies (ZFC_2\models \neg CH)$.

Suppose now that $CH$ is true. Let $X, Y, Z\in P(P(\omega^V))$; again, we have $X, Y, Z\in V$. Since $CH$ holds, by the second-order powerset axiom plus separation we can find a bijection $f$ between two of $X, Y, Z$, so $CH$ holds in $V$. So $\neg CH\implies (ZFC_2\models CH)$.

This shows that $ZFC_2\models CH$ or $ZFC_2\models \neg CH$. The point is that the full power of second-order logic allows $V$ to "ask" certain set-theoretic questions of the "real" underlying universe of sets; these questions include ``Is CH true?" Similarly, it seems to me that they include all questions of the form "Does $V_\alpha\models \phi$ hold?" where $\alpha$ is a computable ordinal and $\phi$ is $\Sigma_1$ over $V_\alpha$ ($\Sigma_1$ is somewhat arbitrary; higher quantifier depth can (I believe) be achieved by passing to larger computable $\alpha$).

I'd imagine that in fact this phenomenon extends much further than what I've outlined, and that a staggeringly large class of sentences of set theory are known to be decided in $ZFC_2$, even if we don't know which way they are decided.