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3 deleted 25 characters in body

François has excellently addressed your question 1; allow me to address question 2. I understand the question to be: what will be the mathematical effects if someone were to show that there are no (weakly) inaccessible cardinals? A similar question would apply to any of several large cardinals. So let me list some consequences.

First, let me note that the existence of a weakly inaccessible cardinal is provably equiconsistent with the existence of a (strongly) inaccessible cardinal, since any weakly inaccessible cardinal is strongly inaccessible in $L$, and so the issue about weakly or strongly inaccessible is entirely irrelevant when it comes to consistency.

Second, let me note that set theorists are not generally satisfied by claims of the sort "the only known proof uses such-and-such," but rather they use the concepts of consistency strength and equiconsistency, which allow for precise claims to be proved about exactly which large cardinals are required to prove which statements. The situation is that for many mathematical assertions, we can prove that any proof must use a certain type of large cardinal or something just as strong, in the sense that the consistency of the statement itself implies the consistency of the large cardinal in question. In this way, we avoid any problematic issue about knowledge concerning whether a better proof is simply not yet discovered.

As a result, if inaccessible cardinals should be refuted, then using the known results we immediately gain an enormous number of positive theorems. So it isn't really a case of losing theorems, but rather gaining.

Theorem. If inaccessible cardinals are inconsistent, then (we can prove that) we can construct a non-Lebesgue measurable set of reals without using the axiom of choice.

This follows from the fact that Solovay and Shelah have proved that the possibility of constructing a non-Lebesgue measurable set of reals (in the context of ZF+DC) without using AC is exactly equivalent to the inconsistency of inaccessible cardinals.

Most people believe that one must use AC in any Vitali-type construction of a non-Lebesgue measurable set, and the theorem above shows that this belief is provably equivalent to the consistency of inaccessible cardinals. Perhaps many mathematicians would find their confidence in the consistency of inaccessible cardinals to increase upon learning of this, and in this sense, this is also an answer to question 1. In any case, many well-known set theorists have emphasized enormous confidence in the consistency of large cardinals, and have stated quite explicitly that if inaccessible cardinals should become known to be inconsistent, then we should expect further inconsistency much lower in ZFC itself or in the low levels of PA.

Theorem. If inaccessible cardinals are inconsistent (and even merely if we can refute infinitely many Woodin cardinals), then (we can prove that) there is a projective set of reals $A\subset\mathbb{R}$ whose corresponding two-person game of perfect information has no winning strategy for either player. In other words, the infinitary de Morgan law $$\neg\forall n_0\exists n_1\forall n_2\exists n_3\cdots A(\vec n)\iff\exists n_0\forall n_1\exists n_2\forall n_3\cdots\neg A(\vec n)$$ will fail for some projective set $A$.

The projective sets of reals are those reals that are definable by a property involving quantification only over real numbers and integers. The reason for the theorem is that projective determinacy is equiconsistent over ZFC with infinitely many Woodin cardinals, and so if we refute the large cardinals in ZFC, then we similarly refute projective determinacy.

Theorem. If inaccessible cardinals are inconsistent (and even if merely measurable cardinals are inconsistent), then (we can prove that) there is an analytic set (a continuous image of a Borel set) that is not determined, not measurable, and etc.

Theorem. If inaccessible cardinals are inconsistent, then we can prove that the full set-theoretic universe is very close to the constructible universe in the sense of covering. In particular, $L$ computes the successors of singular cardinals correctly.

This shocking conclusion follows in this case from Jensen's covering lemma, since refuting inaccessible cardinals implies a refutation of $0^\sharp$.

Theorem. If inaccessible cardinals are inconsistent, then on no set is there a countably complete real-valued measure measuring all subsets of the set and giving points no mass.

This is simply because any real-valued measurable cardinal is measurable and hence inaccessible in an inner model.

Theorem. If inaccessible cardinals are inconsistent, then (we can prove that) there are no uncountable Grothendieck universes and the axiom of universes in category theory is false.

An uncountable Grothendieck universe is exactly $H_\kappa$ for an inaccessible cardinal $\kappa$, and the axiom of universes asserts that every set is in such a universe.

There are many more examples. (I invite any knowledgeable person to edit the answer with additional examples.)

2 added 26 characters in body

François has excellently addressed your question 1; allow me to address question 2. I understand the question to be: what will be the mathematical effects if someone were to show that there are no (weakly) inaccessible cardinals? A similar question would apply to any of several large cardinals. So let me list some consequences.

First, let me note that the existence of a weakly inaccessible cardinal is provably equiconsistent with the existence of a (strongly) inaccessible cardinal, since any weakly inaccessible cardinal is strongly inaccessible in $L$, and so the issue about weakly or strongly inaccessible is entirely irrelevant when it comes to consistency.

Second, let me note that set theorists are not generally satisfied by claims of the sort "the only known proof uses such-and-such," but rather they use the concepts of consistency strength and equiconsistency, which allow for precise claims to be proved about exactly which large cardinals are required to prove which statements. The situation is that for many mathematical assertions, we can prove that any proof must use a certain type of large cardinal or something just as strong, in the sense that the consistency of the statement itself implies the consistency of the large cardinal in question. In this way, we avoid any problematic issue about knowledge concerning whether a better proof is simply not yet discovered.

As a result, if inaccessible cardinals should be refuted, then using the known results we immediately gain an enormous number of positive theorems. So it isn't really a case of losing theorems, but rather gaining.

Theorem. If inaccessible cardinals are inconsistent, then (we can prove that) we can construct a non-Lebesgue measurable set of reals without using the axiom of choice.

This follows from the fact that Solovay and Shelah have proved that the possibility of constructing a non-Lebesgue measurable set of reals (in the context of ZF+DC) without using AC is exactly equivalent to the inconsistency of inaccessible cardinals.

Most people believe that one must use AC in any Vitali-type construction of a non-Lebesgue measurable set, and the theorem above shows that this belief is provably equivalent to the consistency of inaccessible cardinals. Perhaps many mathematicians would find their confidence in the consistency of inaccessible cardinals to increase upon learning of this, and in this sense, this is also an answer to question 1. In any case, many well-known set theorists have emphasized enormous confidence in the consistency of large cardinals, and have stated quite explicitly that if inaccessible cardinals should become known to be inconsistent, then we should expect further inconsistency much lower in ZFC itself or in the low levels of PA.

Theorem. If inaccessible cardinals are inconsistent (and even merely if we can refute infinitely many Woodin cardinals), then (we can prove that) there is a projective set of reals $A\subset\mathbb{R}$ whose corresponding two-person game of perfect information has no winning strategy for either player. In other words, the infinitary de Morgan law $$\neg\forall n_0\exists n_1\forall n_2\exists n_3\cdots A(\vec n)\iff\exists n_0\forall n_1\exists n_2\forall n_3\cdots\neg A(\vec n)$$ will fail for some projective set $A$.

The projective sets of reals are those reals that are definable by a property involving quantification only over real numbers and integers. The reason for the theorem is that projective determinacy is equiconsistent over ZFC with infinitely many Woodin cardinals, and so if we refute the large cardinals in ZFC, then we similarly refute projective determinacy.

Theorem. If inaccessible cardinals are inconsistent (and even if merely measurable cardinals are inconsistent), then (we can prove that) there is an analytic set (a continuous image of a Borel set) that is not determined, not measurable, and etc.

Theorem. If inaccessible cardinals are inconsistent, then we can prove that the full set-theoretic universe is very close to the constructible universe in the sense of covering. In particular, $L$ computes the successors of singular cardinals correctly.

This shocking conclusion follows in this case from Jensen's covering lemma, since refuting inaccessible cardinals implies a refutation of $0^\sharp$.

Theorem. If inaccessible cardinals are inconsistent, then on no set is there a countably complete real-valued measure measuring all subsets of the set and giving points no mass.

This is simply because any real-valued measurable cardinal is measurable and hence inaccessible in an inner model.

Theorem. If inaccessible cardinals are inconsistent, then (we can prove that) there are no uncountable Grothendieck universes and the axiom of universes in category theory if is false.

An uncountable Grothendieck universe is exactly $H_\kappa$ for an inaccessible cardinal $\kappa$, and the axiom of universes asserts that every set is in such a universe.

There are many more examples. (I invite any knowledgeable person to edit the answer with additional examples.)

1

François has excellently addressed your question 1; allow me to address question 2. I understand the question to be: what will be the mathematical effects if someone were to show that there are no (weakly) inaccessible cardinals? A similar question would apply to any of several large cardinals. So let me list some consequences.

First, let me note that the existence of a weakly inaccessible cardinal is provably equiconsistent with the existence of a (strongly) inaccessible cardinal, since any weakly inaccessible cardinal is strongly inaccessible in $L$, and so the issue about weakly or strongly inaccessible is entirely irrelevant when it comes to consistency.

Second, let me note that set theorists are not generally satisfied by claims of the sort "the only known proof uses such-and-such," but rather they use the concepts of consistency strength and equiconsistency, which allow for precise claims to be proved about exactly which large cardinals are required to prove which statements. The situation is that for many mathematical assertions, we can prove that any proof must use a certain type of large cardinal or something just as strong, in the sense that the consistency of the statement itself implies the consistency of the large cardinal in question. In this way, we avoid any problematic issue about knowledge concerning whether a better proof is simply not yet discovered.

As a result, if inaccessible cardinals should be refuted, then using the known results we immediately gain an enormous number of positive theorems. So it isn't really a case of losing theorems, but rather gaining.

Theorem. If inaccessible cardinals are inconsistent, then (we can prove that) we can construct a non-Lebesgue measurable set of reals without using the axiom of choice.

This follows from the fact that Solovay and Shelah have proved that the possibility of constructing a non-Lebesgue measurable set of reals (in the context of ZF+DC) without using AC is exactly equivalent to the inconsistency of inaccessible cardinals.

Most people believe that one must use AC in any Vitali-type construction of a non-Lebesgue measurable set, and the theorem above shows that this belief is provably equivalent to the consistency of inaccessible cardinals. Perhaps many mathematicians would find their confidence in the consistency of inaccessible cardinals to increase upon learning of this, and in this sense, this is also an answer to question 1. In any case, many well-known set theorists have emphasized enormous confidence in the consistency of large cardinals, and have stated quite explicitly that if inaccessible cardinals should become known to be inconsistent, then we should expect further inconsistency much lower in ZFC itself or in the low levels of PA.

Theorem. If inaccessible cardinals are inconsistent (and even merely if we can refute infinitely many Woodin cardinals), then (we can prove that) there is a projective set of reals $A\subset\mathbb{R}$ whose corresponding two-person game of perfect information has no winning strategy for either player. In other words, the infinitary de Morgan law $$\neg\forall n_0\exists n_1\forall n_2\exists n_3\cdots A(\vec n)\iff\exists n_0\forall n_1\exists n_2\forall n_3\cdots\neg A(\vec n)$$ will fail for some projective set $A$.

The projective sets of reals are those reals that are definable by a property involving quantification only over real numbers and integers. The reason for the theorem is that projective determinacy is equiconsistent over ZFC with infinitely many Woodin cardinals, and so if we refute the large cardinals in ZFC, then we similarly refute projective determinacy.

Theorem. If inaccessible cardinals are inconsistent (and even if merely measurable cardinals are inconsistent), then (we can prove that) there is an analytic set (a continuous image of a Borel set) that is not determined, not measurable, and etc.

Theorem. If inaccessible cardinals are inconsistent, then we can prove that the full set-theoretic universe is very close to the constructible universe in the sense of covering. In particular, $L$ computes the successors of singular cardinals correctly.

This shocking conclusion follows in this case from Jensen's covering lemma, since refuting inaccessible cardinals implies a refutation of $0^\sharp$.

Theorem. If inaccessible cardinals are inconsistent, then on no set is there a countably complete real-valued measure measuring all subsets of the set.

This is simply because any real-valued measurable cardinal is measurable and hence inaccessible in an inner model.

Theorem. If inaccessible cardinals are inconsistent, then (we can prove that) there are no uncountable Grothendieck universes and the axiom of universes in category theory if false.

An uncountable Grothendieck universe is exactly $H_\kappa$ for an inaccessible cardinal $\kappa$, and the axiom of universes asserts that every set is in such a universe.

There are many more examples. (I invite any knowledgeable person to edit the answer with additional examples.)