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Noah Schweber
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The answer seems to be "no": since none of the multiverse axioms involve two universes at once, the union of two multiverses is a multiverse. This lets you easily (EDIT: As Joel points out, this is totally false: it is not at all easy, although it may be true) construct counterexamples to the claim.

Meanwhile, if we add the (in my opinion, desirable) axiom that for every pair of universes $M$, $N$ there is a $W$ in which both $M$ and $N$ are countable sets, then the answer is clearly "yes;" and moreover, these two statements are equivalent (given $W$ interpreting $M$ and $N$, just consider a $V$ in which $W$ is countable.

The answer seems to be "no": since none of the multiverse axioms involve two universes at once, the union of two multiverses is a multiverse. This lets you easily construct counterexamples to the claim.

Meanwhile, if we add the (in my opinion, desirable) axiom that for every pair of universes $M$, $N$ there is a $W$ in which both $M$ and $N$ are countable sets, then the answer is clearly "yes;" and moreover, these two statements are equivalent (given $W$ interpreting $M$ and $N$, just consider a $V$ in which $W$ is countable.

The answer seems to be "no": since none of the multiverse axioms involve two universes at once, the union of two multiverses is a multiverse. This lets you easily (EDIT: As Joel points out, this is totally false: it is not at all easy, although it may be true) construct counterexamples to the claim.

Meanwhile, if we add the (in my opinion, desirable) axiom that for every pair of universes $M$, $N$ there is a $W$ in which both $M$ and $N$ are countable sets, then the answer is clearly "yes;" and moreover, these two statements are equivalent (given $W$ interpreting $M$ and $N$, just consider a $V$ in which $W$ is countable.

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Noah Schweber
  • 21.1k
  • 10
  • 110
  • 331

The answer seems to be "no": since none of the multiverse axioms involve two universes at once, the union of two multiverses is a multiverse. This lets you easily construct counterexamples to the claim.

Meanwhile, if we add the (in my opinion, desirable) axiom that for every pair of universes $M$, $N$ there is a $W$ in which both $M$ and $N$ are countable sets, then the answer is clearly "yes;" and moreover, these two statements are equivalent (given $W$ interpreting $M$ and $N$, just consider a $V$ in which $W$ is countable.