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I have been trying to wrap my head around the Continuum Hypothesis. I understand that it is independent of ZFC and so cannot be proven or disproven with the given axioms. What I don't understand is why we don't consider it an axiom to begin with?

I haven't really ever seen a convincing argument for the axiom of choice.

How do we know there isn't some other axiom that we somehow "glossed" over that would allow the continuum hypothesis to be proven/disproven?

Brothersquid
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The Continuum Hypothesis is rarely relevant (nor there is any available intuition as to whether it is preferable than its negation), while the Axiom of Choice is extremely useful in many areas, and fairly natural with the right point of view.

And the negation of the Axiom of Choice gives way more anti-intuitive consequences that the AoC does. So, other than philosophical prejudice, there is little incentive to not use the AoC. Here is a list of results that can appear in mathematics done without the AoC:

  • There is a model of the ZF theory where $\mathbb R$ has an infinite subset without a countable subset. This means that in such model there exists $S ⊂ \mathbb R$ and $a ∈ \overline S$ such that there is no sequence $\{x_n\} ⊂ S$ with $x_n \to a$.

  • In the same model as above, there is a subset $S ⊂ \mathbb R$ that is neither closed nor bounded, but such that every sequence in it admits a convergent subsequence.

  • There exists a model of ZF where there is a vector space with no basis.

  • There is a model of ZF in which there is a field with no algebraic closure.

  • There exists a model of ZF—actually, all known models with no choice—in which there is a set $S$ and an equivalence relation $\sim$ on $S$ such that there is no injection $S/\!\sim\, \to S$. That is, the set $S$ can be partitioned in a larger number of classes than the number of elements of the set.

  • There is a model of ZF where $\mathbb R$ is uncountable but it can be written as a countable union of countable sets.

  • There is a model of ZF where there is a vector space with two basis of different cardinalities. Hence, the notion of dimension does not work.

  • There is a model of ZF where all subsets of $\mathbb R^n$ are measurable. In such model, though, it is possible to show that $\mathcal P(\mathbb N)$ admits an equivalence relation where the quotient has larger cardinality than $\mathcal P(\mathbb N)$. And more importantly from a measure theory point of view, Lebesgue measure is no longer σ-additive.

Martin Argerami
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    That first bullet is strange. If $S\subset\mathbb R,$ what would it mean for a sequence of reals to converge to $S?$ – Thomas Andrews Dec 04 '22 at 23:16
  • My bad, I missed the closure. – Martin Argerami Dec 04 '22 at 23:18
  • It still doesn't mean anything for a sequence to converge to a set. – Thomas Andrews Dec 04 '22 at 23:19
  • Indeed. It was $a$, my bad again. – Martin Argerami Dec 04 '22 at 23:20
  • Sorry to keep beating on that first bullet but I think you mean infinite subset without a countable subset. – blargoner Dec 04 '22 at 23:23
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    I don't really have a problem with any of the bullets after the first two. I see no reason to expect an algebraic closure of every field, nor a basis for every vector space. It is certainly convenient to have these results, but it never struck me as counter-intuitive if we di not have them. – Thomas Andrews Dec 04 '22 at 23:23
  • As for the existence of models, maybe we could find a less sweeping axiom that would protect against the first bullets, eliminating those. Godel's incompleteness says any strong-enough system for set theory will have some models with counter-intuitive results. – Thomas Andrews Dec 04 '22 at 23:26
  • I disagree about the first paragraph. CH has significant applications in analysis; just in this field its relevance is apparent in all sorts of areas, from automatic continuity theory to the study of corona algebras. There are also significant applications of strong forcing axioms (a strong form of the negation of CH), for instance on the theory of C${}^*$-algebras. And a reasonable argument can be made about one of these sides being preferable to the other. – Andrés E. Caicedo Mar 19 '23 at 14:57