Why do we work in ZFC, and not ZF+CH? 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?
 A: 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:

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*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.
