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I'm reading "A First Look At Rigorous Probability" by Jeffrey S. Rosenthal. On chapter one there is a proof which I can't fully understand.

Suppose, to the contrary, that $P(A)$ could be so defined for each subset $A \subseteq [0. 1]$. We will derive a contradiction to this. Define an equivalence relation on [0. 1] by: $x~y$ if and only if the difference $y - x$ is rational. This relation partitions the interval $[0. 1]$ into a disjoint union of equivalence classes. Let $H$ be a subset of $[0, 1]$ consisting of precisely one element from each equivalence class (such $H$ must exist by the Axiom of Choice). ....

My question is this: If we assume the Power Set Axiom, why do we need Axiom of Choice here. If the power set of $A$ exists, then its elements also exist. I mean, how can a set contain non-existent objects? And $H$ is clearly in the power set of $A$.

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Yes, we don't need the Axiom of Choice in order to be able to talk about $\mathcal{P}\bigl([0,1]\bigr)$. But we do need it when, after defining an equivalence relation on $[0,1]$, you want to work with a set with one and only one element of each equivalence class. This is a straightforward application of the Axiom of Choice: we have a family of sets (the equivalence classes) and we want to have a set with one and only one element from each set of that family.

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  • $\begingroup$ To the proposer: Let $E$ be the set of equivalence classes. AC is needed for the existence of a function $f:E\to \Bbb R$ such that $\forall e\in E\,(f(e)\in e).$ $\endgroup$
    – DanielWainfleet
    Jan 26, 2019 at 17:06
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$P(A)$ is not the power set operator. It is the probability operator, I am guessing they mean the standard uniform one.

Indeed, Solovay proved that it is consistent that the probability measures all sets. So the axiom of choice is really needed.

The point is that just like $\sqrt2$ or $\pi$ are not members of every field, the power set operation is not fully decided by the existence of a power set. In particular, in different set theoretic universes there will be power sets that disagree with each other.

What choice allows you, is to prove that there are non-measurable sets. The fact that there are mathematical universes where all sets are measurable means that without choice it is indeed impossible to prove such sets exists: they don't exist in some models of $\sf ZF$.

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  • $\begingroup$ Yes, it is not the power set operator. But in general, why do we need axiom of choice to construct (maybe "construct" is not the right word) $H$, if it is already in the power set of $A$? $\endgroup$
    – ashot4490
    Jan 23, 2019 at 14:56
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    $\begingroup$ Why is it necessarily "in the power set"? $\endgroup$
    – Asaf Karagila ♦
    Jan 23, 2019 at 14:57
  • $\begingroup$ @AshotJanibekyan Sure, you can set up the criterion "$H$ contains one element from each equivalence class", so one can easily check whether a given set is like that or not without the axiom of choice. But without choice (or something similar) we can't guarantee that such a set exists. $\endgroup$
    – Arthur
    Jan 23, 2019 at 14:59
  • $\begingroup$ @AsafKaragila Because the power set is the set of all subsets :/ I think you mean we need Axiom of Choice to say that $H$ is a "subset" of $A$? $\endgroup$
    – ashot4490
    Jan 23, 2019 at 14:59
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    $\begingroup$ @Ashot: I have edited my answer to help and clarify. $\endgroup$
    – Asaf Karagila ♦
    Jan 23, 2019 at 15:01

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