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An Vitali set $V$ is constructed in https://en.wikipedia.org/wiki/Vitali_set by applying $\textsf{AC}$ to pick a single element from each of the countable equivalence classes in $\mathbb R / \mathbb Q$, restricted to the interval $[0,1]$.

The interval $[0,1]$ is then reconstructed (and additionally as a by-product of the method of construction, with extra elements in the range $[-1,2]$), from these equivalence classes by forming for each element $q_k$ $\in$ $\mathbb Q$ restricted to $[-1,1]$, the disjoint sets $V_{k}$ defined as:

$$V_{k}=V+q_{k}=\{v+q_{k}:v\in V\},$$ where $V$ is the Vitali set, and

$$[0,1]\subset\bigcup V_k \subset [-1,2].$$

By usage of $\sigma$ additivity and invariance under translation of the measure $\mu$

$$\lambda (V_k) = \lambda (V),$$

we get \begin{equation}1\le\sum_k \mu(V)\le 3\label{1}\tag{1}\end{equation}

However, if in $(\ref{1})$, the countable sum is replaced by $\omega$, then $(\ref{1})$ could become, for example:

\begin{equation}1\le \omega * \mu(V)\le 3\label{2}\tag{2}\end{equation}

or

\begin{equation}\displaystyle \frac{1}{\omega} \le \mu(V)\le \frac{3}{\omega}\label{3}\tag{3}\end{equation}

As creation of a Vitali set via the use of the axiom of Separation isn't possible, then 'in effect' $\textsf{AC}$ is providing a 'supercharged' axiom of Separation to create the Vitali set.

Does this use of $\textsf{AC}$ mean an uncountable $\mathbb R $ sized "equivalent expression" could be thought of as being used in the axiom of Separation, which means in $(\ref{3})$ it can lead to the knowledge of very small number types (e.g. Surreal Numbers etc.) or to an extension of the Lebesgue measure to include non-standard number types?

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    The Axiom of Choice is just a sentence in the language of set theory. Much like Jon Snow, it knows nothing. – Asaf Karagila Jan 03 '20 at 13:23
  • And there was me thinking that AC was like Lord Varys who knows everything (and much more than the axiom of Separation). –  Jan 03 '20 at 20:30

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What you are doing makes no sense. Cardinals and ordinals are not real numbers. If you want to think about them as surreal numbers, then you need to give up their standard arithmetic (in the case of cardinals you need to forfeit everything, in the case of ordinals you need to resort to "natural/Hessenberg sums" instead which tend to disagree with the standard arithmetic in most cases).

So there is no sense in moving from the sum to multiplying a constant by $\omega$. You're just mixing up concepts in a way that makes no sense.

Asaf Karagila
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  • @ Asaf Karagila : Lectures on the Hyperreals, R. Goldblatt, 1998, Section 3.8, page 27, derives the expression: 1/ω=ϵ where ϵ is an infinitesimal and ω is an unlimited number. It says "The properties observed of ϵ and ω show that ∗R is a proper extension of R and hence a new structure". So in a hyperreal number system the expression (3) has meaning when μ(V) is replaced by ϵ. Similarly hyperfinite sums can exist. As a result does this not explain why a Vitali set is not Lebesgue measurable because use of AC for Vitali sets means that infinitesimal numbers are 'effectively' created ? –  Jan 03 '20 at 20:05
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    As I said, you're mixing concepts. (This is my gripe with modern internet-based education (see also this.)) The $\omega$ used in the sentence you're citing is not the standard use of $\omega$ in set theory, and you would need to specify the context for it. Moreover, it's still not the same as just infinite summation and you're just mixing up concepts that have no business being mixed up. – Asaf Karagila Jan 03 '20 at 20:53
  • @ Asaf Karagila. The use of hyperreals in Vitali sets is discussed here : https://math.stackexchange.com/questions/257655/hyperreal-measure. –  Jan 03 '20 at 21:19
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    And as you can see, it's not a very well form idea to begin with. – Asaf Karagila Jan 03 '20 at 21:22