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I am following this answer by Asaf to a question, whether every set has a group structure. The main part is the following:

To prove that "every set can be made into a group $\Rightarrow$ Axiom of choice"

  • Given an infinite set $X$ we define $H(X)$ to be the least ordinal $\alpha$ that there is no injection $g:\alpha\to X$ (this is known as the Hartog number of $X$)
  • If $X$ can be injected into $H(X)$ then $X$ can be well ordered, since being injected into an ordinal means that $X$ inherits a well order.
  • Using the assumption that every set can be given a group structure we give a group structure to $X\cup H(X)$, and from this we deduce that there exists an injection from $X$ into $H(X)$.
  • Therefore if every set can be given a group structure, every set can be well ordered and therefore the axiom of choice holds.

With this (part o answer), I have a very simple question, which came to me while considering necessaties of the objects used in the arguments.

What is the necessity of taking "$H(X)$, least ordinal etc", in the arguments?

I mean, can we take simply $\mathcal{P}(X)$, the power set of $X$, instead of these objects "$H(X)$, least ordinal " in the arguments, and carry proof?

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p Groups
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    $H(X)$ is well-ordered; ${\cal P}(X)$ not. That is used in the next step. – Magdiragdag Sep 26 '16 at 06:35
  • Oh, this means, "we do not know whether $X$ is well ordered, so we can not say whether $\mathcal{P}(X)$ is well ordered", do you want to say this? – p Groups Sep 26 '16 at 06:36
  • I mean: $H(X)$ has a natural well-ordering, whereas ${\cal P}(X)$ does not (and, without further assumptions, may not even be well-orderable). – Magdiragdag Sep 26 '16 at 06:41

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You don't have to take the least ordinal, but you have to take an ordinal at least large as the least ordinal (which exists). So might as well be explicit and take the least ordinal.

The reason is that we want to end up showing there is an injection from $X$ into an ordinal. It is not going to help us to show that there is an injection from $X$ into its power set, we already know that: $x\mapsto\{x\}$.

Using the group structure we can conclude there is either an injection from $X$ into $H(X)$ or there is an injection in the other direction. But there cannot be an injection from $H(X)$ into $X$, so there is an injection from $X$ into an ordinal.

Asaf Karagila
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