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What is the difference between ${^\sigma}\mathcal{P}(\mathbb{R})$, ${^\ast}\mathcal{P}(\mathbb{R})$, and $\mathcal{P}({^\ast}\mathbb{R})$? I know that $\mathcal{P}({^\ast}\mathbb{R})$ is the powerset of the hyperreals, and I know that ${^\sigma}\mathcal{P}(\mathbb{R})$ is the element-wise translation of $\mathcal{P}(\mathbb{R})$ into the hyperreals, but why is $\mathcal{P}({^\ast}\mathbb{R})$ a proper superset of ${^\sigma}\mathcal(P)\mathbb{R}$?

I stumbled across this in section 4 of, in case anyone wants to see their definitions.


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I think you intend to say "proper superset" rather than "proper subset", as I explain in my answer. – Joel David Hamkins Nov 24 '12 at 22:37
Oh you're right I typed those backwards – Brian Nov 24 '12 at 23:20
up vote 4 down vote accepted
  1. $\cal{P}({}^\ast\mathbb{R})$ is the full standard power set of the nonstandard reals, the set of all subsets of ${}^\ast\mathbb{R}$. This power set includes the subsets consisting solely of infinitesimals, solely of the standard integers, and so on, since these are subsets of ${}^\ast\mathbb{R}$; these particular subsets do not exist in the nonstandard universe.

  2. ${}^\ast\cal{P}(\mathbb{R})$ is the nonstandard version of the power set of $\mathbb{R}$, the set of all sets of reals as it is seen in the nonstandard universe, which thinks the reals are Archimedean and so on (with respect to its nonstandard ${}^\ast\mathbb{N}$). This version of the power set, accordingly, does not include the collection of infinitesimals or the set of standard integers, since these would reveal the reals to be incomplete or non-Archimedean. But every set that is thought to be a set of reals in the nonstandard realm is really a subset of ${}^\ast\mathbb{R}$, and so ${}^\ast\cal{P}(\mathbb{R})\subset \cal{P}({}^\ast\mathbb{R})$.

  3. Finally, as I understand the notation, ${}^\sigma\cal{P}(\mathbb{R})$ consists only of the nonstandard analogues ${}^\ast X$ of standard subsets $X\subset\mathbb{R}$ obtained by the transfer principle. These are all genuine subsets of ${}^\ast\mathbb{R}$, and indeed are thought to be sets of reals in the nonstandard realm. But not every nonstandard set in the nonstandard universe is the transfer of a standard set. For example, no nonstandard (pseudo) finite set of reals is ${}^\ast X$ for any standard $X$.

To summarize, we have $${}^\sigma\cal{P}(\mathbb{R})\quad \subsetneq\quad {}^\ast\cal{P}(\mathbb{R})\quad \subsetneq\quad \cal{P}({}^\ast\mathbb{R}).$$ Every ${}^\ast X$ for $X\subset\mathbb{R}$ is a nonstandard set of reals in the nonstandard realm, but not all sets of reals in the nonstandard universe arise this way (because of the nonstandard finite sets), and furthermore, there are standard subsets of ${}^\ast\mathbb{R}$ that do not exist in the nonstandard universe, such as the set of infinitesimals.

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The last sentence of the first point should perhaps end with "these particular subsets do not exist in the standard universe." or am I missing something? – Asaf Karagila Nov 24 '12 at 23:15
They do exist in the standard universe, since they are standard subsets of ${}^\ast\mathbb{R}$, but they do not exist in the non-standard universe, since they would reveal to that universe that the reals are incomplete or non-Archimedean. – Joel David Hamkins Nov 24 '12 at 23:20
Asaf, the situation is that we have the standard universe $V$, which can see everything, and then the transfer principle map $j:V\to {}^\ast V$, which is an elementary embedding. Since $V$ sees this map and all of ${}^\ast V$, it has ${}^\ast\mathbb{R}$ and all its subsets, but the nonstandard universe ${}^\ast V$ only has the subsets that are visible in that universe. With this notation, what we have is ${}^\sigma\cal{P}(\mathbb{R})=j''\cal{P}(\mathbb{R})$, ${}^\ast\cal{P}(\mathbb{R})=j(\cal{P}(\mathbb{R}))$ and ${\cal{P}}({}^\ast\mathbb{R})=P(j(\mathbb{R}))^V$. – Joel David Hamkins Nov 25 '12 at 2:29
I finally got a chance to read Robinson's book. Now this makes complete sense to me. Thanks so much! – Brian Sep 1 '15 at 1:30

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