Axiom Schema of Restricted Comprehension and Big O notation

Question

In this blog post about ultrafilters and nonstandard analysis, it is stated that "if we attempt to formalise this by trying to create the set $A := \{ x \in {\Bbb R}: x = O(1) \}$ of all bounded numbers, and asserting that this set is then closed under addition and multiplication, we are speaking nonsense; the O() notation cannot be used within the axiom schema of specification, and so the above definition of A is meaningless."

Why is this true? I thought that the defintion of $O(g(x))$ was $O(g(x)) := \{ f(x) \in {\Bbb R^{\Bbb R}}: \limsup\limits_{x\rightarrow \infty} |\frac{f(x)}{g(x)}| <∞\} $, which requires the axiom schema of specification.

Answer 1

The hypothetical notation $\{x\in\mathbb R : x=O(1)\}$ seems to be used in the context of a model created as an ultraproduct. It tries to distinguish $x$ within the non-standard model by referring to the growth rate of the function at the metalevel that represents $x$.

However, the comprehension axiom demands that the property you select a subset by is definable speaking only about concepts within the non-standard model. The language this property must be phrased in does not know that the "reals" of the model are really functions at the metalevel, so it makes no sense to apply big-O notation to them.

Looking in from the metalevel, we can of course form the collection of "reals-in-the-model" that are represented by bounded functions. (Or can we? There's a quotienting step happening on the way to the ultraproduct model, so some numbers will be represented both by some function that are $O(1)$ and some functions that aren't. But we could form a collection of reals-in-the-model that have at least one bounded representative). But this collection does not correspond to something the non-standard model considers a "set".

Answer 2

As you said, usually the notation $x=O(1)$ is connected to some limiting process. Without that process, the set $\{x:x=O(1)\}$ as the set of somehow "bounded" numbers is either non-sensical or just is the full set $\Bbb R$, since trivially $|x|\le C$ with $C=|x|$.

What the article is alluding to is probably the idea that elsewhere is called the "appreciable numbers" $x$ with $\exists^{st}N:|x|<N$. Note that there can be no set of appreciable numbers as subset of $\Bbb R$, as there can be no supremum of it. The same goes for the idea of a set of infinitesimals.

Answer 3

I think the problem is misidentified in the quoted passage:

"But if we attempt to formalise this by trying to create the set $A := \{ x \in {\Bbb R}: x = O(1) \}$ of all bounded numbers, and asserting that this set is then closed under addition and multiplication, we are speaking nonsense; the $O(\,)$ notation cannot be used within the axiom schema of specification, and so the above definition of $A$ is meaningless."

The problem is not that the $O(\,)$ notation can't be used with the axiom schema of specification; $O(\,)$ is a perfectly well-defined notion, and you can use it, for example, to create a set such as $\{f\colon\mathbb{R}\to\mathbb{R} \mid f(x)=O(x^2)\},$ using the axiom schema of specification. The problem is that the big $O$ notation applies to functions, not to numbers, so the string of characters $"\!x=O(1)\!"$ (where $x\in\mathbb{R})$ may look at first glance like a formula but doesn't actually mean anything.