I am trying to study ultraproducts and non-standard analysis, and I have the following question.

Fix a non-principal ultrafilter $\mathcal{F}$ to construct the hyperreal field ${}^*\mathbb{R}$. An element $x \in {}^*\mathbb{R}$ is said to be bounded if there exists $M \in \mathbb{R}$ such that $\lvert x \rvert < M$.

Let $f:{}^*\mathbb{R} \to {}^*\mathbb{R}$ be an internal function (that is, it is itself an ultraproduct of functions from $\mathbb{R}$ to itself).

Let $f$ be such that for every $x \in {}^*\mathbb{R}$, $f(x)$ is a bounded element in ${}^*\mathbb{R}$. Then is it true that $f$ is uniformly bounded? That is, does there exist $M>0$ such that for all $x \in {}^*\mathbb{R}$, $\lvert f(x) \rvert < M$?

Essentially, I am asking if for internal functions from ${}^*\mathbb{R}$ to itself, pointwise boundedness implies a uniform bound. My intuition is that it should be true, as if it weren't the case, we should be able to construct (using a diagonal argument, maybe?) an element $x$ such that $f(x)$ is an infinity.

However, my intuition and expertise in this area are limited, and I am unable to satisfactorily get an answer. Thanks in advance.


Yes, this is true. In $\mathbb{R}$, every nonempty set which is bounded above has a least upper bound, and thus the same is true for any internal subset of ${}^*\mathbb{R}$ (concretely, just take the least upper bound on each coordinate of the ultrapower). In particular, the image of $|f|$ is bounded above in ${}^*\mathbb{R}$ (by any unbounded element of ${}^*\mathbb{R}$), so it has a least upper bound $M\in {}^*\mathbb{R}$. Since $M$ is the least upper bound, there exists $x$ such that $|f(x)|>M-1$. Thus $M-1$ is bounded, which implies $M$ is also bounded.

More generally, the same argument shows that any internal subset which consists only of bounded elements must be uniformly bounded (this is sometimes called "overspill", since it means an internal set that contains arbitrarily large bounded elements must "spill over" and also have unbounded elements).

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