# Uniformly Bounded Functions of Hyperreals

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.