# Proof uniform convergence -> continuity: Why use hyperhyperreals?

1. Feb 15, 2014

### sal

A uniformly convergent sequence of continuous functions converges to a continuous function.
I have no problem with the conventional proof. However, in Henle&Kleinberg's Infinitesimal Calculus, p. 123 (Dover edition), they give a nonstandard proof, and they use the hyperhyperreals to do it. I don't understand why that's needed.

The proof is very short; here's an outline:

Let '~' mean infinitesimally close to and '~~' mean hyperinfinitesimally close to (i.e., infinitesimal as viewed from within the hyperreals, using the hyperhyperreal system constructed on top of the hyperreals).

Given a functions fn → (uniformly) f, and given any real x, THEN for any h ~~ x, and for any infinite N, we'll have:
fN(x) ~~ fN(h) (by assumption),
fN(x) ~ f(x) (by assumption),
and fN(h) ~ f(h) (by assumption).

Then we have f(x) ~ fN(x) ~~ fN(h) ~ f(h), and so f(x) ~ f(h), and f is continuous at x.

I don't understand why the proof doesn't work equally well if we just use the hyperreals, and replace the '~~' with '~' everywhere. Any clue would be appreciated.

2. Feb 19, 2014

### sal

So here it's been four days since I posted the question, 120 people have looked at it, and nobody's attempted an answer. Fooey. In the mean time, though, I think I may have come to a dim understanding of why it must be done this way:

In order to even pose the question in a way that makes it possible to find a nonstandard proof, we must have the sequence of functions $\{f_n\}$ defined for n infinite as well as for all finite n. And then, to make that work, we need to talk about continuity of a native hyperreal function defined natively over the hyperreals, which is not just an extension to nonstandard values of a function which was originally defined over the reals. And that, in turn, leads to the need to talk about continuity of a hyperreal function, which is what leads to the need for the hyperhyperreals.

At least I think that's what's going on. I would love it if someone who has more of a clue about this than I do would comment, however.

3. Apr 10, 2014

### GenePeer

I'm only just now beginning to properly understand hyperreals so I'll take a stab at it.

$f_N(x)$ is not a real function so we have yet to define what it being continuous means. Suppose we use the same definition as we used for real function. Then $h\sim r \implies f_N(h)\sim f_N(r)$, right?

This is problematic since it fails for $f_N(x) = x^N$. Since all $x^n$ are continuous for all real integers $n$, then by extension principle, $x^N$ should be continuous. But $s = \frac{1}{\sqrt[N]{2}} \sim 1$ yet $f_N(s) = \frac{1}{2} \not\sim 1 = f_N(1)$. Our definition failed because $s$ and $1$ are not close enough in terms of hyperreal numbers. So we need numbers smaller than positive hyperreals to define continuity on them.

I guess you can get away with using $f_N(r) \sim f_N(h)$ since $a \approx b \implies a \sim b$. I'm not sure. But the assumption that $h \approx r$ is necessary for the proof.

4. Apr 10, 2014

### HallsofIvy

Well, your question was "why are hyper-reals needed" and the simple answer is that they are NOT "needed". This particular proof happens to use them. But the statement that "uniform continuity implies continuity" can be easily proven without using hyper-reals. It looks to me like this proof is intended to show that "uniform continuity implies continuity" is true even if you extend to the hyper-reals. That is, the "hyper-reals" are part of the hypotheses.

5. Apr 10, 2014

### sal

HallofIvy: No, you missed the point. The question was why the proof which uses hyperreal calculus requires the use of hyperhyper reals (there are two "hypers" there).

6. Apr 11, 2014

### GenePeer

And the only way out of using hyperhyper reals is to use the epsilon-delta definition of continuity on hyperreal functions, which defeats the purpose of using nonstandard analysis in the first place. So once you've committed to hyperreals, you do need hyperhyper reals for this proof.