How to distinguish convergence with limited fluctuation in a non-standard setting?

Question

I'm reading Edward Nelson's Radical Probability theory, and got confused on two concepts, convergence and limited fluctuation in a non-standard setting.

On page 21-22(see here):

Let $T$ be a subset of $\bf R$, and let $\xi: T \to \bf R$. We say that $\xi$ admits $k$ $\epsilon$-fluctuation, in case, there exists elements $t_0 < t_1 \ldots < t_k$ of $T$ with $$|\xi(t_0)-\xi(t_1)| \geq \epsilon, |\xi(t_1)-\xi(t_2)| \geq \epsilon \ldots |\xi(t_{k-1})-\xi(t_k)| \geq \epsilon$$

We say that a infinit sequence is convergent, iff for all $\epsilon > 0$, there exists a $k$ such that the sequence doesn't admit $k$ $\epsilon$-fluctuation.

We say that a finite or infinit sequence is of limited fluctuation, in case for all $\epsilon \gg 0$ and all $k \simeq \infty$, it doesn't admit $k$ $\epsilon$-fluctuation.

$\epsilon \gg 0$ means that $\epsilon$ is positive but not a positive infinitesimal. $k \simeq \infty$ means that $\frac{1}{k}$ is a positive infinitesimal.

To highlight the difference of two concept, the author provide an example which I don't follow:

Let $i \leq \nu$ be unlimited, and let $x_n = 0$,for $n \leq i$, and $x_n = 1$,for $i < n \leq \nu$. Then this seqence is of limited fluctuation, but not convergent.

I can't understand why this is the case. This sequence only takes on two value, for $k = 2$, either $|\xi(t_0)-\xi(t_1)|$ or $|\xi(t_1)-\xi(t_2)|$ has to equal zero. Thus it doesn't admit $k$ $\epsilon$-fluctuation, when $k \geq 2$. It should be convergent. What's wrong?

Answer 1

You've misquoted Nelson slightly. He says "Now an infinite sequence is convergent if and only if for all $\epsilon > 0$ there exists $k$ such that the sequence does not admit $k$ $\epsilon$ fluctuations". But the example you describe is not an infinite sequence: note the third paragraph on p20 where he explicitly contrasts sequence whose indices are in $[1,\ldots.,\nu]$ and "infinite" sequences. $\nu$ is finite (being a natural number), hence so is the sequence, it just happens to be very large.

Of course using the definition that $x_*$ is convergent to $x$ if for all illimited $r \leq \nu$, $x_r \simeq x$ you easily get nonconvergence. $x_*$ is not convergent because the only possible limit is $1$, whereas for any illimted $m < i < \nu$ we have $x_m=0$.

Answer 2

What one needs to understand here is the distinction Nelson makes between internal and external objects (your summary does not reproduce this aspect of Nelson's discussion which is essential). I have the impression that the definition of k epsilon-fluctuations can no longer be used to characterize convergence of sequences that are external. I personally find this terminology a bit confusing. I would guess that what is going on is that for standard sequences (i.e. natural extensions of real sequences in Keisler's terminology), convergence can be defined in terms of k epsilon-fluctuations. On the other hand, for more general internal sequences, such as the example Nelson gives, convergence can no longer be characterized in terms of fluctuations. Thus, in his example the sequence takes the value 0 infinitely many times, yet after that takes also the value 1. One doesn't expect a convergent sequence to do that. But I don't fully understand Nelson's terminology.