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

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?

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 $x_m=0$.
• Thank you for your answer. But I should admit that I can't understand it. It seems to me the example can be easily adapted to an infinite sequence which satisfies the characterization of $k$ $\epsilon$ fluctuations but fail to be convergent. We could add more elements (larger than $\nu$)into $T$ to make it "big" enough such that ${}^{\ast} \Bbb N$ can be injected into and assigns these extra element to $1$, if I understand the meaning of "infinite" correctly. – Metta World Peace Jun 26 '13 at 13:11
• I don't understand what you mean by making $T$ big enough such that $^*\mathbb{N}$ can be injected into it. First of all, $^*\mathbb{N}$ isn't meaningful here. There is only $\mathbb{N}$, which happens to contain nonstandard elements. Secondly, any segment $[1,\nu]$ for $\nu \in \mathbb{N}$ is a finite set so $\mathbb{N}$, being infinite, can't be injected into it. – Matthew Towers Jun 26 '13 at 13:31
• Besides, as Nelson says, an infinite sequence is convergent if for any $\epsilon>0$ there is a $k$ such that the sequence doesn't admit $k$ $\epsilon$-fluctuations. For such a sequence is Cauchy... – Matthew Towers Jun 26 '13 at 13:33