Equivalence relationship defined over membership of a in halo of b. (Non-Standard Analysis, Monads, Halos)

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Let's say we define an equivalence relationship such that $a \sim b \iff b\in \mu(a), \: \mu(a)$ is the halo\monad of $a$. By definition this would include all points an infinitesimal distance away from $a$, and this should be a properly defined equivalence relationship.

However, if we were to define a path by a sequence of points such that each subsequent point were related to the last, would some infinite sequence of these points go any where. Could you define a path between two points $st(a)\not=st(b)$ via such a sequence? Via the transitive property established by this being an equivalence relationship it is such that any point in this sequence would be related to the last; however, if $st(a)\not=st(b)$ then surely $a \sim b$ should hold false.

Please advise. I'm new to this if it isn't terribly obvious.

P.S. I have a feeling that no path defined in this fashion could go from $a$ to $b$ such that $st(a)\not=st(b)$; however, at the same time I feel that some $2^{\aleph_{0}}$ points could go from one point to another.

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Oh yeah, not only is there such a (hyper)sequence, but there's even a (hyper)finite one!

Throughout this post, I'll primarily work in the formalism of Internal Set Theory. I'll provide some minimal context in parentheses to explain how the same things work in the ordinary Robinsonian/superstructure framework of nonstandard analysis, but if you're studying the Robinsonian version, this gets much less accessible: not only will you have to replace "finite" with "hyperfinite" everywhere and replace $\mathbb{N}$ with $~^\ast \mathbb{N}$ in a bunch of places, but you'll probably have to work through the last three chapters of Goldblatt, especially the part on hyperfinite approximation, before you can make any kind of sense of it.

I. Given a standard set $S$, we can find a (hyper)finite set $H$, often called the (hyper)finite approximation of $S$, which contains every standard element of $S$.

In particular, this means we can pick a (hyper)finite set $H$ which contains every standard element of the interval $[0,1]$. Thanks to the finiteness of $H$, we can enumerate the elements of $H$ in a sequence $h: \mathbb{N} \rightarrow [0,1]$ defined by $$h_n = \begin{cases} \text{the $n$th element of $H$} & \text{if $n < |H|$} \\ 1 & \text{otherwise} \end{cases}$$

Now, for every $n\in \mathbb{N}$, we have that $h_n \sim h_{n+1}$. Why? Well, if the distance is not infinitesimal, i.e. $\exists^{st} \varepsilon > 0. |h_{n+1} - h_n| > \varepsilon$, then we can define $r = h_n + \frac{\varepsilon}{2}$ and take its shadow / standard part. This will satisfy $h_n < \mathrm{sh}(r) < h_{n+1}$. But since $\mathrm{sh}(r) \in [0,1]$ is standard, $H$ should have contained $\mathrm{sh}(r)$, a contradiction.

II. Okay, but can't we then use the set $H$ or the sequence $h_{\bullet}$ to show that $0 \sim 1$? No! This is because $\sim$ itself is not, strictly speaking, a relation: it's a predicate in the language of Internal Set Theory. In a Robinsonian framework, we would say it constitutes an external set, not an internal set.

You cannot apply induction to an arbitrary predicate (external sets of pairs), only to equivalence relations that actually form sets (internal sets) to conclude that all the elements in $h$ are related. All that induction would give you in this case is that $\forall^{st} n \in \mathbb{N}. \forall^{st} m \in \mathbb{N}. h_n \sim h_m$. But while $h_0 = 0$ occurs at a standard index, $h_{|H|} = 1$ very much does not, so you can't conclude $0 \sim 1$. In fact, the standard indices of the sequence never leave the halo of $0$: as Nicholas Todoroff said in the comment above, "such a sequence does not get very far", at least if you restrict yourself to standard finite length.

This might seem weird, but it's the same observation that you cannot apply induction to conclude that all elements of $\mathbb{N}$ (in Robinsonian terms, $~^\ast \mathbb{N}$) are standard, even though if $n \in \mathbb{N}$ is standard, so is $n+1$.

IV. A benign act of self-promotion: if you're interested in $\sim$ and other such "nearness predicates" on topological spaces, keep in mind that section 1.3 of my PhD thesis studies them in detail, although it probably won't be accessible to you at this point.

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The equivalence you mentioned holds only if there is a finite number of intermediate terms. Leibniz already would consider a partition of, say, $[0,1]$ into an infinite number, say $H$, of infinitesimal subintervals, with division points $0=x_0<x_1<\ldots<x_H=1$. The points $x_i$ form a "sequence" of sorts but it is not indexed by the natural numbers $\mathbb N$; instead, it is indexed by the hypernatural numbers $\mathbb N^\ast$. In particular, some of the indices are hyperfinite (infinite) numbers. This answer is meant as an introduction to Z. A. K.'s.