User:IssaRice/Subsets of the reals that can define their own convergence

(probably not very important)

Something that has bugged me about the Cauchy sequence construction of the reals is that in the definition of Cauchyness/convergence, a rational $\epsilon >0$ is used, rather than a real one (after all, we don't have the real numbers yet, so we can't let $\epsilon >0$ be an arbitrary real number). Of course, it turns out that this doesn't matter, in the sense that both meanings of convergence are equivalent to each other. But I still have this sense of, like, "what if we built the rationals wrong?" or like "what if we could have built the reals to be something different?" (I have trouble articulating exactly what it is that bugs me).

But anyway, I started thinking about other number systems. It occurred to me that the integers can also "define their own convergence" in the sense that if we replace "for each real $\epsilon >0$" with "for each integer $\epsilon >0$" in the definition of convergence, the definitions end up being equivalent. Two things to note here: (1) an integer sequence that converges is just a sequence that becomes constant after some point, so it's not very interesting; (2) it becomes important that the inequality $|a_{n}-L|<\epsilon$ in the definition is strict, because if it wasn't, an oscillating sequence like $1,2,1,2,1,2,\ldots$ would be considered "convergent" under the integer definition.

I then started thinking about arbitrary subsets of the reals that have this property of "being able to define its own sequence convergence". Some sets, like $S=\{2,3\}$, cannot define their own convergence (the sequence $2,3,2,3,\ldots$ is "convergent" according to $S$, but not according to $\mathbf {R}$).

Which subsets of $\mathbf {R}$ have this property? Is it possible to provide a sufficient condition or some intuitive characterization?

I started thinking that the key property is that all positive distances between elements in the set must be in the set. In other words, for all $x,y\in S$ such that $x>y$, we must have $x-y\in S$.

Real convergence implies convergence with respect to $S$ since $S\subseteq \mathbf {R}$.

So suppose $(a_{n})$ converges with respect to $S$. Let $\epsilon >0$ be real. If there is some $\epsilon >\epsilon '>0$ in $S$ then use this to find an $N$. Then for all $n\geq N$ we have $|a_{n}-L|<\epsilon '<\epsilon$.

What if there is no such $\epsilon '$? Maybe use something like $\inf\{a_{j}-a_{k}:a_{j}>a_{k}\}$? If this is 0 then there must be an $\epsilon '$ like the one above (?). Otherwise the sequence doesn't get any closer than this value, so real convergence should equal convergence with respect to $S$. But what if $\inf\{a_{j}-a_{k}:a_{j}>a_{k}\}\notin S$?

Also if $L\notin S$: consider $(a_{n})=(3,3,3,3,\ldots )$ with $S=\{1,2,3\}$ and $L=2.5$. Then $S$ satisfies the subtraction closure property, $(a_{n})$ converges with respect to $S$, but $(a_{n})$ doesn't converge with respect to $\mathbf {R}$ (because $3\neq 2.5$).

Maybe just think about Cauchyness instead of convergence.