First the proposed theory:
Here $0 \in \mathbb N$.
Definition: A non-constant function $f: \mathbb N \to \mathbb N$ is called a ∞-BinFraction if it satisfies the following:
$\tag 1 \forall n \; f(n+1) = 2f(n) \text{ or } f(n+1) = 2f(n) + 1$
Definition: Let $f$ be a a ∞-BinFraction and suppose for some $k$, $f(n+1)$ is odd for all $n \ge k$. Then we call $f$ overloaded.
If $f$ is overloaded there is a least $k$ (that might be $0$ with $f(0)$ also odd) so that $f(m)$ is odd for all $m \gt k$. We define a function $g$ that agrees with $f$ for $n \lt k$, has $g(k)=f(k)+1$ and for $m \ge k$, we recursively define $g(m + 1) = 2 * g(m)$. This function $g(x)$ is a ∞-BinFraction that is not overloaded.
Using this transformation, we can define an equivalence relation $\rho$ on the a ∞-BinFractions, and each block has a representative that is not overloaded. The partition into blocks is very fine grained - the blocks are either singletons or have two elements. When a block has two elements, one is an overloaded ∞-BinFraction and the other is the one the transformation equates it to.
Example of a block with two elements:
$\quad \{(5,10,21,42,85,171,343,687,1375,\dots),\, (5,10,21,43.86,172,344,688,1376,\dots)\}$
Let $\mathcal M$ represent the set ∞-BinFractions modulo $\rho$.
Definition: A mapping
$\tag 2 \mathtt L: \mathbb N \to \mathcal M$
is said to eventually become stable if for every $n \in \mathbb N$ another integer $N \in \mathbb N$ can be found such that if $\mu \ge N$ and $\nu \ge N$
$\tag 3 \mathtt L(\mu)(k) = \mathtt L(\nu)(k) \; \text{ whenever } k \le n$
For this definition to make sense we insist that $\mathtt L$ 'selects' a ∞-BinFraction that is not overloaded.
Now if $\mathtt L$ eventually becomes stable, it defines a unique element in $f \in \mathcal M$. We write
$\tag 4 \quad f = \lim\limits_{\mu \to +∞} L(\mu)$
Note that $f$ might be overloaded, but just resort back to the 'even/finite/rational' representative in the 2-element block.
The idea for this came from
and reading the wiki article
Question: Is the sequential closure operator defined by $\text{(4)}$ idempotent (i.e. does it define a topological space)?
My Work
I now think of these ideas as a way of looking at the work of Simon Stevin with a modern twist. If the proposed theory described has no flaws (it might need tweaking), the answer to the question is most likely yes by relying on modern techniques. But then we have a brain teaser - provide a direct answer to the question on this elementary logical platform.
When I look at this it seems obvious - any new functions obtained by the sequential closure are not needed in 'converging' to anything else, since the restrictions to the initial segments can be built from the 'starter set'. But that doesn't qualify as a precise mathematical argument.
