$\Bbb Q_2$ - sending $2^n:n\in\Bbb Z$ to $n\in\Bbb N$. Is this the same space, but with more convergent sequences?

Let $$S_n$$ be a sequence comprising the sum of two sequences, one of which converges in $$\Bbb Z_2$$ and the other diverges in $$\Bbb Q_2$$ but converges in $$\Bbb Q$$.

An example is $$x_{n+1}=4x_n+5\cdot2^{\nu_2(x_n)-2}$$ where $$2^{\nu_2(x_n)}$$ measures the highest power of $$2$$ that divides $$x$$.

When $$\nu_2(x_0)$$ is even, this is the sum of a sequence converging to $$-\frac13$$ in $$\Bbb Z_2$$ and another converging to $$\frac13$$ in $$\Bbb Q$$, the sum of which converges to $$0$$.

When $$\nu_2(x_0)$$ is odd, this is the sum of a sequence converging to $$-\frac23$$ in $$\Bbb Z_2$$ and another converging to $$\frac23$$ in $$\Bbb Q$$

EDIT FOR CLARITY: let $$S_n=a_n+b_n$$ and let $$a_0=1$$ and let $$b_0=0$$

Then let $$a_n$$ be the binary number to the left of point and $$b_n$$ the part to the right.

Then proceed: $$a_n=1,5,21,\ldots$$ and $$b_n=\frac12,\frac58,\frac{21}{64}\ldots$$

The example given above does not converge in $$\Bbb Q_2$$ due to the component $$\frac12,\frac58,\frac{21}{64}\ldots$$ that converges in $$\Bbb Q$$. However if instead of $$2^{\nu_2(x)}$$ we take the power of $$2$$ that divides $$x$$ and then take either of the obvious morphisms $$\phi:\Bbb Z\to\Bbb N$$ and use the distance:

$$d(x,y)=2^{-\phi(\nu_2(x))}$$

Then the sequences above converge in the completed (ring?) of $$\overline {\Bbb Z},d$$ which is isomorphic to $$\Bbb Z_2$$.

It would seem therefore that $$\Bbb Q_2$$ embeds in this space, but in such a way that some sequences (e.g. the example given above) will converge where they do not converge in $$\Bbb Q_2$$. Is this correct?

• Is "a sequence comprising the sum of two sequences" deliberately confusing terminology? Because when I hear "sum of two sequences", I think of some $c_n = a_n+b_n$ with explicitly given sequences $a_n, b_n$, and then one has the usual limit laws etc. But in your example, a sequence is given recursively, and it just so happens that the recursion formula is a sum of two terms. I doubt that anything can be inferred from that. E.g. look at $x_{n+1} =x_n +17$. Do you think one can make non-trivial statements about this from knowing the constant sequences $x_{n+1}=x_n$ and $x_{n+1}=17$? – Torsten Schoeneberg Oct 9 at 18:07
• Besides, I don't even see what "summand" of your sequence is supposed to converge to $-1/3$ resp. $1/3$ in any metric, and what it would mean to converge "in $\mathbb Z_2$" resp. "in $\mathbb Q$" (both $\pm 1/3$ are in both sets), so I stopped reading there and vote to close. – Torsten Schoeneberg Oct 9 at 18:12
• @TorstenSchoeneberg I meant it in the meaning you first thought of. I would never be deliberately confusing. If it's unclear how I've partitioned the sequence into a sum of two, I've separated them at the binary point into a sum of the part to the left of the point plus the part to the right. I hoped that would be clear where I indicate the component in $\Bbb Q$ is $\frac12,\frac58,\frac{21}{64}\ldots$ – samerivertwice Oct 9 at 18:15
• @TorstenSchoeneberg with respect to $\frac13$ being in $\Bbb Q_2$ you literally clarified for me in the question I asked yesterday (or this morning depending on time zone) that $\frac12,\frac58,\frac{21}{64}\ldots$ diverges in $\Bbb Q_2$ despite the fact that $\frac13$ is in the set. – samerivertwice Oct 9 at 18:16
• @TorstenSchoeneberg I've now written out the sequences $s_n=a_n+b_n$ which hopefully helps. – samerivertwice Oct 9 at 18:29