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Let $S$ be a set of finitely many prime numbers. Then, define $\left|\cdot\right|_{S}:\mathbb{Q}\rightarrow\left[0,\infty\right)$ by: $$\left|x\right|_{S}\overset{\textrm{def}}{=}\prod_{p\in S}\left|x\right|_{p}$$ where, for each prime $p$, $\left|x\right|_{p}$ is the usual $p$-adic absolute value of $x$. Then, let $d_{S}:\mathbb{Q}\times\mathbb{Q}\rightarrow\left[0,\infty\right)$ be defined by: $$d_{S}\left(x,y\right)\overset{\textrm{def}}{=}\left|x-y\right|_{S}$$ As defined, $d_{S}\left(x,y\right)=d_{S}\left(y,x\right)$, and $d_{S}\left(x,y\right)\geq0$, with equality if and only if $x=y$. However, $d_{S}$ DOES NOT satisfy the triangle inequality. For example, letting $S=\left\{ 2,3\right\}$, we have: $$6=\left|\frac{1}{6}\right|_{\left\{ 2,3\right\} }=\left|\frac{1}{2}-\frac{1}{3}\right|_{\left\{ 2,3\right\} }>\left|\frac{1}{2}\right|_{\left\{ 2,3\right\} }+\left|\frac{1}{3}\right|_{\left\{ 2,3\right\} }=5$$

I find myself considering this construction because it seems to me to be the “natural” way to assign a notion of convergence in the situation I am working in. As an example, let $x_{0}=1$, and let $x_{n+1}=a_{n}x_{n}$, where, for each $n$, $a_{n}$ is a random variable taking the values $2$ and $3$, each with probability $1/2$. I am studying the sum: $$X\overset{\textrm{def}}{=}\sum_{n=0}^{\infty}x_{n}$$ For any trial run for which $a_{n}=3$ only finitely many times, observe that the $2$-adic magnitude $\left|x_{n}\right|_{2}$ will tend to zero as $n\rightarrow\infty$, which is then sufficient to guarantee that the series defining $X$ converges in $\mathbb{Z}_{2}$ (the $2$-adic integers) to a well-defined limit. On the other hand, if $a_{n}=2$ only finitely many times, then $X$ won't converge to a well-defined $2$-adic limit, since, in this situation, the $x_{n}$s do not converge to $0$ in $2$-adic absolute value. Nevertheless, $a_{n}=2$ only finitely many times implies $a_{n}=3$ infinitely many times, and hence, that $X$ converges to a well-defined limit in $\mathbb{Z}_{3}$.

As is very well-known, for distinct primes $p$ and $q$, the $p$-adic and $q$-adic topologies are completely different. In order to circumvent this, the $\left(2,3\right)$-adic “absolute value” $\left|\cdot\right|_{\left\{ 2,3\right\} }$ strikes me as the most natural way to deal with the question of the convergence of $X$. My (admittedly naïve) hope is that using this construction will allow me to preserve the fact that a series $\sum_{n=0}^{\infty}c_{n}$ converges in $\mathbb{Z}_{p}$ if and only if $\left|c_{n}\right|_{p}\rightarrow\infty$. However, I am extremely apprehensive about working with $\left|\cdot\right|_{\left\{ 2,3\right\} }$ and, more generally, $\left|\cdot\right|_{S}$, due to their failure to satisfy the triangle inequality.

Consequently, I am looking for an explanation and/or reference as to whether or not the $\left|\cdot\right|_{S}$ “absolute value” can be used to obtain a topological extension and/or completion of $\mathbb{Q}$ in a manner comparable to how the $p$-adic metric value furnishes a metric completion of $\mathbb{Q}$—if such a thing can even be done.

Edit: To be clear, it's obvious to me that any positive "answer" to this question will not be a metric topology on $\mathbb{Q}$. What I want to know is whether or not there is a non-trivial topology that can be wrung out from this, one which admits a meaningful notion of convergence for infinite series, and hence, is compatible with whatever the algebraic structure is that we get on this space. (I'm not an algebraist, so I have no idea about the details.)

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    $\begingroup$ Why not the sum, rather than product? $\endgroup$ Commented Jun 10, 2021 at 0:09
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    $\begingroup$ yes, but that feature is what makes the triangle inequality hold... I think you've got a fundamental conflict in desiderata... $\endgroup$ Commented Jun 10, 2021 at 0:42
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    $\begingroup$ Leaving as a comment in case it is already known to the OP: The only (non-trivial) absolute values on $\mathbb{Q}$ are the standard (Archimedean) one and the $p$-adic ones; this is the content of Ostrowski's Theorem. So, I think finagling a metric completion of the rationals based on your "absolute value" will not be easy . . . $\endgroup$ Commented Jun 10, 2021 at 1:17
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    $\begingroup$ For any integers $m,n$, $2^m$ and $3^n$ are coprime, so you can write $2^m u+3^n v=1$. If you want to have a basis of neighborhoods of 0 stable by addition, then you will be forced to have 1 in all those neighborhoods. I am not sure what kind of structure you would like to preserve but this looks quite problematic to me. $\endgroup$ Commented Jun 10, 2021 at 9:18
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    $\begingroup$ You can also consider the supremum of the $2$-adic and the $3$-adic absolute value. This gives a non-multiplicative nonarchimedean ring norm on $\mathbf{Q}$, and the completion is $\mathbf{Q}_2\times \mathbf{Q}_3$. $\endgroup$ Commented Jun 10, 2021 at 11:27

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