A metric space in which $3^\infty=2^\infty=0$

Question

I want a space containing all the positive integers in which $3^nx+3^n-2^n\to0$ as $n\to\infty$

Perhaps paradoxically, numbers not factorisable by $2,3$ would be a sufficient set for me (in case that helps).

My rudimentary knowledge says that a sum of two metrics is a metric and therefore I can just take $d(x,y)=\lvert x-y\rvert_2+\lvert x-y\rvert_3$

Am I going about that the right way?

Is this space going to have reasonable properties?

You can construct a metric in which all sequences tending to infinity tend to zero. Put the natural numbers in a circle. — Qiaochu Yuan, Jun 14 '19 at 18:50
Do you need this metric to have any special properties? The metric $d(m,n) = 1/m + 1/n$ has the property that @QiaochuYuan describes in their comment (modulo, at least, an interpretation of what you might mean by "$\to0$", which is ambiguous in this context). — Greg Martin, Jun 14 '19 at 18:51
@QiaochuYuan thanks, that's insightful for me. In this instance I think a metric which minimises to some degree the convergent superset of the example I give would be more useful. — it's a hire car baby, Jun 14 '19 at 18:52
Your metric won't have either going to zero, since neither $|3^n|_2$ nor $|2^n|_3$ converges to zero. — Thomas Andrews, Jun 14 '19 at 18:53
@user334732 Are you playing around with the Syracuse function? This looks like very familiar territory for me, lol. I remember playing with it, p-adics, and even the metric proposed by Umberto P. below. Those were good times, back when I was in grad school. Good luck! — SlipEternal, Jun 14 '19 at 19:05
@GregMartin that was a poor comment; this is perhaps a better one... I'd like a completion of the $5$-rough natural numbers $\Bbb N_5$ such that $2^\infty\Bbb N_5=0$ and $3^\infty\Bbb N_5=0$ — it's a hire car baby, Jun 14 '19 at 19:25

score 5 · Answer 1 · answered Jun 14 '19 at 18:52

5

You can define a metric $d$ on $[0,\infty)$ by \begin{align*} d(x,y) &= \left| \frac 1x - \frac 1y \right|,\quad x,y > 0 \\ d(x,0) = d(0,x) &= \frac 1x ,\quad x > 0 \\ d(0,0) &= 0.\end{align*}

Then $d(3^n x + 3^n - 2^n,0) \to 0$.

answered Jun 14 '19 at 18:52

Umberto P.

52,165

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Yes, we can define limits of sequences using this topology. A sequence is a function of $\mathbb N^+.$ It limit $L$ if $a$ can be extended continuously by setting $a(0)=L.$ In this topology, $0$ is functioning more like $\infty.$ – Thomas Andrews Jun 14 '19 at 19:00
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With a slight variation of the idea, one can also let the OP's sequence converge to $17$. Which maybe should be a warning about how much insight one can hope to derive from such constructions. – Torsten Schoeneberg Jun 15 '19 at 02:34

A metric space in which $3^\infty=2^\infty=0$

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