In the p-adic metric space is $f(x)=3x+\frac{1}{\lvert x\rvert_2}$ consistently closer to $2^{\infty}$ than $x$ is for $x\in\mathbb{N_{>4}}$? Or does it vary?
Here is what I have:
$2^{\infty}=0\implies$ the question is equivalent to whether $\lvert x\rvert_2>\lvert f(x)\rvert_2\quad\forall x$
If we break $x$ down into its odd and even factors $$x=a\cdot2^n:n\in\mathbb{N}_{\geq0}\land a\in \{2\mathbb{N_{>0}}-1\}$$ then we have:$$f(x)=3a\cdot2^n+2^n=3(a+1)\cdot 2^n$$
We know $a$ is odd so if we factor $a+1$ into its odd and even factors we have $a+1=a_1\cdot2^m :m>0$
$\therefore f(x)=a_1\cdot2^{n+m}$
$$\implies\lvert f(x)\rvert_2=\frac{\lvert x\rvert_2}{2^m}:m>0$$
You are correct. I like to work with p-adic valuations, where $v_p(a) = n$ where $a = p^n x/y$. The norm of $a$ is defined as $p^{-v_p(a)}$.
Let $x\in\mathbb{N}_{>4}$ such that $v_2(x)=n$.
Then $v_2(f(x)) = v_2(2^n\cdot3a+2^n) = v_2(2^n(3a+1)) = n + v_2(3a+1) > n$ $($because $2|3a+1)$.
Hence $\Vert f(x)-2^\infty \Vert_2 < \Vert x-2^\infty \Vert_2 $.