Suppose a valuation $v(x)$ on $\Bbb N_{>0}$ assigns $1$ if $x$ is even and $-1$ if $x$ is odd.
Let $v_2(x)$ be some natural or obvious extension of this valuation that continuously measures oddness vs evenness.
Is there a continuous extension of this valuation on $\Bbb R$ that generates a metric space either on $(\Bbb R,d)$, or $(\Bbb R/\Bbb Z,d)$?
Background, and my thoughts on the problem:
The motivation is Collatz-related.
My attempts:
A. I can come up with various continuous extensions of the valuation. Something like $v(x)=4(\lfloor x/2\rfloor-x/2)+1$, assigns a real number in the interval $[-1,1)$.
I was originally thinking of simply $d(x,y)=\lvert v(x)-v(y)\rvert$
But now I think $d(x,y)=\lvert v(x-y)+1\rvert$ yields a metric space on $\Bbb R/2\Bbb Z$
B. For a full metric space on $\Bbb R$ my thoughts so far are: $v_2=\frac{v_1+1}2$ assigns a valuation in $[0,1]$
But I'm not sure how to make a metric space. It would seem:
$d(x,0)=\frac1{v_2}$ has a good chance but I'm struggling to consider how to treat numbers differing by an odd number, whose valuation would be infinite. A possible solution would seem to be to set their distance to zero $0$, but then the metric space would set all odd numbers equal.
C. Another option is a generalisation such as $v_2(x)=\sin^2\frac{\pi\cdot x}2$ which gives a continuous valuation for $x\in \Bbb [0,1]$ with the possible advantage of being differentiable throughout $\Bbb R$, but again, I don't know how (or whether there's a way) to translate the valuation into a metric which would make $\Bbb R$ a metric space.
Given a function $v$, the two-variable function $d(x,y) = |v(x)-v(y)|$ can't possibly be a metric unless $v$ is injective (since otherwise there will be distinct points at "distance" $0$ from each other). Since you want $v$ to take the same value on all integers of a given parity, no extension of this function can yield a metric under your definition.