# Is $\left(2\mathbb{N}^+-1)\cup \{0\}\cup\{-\frac{1}{3}\}\right)\times\langle2\rangle$ a complete space under the 2-adic metric

Take the set: $$\left(2\mathbb{N}^+-1)\cup \{0\}\cup\{-\frac{1}{3}\}\right)\times\langle2\rangle$$

Where $$\langle2\rangle=\{\ldots\frac{1}{4},\frac{1}{2},1,2,4\ldots\}$$

Now create the equivalence classes from this set by the relation $$x\sim y\iff 2^zx=y$$

Is this set of equivalence classes a complete space under the 2-adic metric?

These classes are all uniquely indexed by either a positive odd number, $$0$$, or $$-\frac{1}{3}$$. Their 2-adic distance is measured only at this index. So $$\lvert 10-6\rvert=\lvert 10-48\rvert=\lvert5-3\rvert_2=\frac{1}{2}$$

For example. Consider all odd integer sequences defined by iteration of the function $$f(x)=4x+1$$ such as $$1,5,21,85,341,\ldots$$. These all converge to $$-\frac{1}{3}$$ which is in the space.

A little background: I'm trying to extend Sharkovskii's theorem to the Collatz conjecture. I think the result I need is that either this is a complete space or the sets $$\{0\}$$ and $$\{-\frac{1}{3}\}\times\langle2\rangle$$ can be considered endpoints of a line segment and that the segment is complete in-between.

The "continuous function" required by Sharkovskii's theorem to map the theorem to the Collatz conjecture is $$g(x)=3x+2^{v_2(x)}$$

As you say yourself, your equivalence class construction has nothing to do with the metric, and you are basically asking whether the set $(2\mathbb{N}^+-1)\cup \{0\}\cup\{-\frac{1}{3}\}$ with the $2$-adic metric is complete. (Equivalently, if it's a closed subspace of $\mathbb{Q}_2$, or $\mathbb{Z}_2$.)
It is not. Actually, every element of $\mathbb{Z}_2^*$ (that is, every $2$-adic number with $2$-adic absolute value 1) is an accumulation point of that set. That is kind of obvious from the inverse limit definition of the $p$-adics: If $x\in \mathbb{Z}_2^*$, for each $n \in \mathbb{N}$ you can choose an odd positive integer $x_n$ as a representative of the residue class of $x$ in $\mathbb{Z}_2/2^n \simeq \mathbb{Z}/2^n$. Then $2$-adically, the sequence $x_n$ goes to $x$, more or less by definition.
As a concrete example, the geometric series $s_n := \sum_{i=0}^n 2^i = 2^{n+1}-1$, i.e. $1, 3, 7, 15, 31, 63, ...$, in the $2$-adic metric, converges to $-1$. (Whereas your recursive example $x_n = 4x_{n-1}+1$ has closed expression $x_n = 4^n x_0 +\frac{1}{3} (4^n-1)$ and thus converges to $-\frac{1}{3}$ regardless of the start value $x_0$, so that's ok.)
I leave it to you -- as a good exercise in $2$-adics -- to find sequences of odd positive integers which converge, in the $2$-adic metric, to: a) -43; b) $\frac{13}{5}$; c) a square root of 17.