Can $\mathbb{Z}_2^{\times}$ be constructed as the closure of $4\mathbb{Z}+1$?
I understand $\mathbb{Z}_2^{\times}$ is the closure of $2\mathbb{Z}+1$ in the 2-adic metric space.
I'm interested in whether the entirety of $2\mathbb{Z}+1$ is required. In particular:
A) What's $X$, the closure of $4\mathbb{Z}+1$ in the 2-adic metric space? And,
B) if $X\subsetneq\mathbb{Z}_2^{\times}$ then can $X$ be extended to $\mathbb{Z}_2^{\times}$ by permitting the square root operation?
I'm motivated by the ideas of a) whether there's some way of "tidying up" the nontrivial kernel of the base 2 logarithm (a concept I only understand superficially), and b) since taking square roots in the 2-adics naturally extend numbers $\equiv1\mod 4$ to numbers $\equiv3\mod 4$; to what extent this process can generate the remainder of the 2-adic units from the half which are $\equiv1\mod4$.
In a $p$-adic ring, every set of the form
$$ a + m\mathbb{Z}_p = \{ x \in \mathbb{Z}_p \mid x \equiv a \bmod m \} $$
is both closed and open. (without loss of generality, we can restrict to $m$ being a power of $p$)
Thus, in $\mathbb{Z}_2$, you have $\overline{1 + 4 \mathbb{Z}} \subseteq 1 + 4 \mathbb{Z}_2$. It's not hard to show this is an equality.
Since $ \mathbb{Z}_2^\times \cong \mu_2 \times (1 + 4 \mathbb{Z}_2) $, where $\mu_2 = \{ 1, -1 \}$, you can deduce that
$$ u \in \mathbb{Z}_2^\times \Longleftrightarrow u^2 \in 1 + 8 \mathbb{Z}_2 $$
(in fact, this equivalence can even pick out $\mathbb{Z}_2^\times$ as a subset of $\mathbb{C}_2$)