A strange definition for a strange set. Given a ring $A$ we know that its center $C=\{x : yx= xy \;, \forall y \in A\}$ is a well defined subset of $A$. Now I want define a set that, intuitively, is '' The set of all elements that commute between them'', but I have some trouble to write a good definition. I have written: $ B=\{x : xy=yx \;, \forall y \in B\}$, but this does not seems to me a good definition because the set that I want define is present also in the definition at left side.
I suppose that, if $B$ can be defined, then $C \subset B \subset A$,  and I am interested to know in what cases those inclusions become equalities. But, Is the set $B$ really definable?

Added after the comments.
For a finite set of elements $\{ a_i\}$ with $ 1\le i \le n$ we say that the $a_i$ commute between them if $ [a_i,a_j]=0\;,\forall i,j\,: 0\le i,j \le n$. It seems to me that this definition is good (and, maybe, it can be extended to a countable set of indices) but the problem is to extend it to a possibly not countable set $B$.
 A: If I understand you correctly, you're looking for a set $B \subset A$ such that $$\forall x,y \in B, xy = yx, \tag{$\star$}$$ and such that $B$ is in some sense "maximal". Let
$$P = \{ B \subset A : \forall x,y \in B, xy = yx \}.$$
This set $P$ is partially ordered by inclusion (ie. we say $B \le B' \iff B \subset B'$). It is also not empty, because $\{1\} \in P$. Suppose that you have a chain $C$ in $P$, that is a totally ordered subset of $P$. Then
$$\mathcal{B} = \bigcup_{B \in C} B$$
is an element of $P$. Indeed, let $x, y \in \mathcal{B}$, then $x \in B$ and $y \in B'$ where $B, B' \in C$. Since $C$ is a chain, either $B \subset B'$ or $B' \subset B$; in both cases, $x$ and $y$ commute. It's also clear that $\mathcal{B}$ is an upper bound for $C$.
Thus every chain has an upper bound. So by Zorn's lemma*, $P$ has at least one maximal element. In other words, there exists a subset $B_0 \subset A$ such that:


*

*$\forall x,y \in B_0, xy = yx$;

*$\forall z \not\in B_0, \exists x \in B_0 \text{ st. } xz \neq zx$.


It's possible that $B_0$ is bigger than the center of $A$.
For example, take $A = \Bbbk\{x,y\}$, the free associative algebra on two generators. Then the center of $A$ is just $\Bbbk$, but the subset $\Bbbk\{x\} \subset A$ of polynomials in $x$ satisfies $\forall P, Q \in \Bbbk\{x\}, PQ = QP$. And as you can see, $\Bbbk\{y\}$ satisfies the same property: it's possible that there are different maximal $B_0$ satisfying the condition $(\star)$. You cannot define a single subset $B_0$ of "elements that commute with each other" in general.

(Prompted by the comments.) Such a $B_0$ is necessarily a subring of $A$.
Since $1x = x1$ for all $x$, it's clear that $1 \in B_0$ (otherwise $B_0 \subsetneq B_0 \cup \{1\}$ is not maximal). Similarly, if $x,y \in B_0$, then for all $z \in B_0$,
$$\begin{align}
(x+y)z & = xz + yz = zx + zy = z(x+y) \\
(xy)z & = xzy = z(xy),
\end{align}$$
and so $x+y \in B_0$ and $xy \in B_0$ (otherwise $B_0$ is not maximal: $B_0 \subset B_0 \cup \{xy, x+y\}$). Thus $B_0$ is a subring.
It is also true that $A$ is the union of all the possible $B_0$.
If $a \in A$, consider $P_a = \{ B \in P \mid a \in B \}$. This set is again not empty (eg. $\{a\} \in P_a$: it is true that $a \cdot a = a \cdot a$!), and every chain still has an upper bound by the previous argument. Thus $P_a$ has at least one maximal element $B_a$, which contains $a$. It's also clear that $B_a$ is a maximal element of $P$, ie. it is "a $B_0$". So every element of $A$ is in one of the $B_0$.

* Yes, this argument requires the axiom of choice. I don't know enough about set theory to say what the precise relation between AC and this is, though.
