Wikipedia article for matroids says
For all elements $a$, and $b$ of $E$ and all subsets $Y$ of $E$, if $a\in\operatorname{cl}(Y\cup b) \setminus Y$ then $b\in\operatorname{cl}(Y\cup a) \setminus Y.$ ... (it) is sometimes called the Mac Lane–Steinitz exchange property.
Another article for closure says:
the exchange property: If $x$ is in the closure of the union of $A$ and $\{y\}$ but not in the closure of $A$, then $y$ is in the closure of the union of $A$ and $\{x\}$.
As far as I can tell they seem not equivalent, because the first one writes $a$ not in $Y$, and second one writes $x$ not in the closure of $A$. So I wonder if one of them is wrong for defining the exchange property, or I fail to see their equivalence?
I am also not able to see if the following two definitions of the anti-exchange property for a closure operator are equivalent: from antimatroid at Wikipedia:
anti-exchange axiom: If neither $y$ nor $z$ belong to $τ(S)$, but $z$ belongs to $τ(S \cup \{y\})$, then $y$ does not belong to $τ(S\cup \{z\})$.
from closure at Wikipedia:
the anti-exchange property: If $x$ is not contained in the union of $A$ and $\{y\}$, but in its closure, then $y$ is not contained in the closure of the union of $A$ and $\{x\}$.
Thanks and regards!
Question 1: The first of the quoted exchange properties is wrong as it stands. It would become correct (and equivalent to the second version) if it were restricted to closed sets $Y$.
Question 2: The first quoted version is again wrong, but only because it neglects to exclude the possibility that $y=z$. If you add $y\neq z$ to the hypotheses, then it becomes correct. The second version is also wrong (consider the case where $x$ and $y$ are distinct points in the closure of $A$), but it becomes correct if you add the hypothesis that $A$ is closed.