**A revision:**


**According to the comment of Nate Eldredge, in order to avoid the triviality, we revise the property $P$.**


Assume that $A$ is a commutative unital Banach algebra. Its maximal ideal space is denoted by $\Delta\_{A}$. For $a\in A$, the Gelfand transform of $a$ is denoted by $\hat{a}$. An element $a\in A$ is called an injective (constant) element if $\hat{a}$ is an injective(constant) function from $\Delta\_{A}$ to $\mathbb{C}$. Consider the following property "$P$" for a commutative unital Banach algebra: 



> 
> $A$ satisfies $P$ if $A$ has at least one injective element and the limit of every sequence of injective elements is either injective or constant.
> 
> 
> 


**The main motivation for this property (and this post) is that the algebra of all bounded holomorphic functions on a connected open subset of the plane, satisfies this property**


**Some questions:**


1.What is an example of a commutative $C^{\*}$ algebra, different from $\mathbb{C}$, with this property?


2.Does The disc algebra satisfies $P$?(We know that a limit of a sequence of injective holomorphic maps defined on an **open** connected set is either injective or constant. So I think that we should be carefull about the boundary points of the disc ). If the answer to this question is negative, is it true to say that every infinite dimensional banach algebra with this property is isomorphic to $Hol(U)$, the algebra of bounded holomorphic functions on an open subset of the plane? 


3.Is there any relation between "existence of injective element in $A$" and the real or topological stable rank of $A$?



Injective element of a commutative Banach algebra
$A(D)$ does not satisfy P. I will actually work on the upper semi-disk, but of course one could map things back to $D$. Consider $f\_{\epsilon}(z)=1-z^2-i\epsilon z$. The other point with the same image as $z$ is $w=-z-i \epsilon$, which is never in the (closed) semi-disk if $z$ is, so all $f\_{\epsilon}$ are injective and they converge to $f(z)=1-z^2$, which is not.


