topological divisor of zero

Let $A$ be a normed ring.  An element $a\in A$ is said to be a left topological divisor of zero if there is a sequence $a_n$ with  $\parallel a_n\parallel =1$  for all $n$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel a{a}_n\parallel =0.$$

Analogously, $a$ is a if

$$\underset{n\to \mathrm{\infty }}{lim}\parallel b_{n}a\parallel =0,$$

for some sequence $b_n$ with  $\parallel b_n\parallel =1$.  The element $a$ is a topological divisor of zero if it is both a left and a topological divisor of zero.

Remarks.

• Any zero divisor is a topological divisor of zero.

• If $a$ is a (left) topological divisor of zero, then $ba$ is a (left) topological divisor of zero. As a result, $a$ is never a unit, for if $b$ is its inverse, then $1=ba$ would be a topological divisor of zero, which is impossible.

• In a commutative Banach algebra $A$, an element is a topological divisor of zero if it lies on the boundary of $U(A)$, the group of units of $A$.