Let $A$ be a Banach Algebra. An element $a\in A$ is said to be a left topological divisor of zero if there exists a sequence $\{x_n\}$ in $A$ with $\|x_n\|=1$ such that $ax_n\to 0$ as $n\to \infty$. We denote the set of all left topological divisor of zero by $Z^{l}$. My question is: Is $Z^{l}$ connected?
I think it is connected. Moreover it is path connected.
My attempt:
Note that $0\in Z^{l}$. Let $a\in Z^l$. Let $0\leq t \leq 1$. Since $a\in Z^l$ we get a sequence $\{x_n\}$ in $A$ with $\|x_n\|=1$ such that $ax_n\to 0$ as $n\to \infty$. Now, \begin{align*} \|(ta+(1-t)0)x_n\| &= \|tax_n\| \\ &=|t| \, \|ax_n\| \to 0 \quad \text{ as } n \to \infty \end{align*} So $ta+(1-t)0\in Z^l$. We get a path between $a$ and $0$ in $Z^l$. If $b\in Z^l$ then combining the path from $a$ to $0$ and $0$ to $b$ we get a path between $a$ and $b$. Hence $Z^l$ is path connected. Is this correct? Similarly set of all right topological divisors of zero is also connected.
Yes, I think you're right: you've shown that the set $Z^l$ is "star-shaped" with centre $0$ ($x \in Z^l$ implies $tx \in Z^l$ for all $t \in [0,1]$) and such sets are path-connected in any TVS.