Kernel of a bounded linear operator on a normed linear space need not be closed or open?

Question

How should be the kernel of a bounded linear operator on a normed linear space as a set? Kernel of a bounded linear operator on a normed linear space need to be closed or open? Or it need not be closed or open?

Answer 1

It should be closed. Let $T:V\to W$ be a bounded linear operator between normed linear spaces. We have that $$ \ker(T) = \{x\in X: Tx =0\} = T^{-1}(\{0\}). $$ Since bounded operators are continuous and $\{0\}$ is a closed set in the norm topology, we see that the kernel is closed.

The kernel is not open unless $T$ is the zero operator. Assume $T\neq 0$. We can take a vector $x\notin \ker(T)$ and approach 0 with the sequence $\{\frac{1}{n} x\}. $ We have that $T(\frac{1}{n}x) \neq 0$ for all $n$, so this sequence is not contained in $\ker(T)$. However, it clearly converges to $0\in \ker(T)$, which shows that $V\setminus \ker(T)$ is not closed.

Answer 2

Every normed linear space is Hausdorff, hence singletons are closed sets in a normed linear space. $T:V\to W$ is bounded iff it is continuous, Now $\{0\}$ is closed in $W$ and $T$ is continuous. So, by continuity of $T$, $$ T^{-1}(\{0\})$$ is closed. Hence $$\ker(T) = \{x\in X: Tx =0\}$$ is closed.