Let $X, Y$ be normed vector spaces and $T$ a closed operator from $X$ to $Y$ with domain $D_T$.
Show that the null space of $T, N_T = \{u\in D_T; Tu = 0\}$ is a closed subspace of $X$.
I have tried to prove and I will appreciate any feedback on this.
Observation: It is not given that $T$ is linear. So I am not sure if something is missing here. Anyways, I will assume it is linear and try to prove the theorem.
My attempt:
I have to show that $N_T$ is a vector subspace and is closed.
To show that it is a subspace, let $a,b$ be scalars and $u,v$ be vector in $N_T$, such that $Tu=0, Tv=0$, then
$T(au+bv)=aTu+bTv=a0+b0=0$ (I have used that $T$ is linear!!)
Also by linearity of $T$ it follows that $T0=0$, this shows that zero vector is also in $N_T.$
Now to show that it is closed:
Let $(x_n)$ be a sequence in $N_T$ with limit say $x$ in $X$, I will have to show that $x \in N_T$.
Since $(x_n)$ is a sequence in $N_T$, then $x_n \in D_T$ and $Tx_n=0,~ \forall n \in \mathbb{N}.$
Since $T$ is a closed operator, there exist a $y \in Y$ such that $x_n \to x$ and $Tx_n\to y$ as $n \to \infty$ and $Tx=y.$
To show that $x \in N_T$, I must prov that $y=0$.
Since $y=Tx=T \lim_{n \to \infty}x_n=\lim_{n \to \infty}Tx_n=\lim_{n \to \infty} 0 =0.$
Is my approach in the right direction?
Your interpretation of a closed operator is not correct. The correct argument is as follows: Let $Tx_n=0$ for all $n$ and suppose $x_n \to x$. We have to show that $Tx=0$. Now $(x_n,T(x_n))$ is a sequence of points on the graph of $T$ which converges to $(x,0)$. Hence $(x,0)$ lies on the graphs which means $Tx=0$.