The normalizer of a torus is a closed subgroup of the Lie group

Question

Let $G$ be a Lie group and $T \subset G$ a torus, show that $N(T)$ is a closed subgroup of $G$. Could somebody give a sketch of the proof?

Answer 1

Say $n_i\in N(T)$ converges to $n$, and let $t\in T$. Then $$n_itn_i^{-1}\rightarrow ntn^{-1}.$$ Since $n_itn_i^{-1}\in T$ we have $ntn^{-1}\in T$ since $T$ is closed.