Let $M_{n}(\mathbb{C})$ be the space of all complex $n\times n$ matrices equipped with the operator norm: $$||A|| := \sup_{||x||\le 1}||Ax|| \quad A \in M_{n}(\mathbb{C})$$
Let $GL(n, \mathbb{C})$ be the subspace of all invertible matrices of $M_{n}(\mathbb{C})$. I would like to prove that (sub)-groups such as $SL(n,\mathbb{C})$ (the set of all $n\times n$ matrices with determinant equals to one), $O(n)$ (set of all matrices $A\in M_{n}(\mathbb{C})$ such $A^{T}A = I$) and $SO(n)$ (subset of $O(n)$ of matrices with determinant one) are closed in $GL(n,\mathbb{C})$.
I suppose that the idea (for all proofs) is to take a sequence $\{A_{n}\}_{n\in \mathbb{N}}$ such that $A_{n}\to A$ and prove that $A$ is in the same set. But in order to prove that $A$ is an element of, say, $SL(n,\mathbb{C})$ or $SO(n)$, I have to prove that $\text{det}A = 1$ and I have no idea of how this is a conseguence of the convergence $A_{n}\to A$.
Any explanations and/or references are helpful!
Actually, it's better to use this approach: $SL(n,\Bbb C)=\det^{-1}(\{1\})$ and so, since $\det$ is continuous and $\{1\}$ is closed, $SL(n,\Bbb C)$ is closed.
With $O(n)$ it's similar; you consider the map $f\colon GL(n,\Bbb C)\longrightarrow GL(n,\Bbb C)$ defined by $f(A)=A^TA$, which is continuous. Then $O(n)=f^{-1}(\{\operatorname{Id}_n\})$, and therefore it is closed.