# Understanding that Lie algebra $\mathfrak{g}$ is a closed subset of $M_n(\mathbb{C})$

I am trying to understand a piece of proof in Brian Hall's book: Lie Groups, Lie Algebras, and Representations.

The Lie algebra $$\mathfrak{g}$$ associated with a given matrix Lie group $$G$$ is defined first as a set:

Definition 3.18. Let $$G$$ be a matrix Lie group (a closed subgroup of $$GL_n(\mathbb{C})$$). The Lie algebra of $$G$$, denoted $$\mathfrak{g}$$, is the set of all matrices $$X$$ such that $$e^{tX}$$ is in $$G$$ for all real numbers $$t$$.

And then $$\mathfrak{g}$$ is given the structure of a Lie algebra.

Theorem 3.20. Let $$G$$ be a matrix Lie group with Lie algebra $$\mathfrak{g}$$. If $$X$$ and $$Y$$ are elements of $$\mathfrak{g}$$, the following results hold.

• (1) $$sX\in\mathfrak{g}$$ for all real numbers $$s$$;
• (2) $$X+Y\in\mathfrak{g}$$;
• (3) $$XY-YX\in\mathfrak{g}$$.

A step proving (3) says the following:

By (1) and (2), $$\mathfrak{g}$$ is a real subspace of $$M_n(\mathbb{C})$$, from which it follows that $$\mathfrak{g}$$ is a (topologically) closed subset of $$M_n(\mathbb{C})$$.

Question: Could anyone elaborate that why the fact that $$\mathfrak{g}$$ is a real vector space implies that it is closed in $$M_n(\mathbb{C})$$?

If $$V$$ is a vector subspace of some $$\mathbb{C}^n$$ (and $$M_n(\mathbb{C})$$ is basically $$\mathbb{C}^{n^2}$$), then it has a basis $$\{e_1,e_2,\ldots,e_k\}$$. Extend it to a besis $$\{e_1,e_2,\ldots,e_n\}$$ of $$\mathbb{C}^n$$. Then$$V=\{\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ne_n\in\mathbb{C}^n\,|\,\alpha_{k+1}=\cdots=\alpha_n=0\}$$and therefore it is a closed subset of $$\mathbb{C}^n$$.

In fact any linear subspace of a finite-dimensional vector space is closed. I think even more general statement is true: finite-dimensional subspace of a Hausdorff topological vector space is closed.