An open subgroup of GL_n is also closed

Idea. The cosets of $H$ partition the group. If $H$ is open, so are all its cosets (translations are homeomorphisms), and the complement of $H$ is a union of other cosets.

Step 1: translations are homeomorphisms. For $g\in GL_n$ the left translation $L_g:X\mapsto gX$ is continuous (matrix multiplication is polynomial in the entries) with continuous inverse $L_{g^{-1}}$. So $L_g$ maps open sets to open sets.

Step 2: cosets are open. $gH=L_g(H)$ is open if $H$ is.

Step 3: the complement is a union of cosets. The left cosets $gH$ partition $GL_n$, with $H=1\cdot H$ one of them. Two cosets are equal or disjoint. Therefore
$$GL_n\setminus H=\bigcup_{g\notin H}gH$$
(a coset $gH$ with $g\notin H$ is disjoint from $H$: if $gh\in H$ then $g\in H$). This is a union of open sets, hence open. So $H$ is closed in $GL_n$.

Example. $GL_n^+=\{\det>0\}$ is an open subgroup (it is open, closed under products and inverses since $\det(AB)=\det A\det B$), and its complement $\{\det<0\}$ is the open coset $gGL_n^+$ for any $g$ with $\det g<0$.

Remark.
- The argument works in every topological group. A consequence: if $G$ is connected, an open subgroup $H$ is open and closed and non-empty, hence $H=G$. We use it in Part 5 to show that $GL_n^+$ is the identity component.
- "Closed in $GL_n$" is *not* the same as closed in $M_n(\mathbb{R})$: the set $GL_n^+$ is not closed in $M_n$, since its closure contains singular matrices.