Below is an example of embedded Lie Subgroup from John Lee's Introduction to Smooth Manifolds example 7.18 (d). In this example, why is the image of $\beta$ a properly embedded Lie subgroup of $GL(2n,\mathbb{R})$? This would require that $\beta$ be a topological embedding and the image of $\beta$, $\beta(GL(n,\mathbb{C}))$ be a closed subset of $GL(2n,\mathbb{R})$. But I can't see why the image of $\beta$ is closed. I would greatly appreciate some help.
The complement consists of things for which the (1,1) entry and the (2,2) entry differ, or the (1,2) entry plus the (2, 1) entry is nonzero, and similar statements about many other pairs of entries.
Each of these conditions (like $m_{11} \ne m_{22}$) describes an open set $U$ in $GL(2n, \mathbb R)$. Why? Consider the map from $GL(2n, \mathbb R)$ to $\mathbb R$ defined by $M \mapsto m_{11} - m_{22}$. It's evidently continuous. And the set $U$ is the preimage of the open set $\mathbb R - \{0\}$ under that map.
Applying this idea to all the various conditions shows that the image is the complement of a union of somewhere around $2n^2$ open sets, which is still an open set. Hence the image is a closed set.