I previously did this exercise:
Prove that $\operatorname{Isom_n}{\mathbb R^n}$ is a matrix group.
where
$$\operatorname{Isom_n}{\mathbb R^n} = \left \{ \left ( \begin{array}{cc} A & 0 \\ V & 1 \end{array} \right ) \mid A \in O(n) (\mathbb K), V \in \mathbb R^n \right \} $$
I forgot to answer whether it is compact so I am answering it now:
Let $D_n \in O(n)$ be the diagonal matrix with entries $d_{11}=n, d_{22}={1\over n}$ and $d_{kk}=1$ for $k > 2$.
Since the determinant of a diagonal matrix equals the product of the entries in the diagonal we have $\det D_n = 1$ hence $D_n \in O(n)$.
Since $\|D_n\|\ge \sqrt{n}$ where $\|\cdot\|$ is the Euclidean norm the sequence $D_n \in O(n)$ is unbounded and hence so is $O(n)$.
My problem now is that the limit of $D_n$ has $d_{22}=0$ and is therefore not in $O(n)$ and so
$$\left ( \begin{array}{cc} D_n & 0 \\ 0 & 1 \end{array} \right ) $$
appears to be a sequence in $\operatorname{Isom_n}{\mathbb R^n}$ such that its limit is not in $\operatorname{Isom_n}{\mathbb R^n}$ contradicting my previous proof that $\operatorname{Isom_n}{\mathbb R^n}$ was closed.
What am I doing wrong?
There is an obvious continuous and surjective map from that group to $\mathbb R^n$.