I am solving some exercises in Tapp's matrix groups for undergraduates and would be very grateful if someone could check my work (it's exercise 4.12):
A matrix $A\in M_n(\mathbb K)$ is called upper triangular if all entries below the diagonal are zero. Prove that $$ UT_n (\mathbb K) = \{A \in GL_n(\mathbb K) \mid A \text{ is upper triangular }\}$$ is a matrix group. Show that $UT_n(\mathbb K)$ is not closed in $M_n(\mathbb K)$. Is $UT_n(\mathbb K)$ compact?
Here are my thoughts on it:
That $UT_n$ is a group is clear: it is a well known fact that the product of upper triangular matrices is again upper triangular. The same holds for inverses. Therefore $UT_n$ is a group.
Furthermore, since convergence is entrywise, the limit of a sequence of upper triangular matrices is again upper triangular.
Hence if we restrict $UT_n$ to matrices of non-zero determinant then it follows that $UT_n $ is closed in $GL_n$ (because $\det$ is continuous so its limit is also non-zero).
Since given $U \in UT_n$ the matrix ${1\over k }U$ is also in $UT_n$ but the limit is zero, $UT$ is not closed in $M_{n}$.
For compactness I would argue that given $U \in UT_n$, $kU$ is also in $UT_n$. But then $UT_n$ is unbounded hence not compact.