Let $k = \mathbb{R}$ and view $\text{M}_m(\mathbb{R})$ as a topological space using the natural identification $\text{M}_m(\mathbb{R}) \cong \mathbb{R}^{m^2}$. Let$$\mathbb{O}_a = \{g \cdot a \cdot g^{-1} : g \in \text{GL}_n(\mathbb{R})\}$$denote the conjugacy class of an element $a \in \text{M}_m(\mathbb{R})$.
Question. Does it follow that the set $\mathbb{O}_a$ is closed in $\text{M}_m(\mathbb{R})$ if and only if the element $a$ is semisimple?
Let $a$ be semi simple and $p(x)$ be its characteristic polynomial; then there is $m\in\mathbb{R}[x]$, with simple complex roots, s.t. $m(a)=0_n$. Let $(u_k)$ be a sequence in $O_a$ that converges to $u\in M_n(\mathbb{R})$. Clearly $m(u_k)=p(u_k)=0_n$ and , by continuity, $m(u)=p(u)=0_n$, that implies $u\in O_a$.
Conversely, let $a$ be not semi simple. There are $D\in M_n(\mathbb{R})$ semi simple, $N\in M_n(\mathbb{R})$ non-zero nilpotent s.t. $a=D+N,DN=ND$. For every positive integer $k$, $u_k=D+1/kN\in O_a$. The sequence $(u_k)$ tends to $D\notin O_a$.