showing that $S=\{A\in M_{n}; x'=Ax \ \text{is hyperbolic}\}$ is open I tried to show that the complement is closed, $S^c=\{A\in M_n, \text{real part of eigenvalues is 0}\}$. So i tried to show that this set is closed by $(\{0\}\times \mathbb{R})$ in the fuction:
\begin{align}
F:M_n&\to \{\text{eigenvalues of M_n}\}\subset \mathbb{C} \\
\end{align}
where $F(M)$ is where the eigenvalues is. 
Is this function continuous? Is there another way to do this?
 A: I would have been able to get your approach better if you clarified what $F$ precisely is,you haven't defined it clearly.In questions like this, set on a "there exists something with this property" premise, it might pose difficulties to actually come up with a nice continuous function, usually, an obvious choice doesn't seem to be readily available.As for this question,I would like to offer a solution taking a different route,exploiting the compactness of the closed unit sphere in $\mathbb{C}^{n}$.You know,in eigenvalue considerations,the spectral norm on $M_{n}$ given by $||A||= sup_{||v||=1}\hspace{0.5mm}||Av||$ is pretty useful.Of course,$v$ is in $\mathbb{C}^{n}$.Since all norms on $M_n(\mathbb{C})$ are strongly equivalent,this won't affect our topological or convergence-related considerations.Observe that in this norm,convergence of linear maps implies uniform convergence on compact sets.I shall prove that the set $S$ of matrices for which there exists at least one eigenvalue with zero real part is closed.Let $\{A_{n}\} \in S$ such that $A_{n} \to A$.Then it needs to be shown that $A \in S$.Consider for each $n$,eigenvectors $v_{n}$ of $A_{n}$ whose corresponding eigenvalues are $ia_{n}$ such that $a_{n} \in \mathbb{R}$ and $||v_n||=1$.Then since $\{v_n\}$ is a sequence in the closed unit sphere, which is compact, it has a convergent subsequence $v_{n_{j}}$.Let $v_{n_j} \to v$ where $||v||=1$.Again,due to uniform convergence of $A_n$ on the unit sphere(or, due to the fact that $||A_nv - Av||\leq ||A_n - A||.||v||$ for all $v$), we observe that $A_{n_j}v_{n_j} \to Av$.Since each $v_{n_j}$ is an eigenvector with eigenvalue $ia_{n_j}$, it can be concluded that $ia_{n_j}v_{n_j} \to Av$.Since every convergent sequence is bounded, and $||v_{n_j}||=1$, we have $C>0$ such that $|a_{n_j}|\leq C$.Thus, $ia_{n_j}$ is a bounded sequence and has a convergent subsequence(which I shall denote with a shortened notation) $ia_{k}$.Since the imaginary axis in $\mathbb{C}$ is closed, it's limit must also be some $x \in \{z:Re(z)=0\}$.The corresponding subsequence of $v_{n_j}$ would also converge to $v$ and hence, we have a subsequence of $a_{n_j}v_{n_j}$ converging to $xv$.Since the sequence itself converged to $Av$, so must every subsequence of it, hence we have $Av = xv$.As $||v||=1$, $v$ is an eigenvector with eigenvalue $x$ having zero real part.Hence, $S$ is closed.Hope this helps.And yes,nice question. 
Edit :- You might have a look here.It gives a shorter solution :-
Hyperbolic Systems ODE
