Openness of the set of hyperbolic linear systems I need to prove that the set $S=\{A\in M_n(\mathbb{R}); x'=Ax\hspace{0.2cm} is\hspace{0.2cm} hyperbolic\}$ is open in $M_n(\mathbb{R})$ the set of real matrices of order $n\times n$.
I have already seen a prove of this question but I hace doubts in the details. This is what I have done:
Pick a sequence $(A_n)$ of hyperbolic matrix, such that $A_n\longrightarrow A$, Lets prove that $A$ is also hyperbolic. As $A_n$ is hyperbolic, there is at least one proper value pure imaginary, so lets pick $y_n\in \mathbb{R}$ for each $n\in\mathbb{N}$ with $det(A_n-iy_nI)=0$. As $A_n\longrightarrow A$, for $\epsilon=1>0$, exists $N\in\mathbb{N}$ with $\parallel{A_n-A}\parallel<1$, so $(A_n)$ is bounded, and then $(y_n)$ is bounded (I don't know if this last statment is correct). So by Bolzano Weierstrass you can find a subequence $({y_n}_k)\longrightarrow y$. As the determinant function in continuous, then $0=det({A_n}_k-i{y_n}_kI)\longrightarrow det(A-iyI)$ when $n\longrightarrow\infty$, so $det(A-iyI)=0$, then $A$ is hyperbolic, becouse we have found a proper value of A with null real part.
 A: I think this proof will work, but you should certainly show that $\{y_n\}$ is a bounded sequence (actually, somewhat more is true; the spectrum is continuous in the operator norm when you make sense of 'continuity' for subsets of the complex plane). 
I'll point you in a possible direction for showing the boundedness of $\{y_n\}$. The existence of an imaginary eigenvalue $i y$ for an operator $B \in M_n(\mathbb{R})$ means that there is some vector $w = u + i v \neq 0$, with $u,v$ real vectors, such that $A w = i y w$. Developing this,
\begin{align*}
A u &= - y v \\
A v &= y u
\end{align*}
So, we get the bounds $ |y v| = |A u| \leq \|A\| \cdot \|u\|$ and $|y u| \leq \|A\| \cdot \|v\|$. Can you see now how to put these together to bound $|y|$ in terms of $\|A\|$, independently of $|u|$ or $|v|$?
A: From $|yv|\leq \parallel A\parallel .\parallel u\parallel$, we have that $|y|\parallel v\parallel \leq \parallel A\parallel .\parallel u\parallel$
and from $|yu|\leq \parallel A\parallel .\parallel v\parallel$, we have that $|y|\parallel u\parallel \leq \parallel A\parallel .\parallel v\parallel$
So $|y|\parallel v\parallel = |yv|\leq \parallel A\parallel \parallel u\parallel\leq\parallel A\parallel\dfrac{\parallel A\parallel\parallel v\parallel}{|y|}$, then $|y|^2\leq \parallel A\parallel ^2$, and $|y|\leq\parallel A\parallel$.
