$U$ be the set of all $2×2$ matrices with real entries such that all their eigenvalues belong to $C - R $, and $X = M_2(R)$. Is $U$ open? Let $U$ be the set of all $2×2$ matrices with real entries such that all their eigenvalues belong to $C - R  $, and $X = M_2(R)$.
How can I prove $U$ is open in $X$? Can anyone please help me by giving any hint? I know this is not a closed set.
 A: If $A \in U$, the eigenvalues or $A$, being in $\Bbb C - \Bbb R$, have non-zero imaginary part.  Since $A \in U \subset M_2(\Bbb R)$, i.e., since the entries of $A$ are real, the characteristic polynomial $\chi_A(x)$ of $A$ has real coefficients.  Indeed, it is quite well-known and also very easy to derive that
$\chi_A(x) = x^2 - \text{Tr}(A) x + \det A, \tag 1$
where $\text{Tr}(A)$ is the trace of the matrix $A$.  Since $\chi_A(x)$ has real coefficients, the eigenvalues, satisfying as they do
$\chi_A(x) = x^2 - \text{Tr}(A) x + \det A  = 0, \tag 2$
must be a complex conjugate pair $a \pm bi \notin \Bbb C -\Bbb R$.  If we apply the quadratic formula to (2) we find that
$a \pm bi = \dfrac{1}{2}(\text{Tr}(A) \pm \sqrt{(\text{Tr}(A))^2 - 4 \det A}); \tag 3$
it follows that we can, without loss of generality, by if necessary reversing the sign of $b \in \Bbb R$, take
$b i = \dfrac{1}{2} \sqrt{(\text{Tr}(A))^2 - 4 \det A}; \tag 4$
since $bi$ is purely imaginary, we must have
$D(A) = (\text{Tr}(A))^2 - 4 \det A < 0 \Longleftrightarrow A \in U; \tag 5$
also, taking
$A = \begin{bmatrix} 0 & -b \\ b & 0 \end{bmatrix}, \; 0 \ne b \in \Bbb R, \tag 6$
we see that
$D(A) = - 4b^2, \tag 7$
which shows that $D(A)$ takes on every negative real value as $A$ varies over $U$.  Now $D(A) = (\text{Tr}(A))^2 - 4 \det A$ is a continuous function of the matrix $A$, and we have seen that in fact
$U = D^{-1}(\{ r \in \Bbb R \mid r < 0 \}) = D^{-1}(\Bbb R_-); \tag 8$
now since $\Bbb R_-$ is open in $\Bbb R$, by the continuity of $D$ we have $U$ is open in $X$.
