Is a certain subset of $M_2(\mathbb{R})$ closed? I am trying to prove that the set M of all matrices in the normed linear space $M_2(\mathbb{R})$ such that both eigen values are real is closed (under metric topology; metric induced by the norm).
Following is my attempt. I recall the property that finite dimensional subspaces of normed spaces are complete. And since complete metric spaces are closed, if I can prove it is a vector subspace of $M_2(\mathbb{R})$, I am done. But I don't know if M is closed under "vector" addition. Is it true that the sum of two "real- eigenvalued" matrices have real eigen values?
 A: $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$
So it's not a subspace.  But to show it's closed, you can produce a continuous function $f : M_2(\mathbb{R}) \to \mathbb{R}$ such that $M = f^{-1}([0,\infty))$.  Hint: look at the characteristic polynomial of a matrix $A$, and think about the discriminant of a quadratic equation.
A: It's not true, as Nate Eldredge counterexemple show.
For your original question, you can look at it this way :
the eigenvalue of $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ are the roots of the polynomial 
$$P(x) = \begin{vmatrix} a-x & b \\ c & d-x \end{vmatrix}$$
This is a 2nd degree polynomial, so it has real roots if the discriminant is positive.
Hence, if $\Delta(a,b,c,d)$ is the discriminant of $P$, the set of matrix with real eigenvalues is 
$$ \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid (a,b,c,d)\in \Bbb R,  \Delta(a,b,c,d) \geq 0 \right\} $$
So if you can prove that $\Delta(a,b,c,d)$ is continuous, you've won (do you see why?)
