Compact subset of space of matrices and compactness verification of a set of eigenvalues Let $M_n(\mathbb R)$ be the vector space of real matrices of size $n$ , identified with $\mathbb R^{n^2}$ ; let $X \subseteq M_n( \mathbb R)$ be a compact set ; let $S \subseteq \mathbb C$ be the set of all numbers that are eigenvalues of at least one element of $X$ , then is $S$ compact ? 
 A: First of all, $S$ is bounded. Indeed, equip $(M_n(\Bbb R)\subset )M_n(\Bbb C)$ with a norm $\|\cdot\|$ subordinate to some norm $|\cdot|$ on $\Bbb C^n$. If $\lambda\in\Bbb C$ is an eigenvalue of $M\in M_n(\Bbb C)$ then $|\lambda|\leq\|M\|$, and thus, since $X$ is bounded, so is $S$.
$S$ is also closed. Indeed, if you take a sequence $s_k$ of elements of $S$ that converges to some scalar $s\in\Bbb C$, then $s\in S$. This is because there are matrices $x_k\in X$ and eigenvectors $v_k$ (which we will take of length one: $|v_k|=1$ for all $k$) such that $$\forall k\in\Bbb N,\,x_kv_k=s_k v_k\,.$$ By compactness of $X$ and the unit sphere of $\Bbb C^n$, we can extract a subsequence $s'_k=s_{\phi(k)},x'_k=x_{\phi(k)},v'_k=v_{\phi(k)}$ such that $x'_k$ converges in $X$ (to some limit $x'\in X$), $v'_k$ converges to some vector of length $1$ (say $v'$), and then, by continuity of $M_n(\Bbb C)\times \Bbb C^n\to\Bbb C^n,(M,v)\mapsto Mv$,
$$x_k'v_k'\longrightarrow x'v'\text{ and }s_k'v_k'\longrightarrow sv'$$
and $x'v'=s'v'$, thus $s\in S$.
A: Ok, if $P$ is a polynomial, note $\|P\|$ the max of the absolute values of its coefficients and $V(P)$ the set of the roots of $P$. Then you have the following results :


*

*if $P$ is of degree $n$, for each $\varepsilon > 0$ there exist an $\eta >0$ such that for each $Q$ of degree $n$, $\|P-Q\| < \eta$ implies $V(Q)\subseteq V(P)+B(0,\varepsilon)$, which gives you the corolary

*if $A\in M_n(\mathbf{R})$, for each $\varepsilon > 0$ there exist an $\eta >0$ such that for each $B\in M_n(\mathbf{R})$, $\|B-A\| < \eta$ implies $\textrm{Spectrum}(B)\subseteq \textrm{Spectrum}(A)+B(0,\varepsilon)$. (You don't care about the vector space norm on $M_n(\mathbf{R})$ as they are all equivalent since $M_n(\mathbf{R})$ is of finite dimension over $\mathbf{R}$.)
Now, $S = \cup_{M\in X} \textrm{Spectrum}(M)$. Let $\varepsilon > 0$. What preceeds implies that for each $M\in X$, there exist $\eta_{M, \varepsilon} > 0$ such that $B\in M_n(\mathbf{R})$, $\|B-A\| < \eta_{\varepsilon}$ implies $\textrm{Spectrum}(B)\subseteq \textrm{Spectrum}(A)+B(0,\varepsilon)$. Now $\left( B_F (M, \eta_{M, \varepsilon})\right)_{M\in X}$ covers the compact set $X$, and has therefore a finite subcover : $X \subseteq \cup_{i = 1}^d B_F (M_i, \eta_{M_i, \varepsilon})$. And then $S$ is by the second property included in $\cup_{i = 1}^d (\textrm{Spectrum}(M_i) + B_F(0,\varepsilon))$. Now each $\textrm{Spectrum}(M_i)$ is finite, so that each $\textrm{Spectrum}(M_i) + B_F(0,\varepsilon)$ is a union of finite closed balls of radius $\varepsilon$, and it is therefore also the case for $\cup_{i = 1}^d (\textrm{Spectrum}(M_i) + B_F(0,\varepsilon))$. That is, $S$ is included in a finite union of finite closed balls of radius $\varepsilon$. And this is true for each $\varepsilon>0$. We have showed that $S$ is precompact. (This is the french word for metric space that is for each $\varepsilon > 0$ coverable by a finite union of finite closed balls of radius $\varepsilon$, I am not sure anymore of the english word for it.)
Now, it is an classic exercise to show that compact is equivalent to complete and precompact. So that to show the compacity of $S$, we need to show it is moreover complete and as $M_n(\mathbf{R})$ is complete, it is equivalent to showing that $S$ is closed. Let $(x_i)_i$ a sequence of point in $S$ converging to a point $x\in\mathbf{C}$. Each $x_i$ is in the spectrum of some $M_i \in X$, so that there exists a subsequence $M_{\varphi(i)}$ converging to a $M\in X$, as $X$ is compact. Denote $P_M( X )$ the degree $n$ characteristic polynomial of $M \in M_n(\mathbf{R}) $. By continuity of the caracteristic polynomial of a matrix, we know that the coefficients of $P_{M_{\varphi(i)}}$ converge to those of $P_M$ and as $x_{\varphi(i)}$ still converges to $x$, we get that $0 = P_{M_{\varphi(i)}} (x_{\varphi(i)}) \rightarrow P_M(x)$ as $i\rightarrow +\infty$, showing that $P_M(x) = 0$, that is, that $x$ is in the spectrum of $S$. $S$ is closed indeed.
For the two points I know this french reference, that is quite followable even if you're not understanding french. For the precompacity and the equivalence (for metric space) compact <=> complete + precompact, I learned that in school in 98' but do not know any reference. (Again, you have this one, in french, but I don't know any book where it is?) I'm sure someone will find one.
