# Closure of the set of degree $n$ polynomials with $n$ real roots

A very beautiful (but maybe classic) exercise:

Consider the set $\Omega_n$ of polynomials of degree $n$ with real coefficients and $n$ distinct real roots.

What is the closure of $\Omega_n$ in $R_n[x]$?

($R_n[x]$ denotes the set of polynomials with real coefficients of degree at most $n$)

The answer is the set of polynomials in $R_n[x]$ all of whose roots are real, plus the constant polynomials. Showing that this set is contained in the closure is not too hard, however the reverse implication is more difficult and I'm wondering what is the most elementary method to prove it (without talking about holomorphic functions for example).

• The topology is that of $\Bbb R^{n+1}$ according to the coefficients? – Hagen von Eitzen Nov 12 '17 at 20:38
• It doesn't matter, all the norms are equivalent in finite dimmension. The distance is the one induced by any norm. – Joshua Benabou Nov 12 '17 at 20:53
• Are you considering only monic polynomials? If P is a polynomial of the type you mentioned and divide it by n doesn't it converge to 0? – Kavi Rama Murthy Nov 13 '17 at 6:23
• No they are not necessarily monic. I don't understand the relevance of your example. – Joshua Benabou Nov 13 '17 at 11:34

Here's a slick proof of the reverse inclusion that uses no more than basic pointset topology and a little theory of projective spaces. The key ideas are to homogenize your polynomials so you can treat polynomials of degree less than $n$ on an equal footing with degree $n$ polynomials and to projectivize your spaces so you can use a compactness argument.
Let $X\subseteq R_n[x]$ be the set of nonzero polynomials all of whose roots are real. We wish to show $X\cup\{0\}\supseteq\overline{\Omega_n}$, and to show this it suffices to show $X\cup\{0\}$ is closed. To show that, it suffices to show $X$ is closed in $R_n[x]\setminus\{0\}$. Now let $P^n$ be the projective space of the vector space $R_n[x]$ and consider the quotient map $\pi:R_n[x]\setminus\{0\}\to P^n$. Since $X$ is closed under nonzero scalar multiplication, $X=\pi^{-1}(\pi(X))$, so since $\pi$ is continuous, it suffices to show $\pi(X)$ is closed in $P^n$.
Now let us homogenize our polynomials, and think of elements of $R_n[x]$ not as polynomials in $x$ of degree at most $n$, but as homogeneous polynomials in $x$ and $y$ of degree $n$. (Explicitly, we replace a polynomial $f(x)$ with $g(x,y)=y^nf(x/y)$.) Note that with this identification, $X$ is now the set of polynomials which are a product of $n$ homogeneous linear polynomials over $\mathbb{R}$. Explicitly, if $f(x)=c(x-a_1)\dots(x-a_k)$ has real roots $a_1,\dots,a_k$, then the corresponding homogeneous polynomial is $g(x,y)=cy^{n-k}(x-a_1y)\dots(x-a_ky)$.
So, $\pi(X)$ is the set of nonzero homogeneous polynomials of degree $n$ (modulo scaling) that are a product of $n$ homogeneous linear polynomials. That is, $\pi(X)$ is the image of the map $\mu:(P^1)^n\to P^n$ which takes $n$ nonzero homogeneous linear polynomials (mod scaling) and multiplies them together to get a homogeneous polynomial of degree $n$. It is easy to see that $\mu$ is continuous. But $(P^1)^n$ is compact, so the image of $\mu$ is compact and thus closed. Thus $\pi(X)$ is closed, as desired.