Geometry of the set of coefficients such that monic polynomials have roots within unit disk We let $\pi$ be the bijection between coefficients of the real monic polynomials to the real monic polynomials. Let $a\in \mathbb R^n$ be fixed vector. Then
\begin{align*}
\pi(a) = t^n + a_{n-1} t^{n-1} + \dots + a_0. \\
\end{align*}
Now denote the set
$$ \Delta = \{ x\in \mathbb R^n: \pi(x) \text{ has roots in the open unit disk of } \mathbb C\}.$$
It can be shown $\Delta$ is a path-connected set by Vieta's formula (maybe slightly modified for the real case). Let us consider a line (1-dim subspace) in $\mathbb R^n$, $L = \alpha b$ where $\alpha \in \mathbb R$ and $b \in \mathbb R^n$ is fixed. Clearly $L \cap \Delta$ is nonempty since $0 \in L \cap \Delta$. 
I am trying to determine the number of connected components of $L \cap \Delta$.
We shall assume $n > 2$. If $n=1$, $L \cap \Delta$ is clearly connected. For $n=2$, if I am not mistaken, the paper by Fell https://projecteuclid.org/euclid.pjm/1102779366#ui-tabs-1 asserts that convex combination of real monic polynomials of the same degree with roots in the unit disk remains in the unit disk (Theorem 4). This allows us to construct a path between any $x \in L \cap \Delta$ and $0 \in \mathbb R^n$ and so $L \cap \Delta$ is connected.

Edit: The question was initially unclear in the sense: I didn't know whether $L \cap \Delta$ is connected.  @Jean-Claude Arbaut gave nice numerical examples and plot to show that the set is not connected. I rewarded a bounty and started a new one to see whether there is some bound of the number of connected components. I would reward the bounty to any bound or if the number of connected components could be unbounded.
 A: Here is an illustration of the comment above, and another example.
The set $L\cap\Delta$ considered here is a subset of a line, or if we consider a parametrization, a subset of $\Bbb R$. The connected components are thus intervals.
I think the main question is: how many intervals are there?

First, a plot of the roots of the polynomial $(x-0.9)^3$ and of the subsequent polynomials when you multiply $a_0\dots a_2$ by $\lambda\in[0,1]$. The roots follow three differents paths, and eventually get closer to zero, but two of them first get out of the unit circle. Note that here I consider only $\lambda\in[0,1]$ and not $\lambda\in\Bbb R$, but for all the examples below, that does not change the number of connected components.


Another example, of degree $9$, with initially three roots with multiplicity $3$ each. The roots are $0.95$, $0.7\exp(2i\pi/5)$ and $0.7\exp(-2i\pi/5)$. The colors show different portions of the paths, so that you can see there will be $3$ connected components (which correspond to the black portions below).


Here is an example with $4$ components.
The initial roots are $0.95$, $0.775\exp(\pm0.8482i)$, $0.969\exp(\pm2.7646i)$ with multiplicity $3,2,2$ respectively. Note the behaviour of "root paths" is highly sensitive to the initial roots.

How to read this ($\lambda$ decreases from $1$ to $0$ in the successive steps):


*

*Initially ($\lambda=1$), all roots are inside the unit circle, and we are on a connected component of $L\cap\Delta$.

*The first roots to get out are in red. The other ones are still inside the unit circle.

*The "red roots" get inside the unit circle: second component.

*Now some of the "blue roots" get out.

*The "blue roots" get in, and all roots are inside the unit circle: third component.

*Some "green roots" get out.

*The "green roots" get in, and all roots are inside the unit circle: fourth and last component, and after that the roots converge to zero as $\lambda\to0$.



Now, could there be more components? I have not a proof, but my guess would be that by cleverly choosing the initial roots, it's possible to get paths that will get outside then inside the unit circle in successive order, and the number of components could be arbitrary. Still investigating...

R program to reproduce the plots (as is, the last plot).
# Compute roots given vector a in R^n and coefficient e
# That is, roots of $x^5 + e a_n x^{n-1} + \cdots + e a_0$
f <- function(a, e) {
  polyroot(c(a * e, 1))
}

# Given vector a and number of points, compute the roots for
# each coefficient e = i/n for i = 0..n.
# Each set of root get a color according to:
# * if |z|<1 for all roots, then black
# * otherwise reuse the preceding color (and change if
#   the preceding was black)
# Return in z the list of all roots of all polynomials,
# and in cl the corresponding colors.
mk <- function(a, n) {
  cls <- c("red", "blue", "green", "yellow")
  z <- NULL
  cl <- NULL
  cc <- "black"
  k <- length(a)
  j <- 0
  for (i in n:0) {
    zi <- f(a, i / n)
    if (all(abs(zi) <= 1)) {
      cc <- "black"
    } else {
      if (cc == "black") {
        j <- j + 1
        cc <- cls[j]
      }
    }
    z <- c(z, zi)
    cl <- c(cl, rep(cc, k))
  }
  list(z=z, cl=cl)
}

# Compute polynomial coefficients from roots
pol <- function(a) {
  p <- c(1)
  for (x in a) {
    p <- c(0, p) - c(x * p, 0)
  }
  p
}

# New plot, and draw a circle
frame()
plot.window(xlim=c(-1.0, 1.0), ylim=c(-1.0, 1.0), asp=1)
z <- exp(2i  *pi * (0:200) / 200)
lines(z, type="l", col="darkgreen", lwd=2)

# The third example given
a <- c(0.95, 0.775 * exp(0.8482i), 0.775 * exp(-0.8482i),
       0.969 * exp(2.7646i), 0.969 * exp(-2.7646i))
# Duplicate roots, compute coefficients, remove leading x^n
a <- head(pol(rep(a, times=c(3, 2, 2, 2, 2))), -1)

# Plot roots
L <- mk(a, 3000)
points(L$z, col=L$cl)

