Consider the following subset of $\mathbb{R}^n$: \begin{eqnarray}V_i:=\{(p_1, \cdots, p_n)\in\mathbb{R}^n|x^n-p_1x^{n-1}+p_2x^{n-2}-\cdots+(-1)^np_n=0\text{ has at least one root with multiplicity at least }i\}\end{eqnarray}
Then one can see that $V_i$ is a real affine subvariety of $\mathbb{R}^n$, and that there is this string of inclusions \begin{eqnarray}V_n\subset V_{n-1}\subset\cdots\subset V_1=\mathbb{R}^n\end{eqnarray}
Question: Is there a finite open cover $\{U_j\}_{j=1}^m$ of $\mathbb{R}^n$ such that the following holds:
For any subset $J\subset\{1, 2, \cdots, m\}$, let $i_J$ be the largest number such that $\left(\bigcap_{j\in J}U_j\right)\cap V_{i_J}\neq\varnothing$. Then for any $i$ satisfying $1\leq i\leq i_J$, $\left(\bigcap_{j\in J}U_j\right)\cap V_i$ deformation retracts to a point $v_{i_J}$ in $V_{i_J}$.
Does such a finite open cover exist if $\mathbb{R}^n$ is replaced by $\mathbb{C}^n$?
For $n=2$ and $\mathbb{R}^2$, the question is easy: in this case, $V_2$ is the parabola $p_1^2-4p_2=0$. We may simply take the open cover $\{U_1:=\mathbb{R}^2\}$ and both $U_1\cap V_1=\mathbb{R}^2$ and $U_1\cap V_2=V_2$ deformation retract to $v_2:=(0, 0)$ in $V_2$.
Yes to both the real and complex cases. More generally, let $\{V_i\}$ be a finite collection of closed semialgebraic sets in $\mathbb R^n.$ I will argue that there is a finite open cover of $\mathbb R^n,$ such that for any non-empty intersection $U$ of sets in this cover, there is a deformation retraction $F:U\times[0,1]\to U$ of $U$ to a point, that restricts to a deformation retraction of each $V_i,$ specifically: $F((U\cap V_i)\times [0,1])\subseteq V_i.$ (So the deformation retract point necessarily lies in all the non-empty sets $U\cap V_i.$)
Embed $\mathbb R^n$ in $S^n\subset\mathbb R^{n+1}$ by stereographic projection from a point $N\in S^n.$ Applying Bochnak-Coste-Roy, Real algebraic Geometry, Theorem 9.4.1 "Triangulability of Semi-algebraic functions", there is a finite simplicial complex $K$ and a homeomorphism $\Phi:|K|\to S^n$ where $|K|$ denotes the geometric realization of $K,$ such that each set $V_i,$ and also the set $\{N\},$ is the image under $\Phi$ of a union of open simplices in $K.$
The rest of the argument is basically: simplicial complexes have good open covers by stars. I'll give a bit more detail to show that we have the slightly stronger property required.
For each vertex $v$ the star $\mathrm{St}(v)$ is defined to be the union of simplices in $K$ that have $v$ as a vertex. I claim that the geometric realization of $\mathrm{St}(v)$ for $v\neq N$ gives the required open cover of $|K|\setminus \{N\}$ i.e. the required open cover of $\mathbb R^n$ after undoing the various homeomorphisms.
The first thing to check is that it's actually a cover. The points of $|K|$ have barycentric coordinates, formal convex combinations $\lambda_1 v_1+\dots+\lambda_k v_k.$ Any point except $N$ will use some $v$ with $v\neq N,$ and will therefore lie in $\mathrm{St}(v).$
Given distinct $v_1,\dots,v_k\neq N,$ the intersection $U=\bigcap_{i=1}^k\mathrm{St}(v_i)$ consists of the points whose barycentric coordinates use $v_1,\dots,v_k$ with a strictly positive coefficient (and possibly other $v$). For this set to be non-empty there must be a simplex $\{v_1,\dots,v_k\}.$ Let $c$ be the formal convex combination $(v_1+\dots+v_k)/k.$ We can define a deformation retraction $F:U\times[0,1]\to U$ to $c$ by $F(u,t)=(1-t)u+tc.$ Note $F(u,t)$ is in the same open simplex as $u$ for $t<1,$ and $F(u,1)=c$ lies in a face of the open simplex containing $u.$ Each $V_i$ is closed in $K$ except at $N,$ so if $U\cap V_i$ is non-empty then $c\in V_i.$ This hopefully veries all the required properties.