Characterizing intersection of zero sets of elementary symmetric polynomials on R^n

Stated simply, the question is:

Consider two elementary symmetric polynomials $\sigma_{k}$ and $\sigma_{k+1}$ on $\mathbb{R}^{n}$ with zero sets $U_{k}$ and $U_{k+1}$. Let $V_{i_{1}i_{2}\dotsb i_{j}}$ be the coordinate linear space $\{x\in\mathbb{R}^{n}: x_{i_{1}} = \dotsb = x_{i_{j}} = 0\}$ and $W_{n-k+1} = \cup_{i_{1} < i_{2} < \dotsb < i_{n-k+1}}V_{i_{1}\dotsb i_{n-k+1}}$, the set of all points in $\mathbb{R}^{n}$ with at least $n-k+1$ coordinates equal to zero. Does $$(\*) \qquad U_{k} \cap U_{k+1} = W_{n-k+1}$$ ? Why?

Note: I am interested in a slightly more specific question. Namely, if $\Gamma_{k}^{+}$ is the component of $\sigma_{k} > 0$ containing the point $(1,1,\dotsb,1)$, then is

$$(\*\*) \qquad \overline{\Gamma_{k}^{+}} \cap \overline{\Gamma_{k+1}^{+}} \subset W_{n-k+1}$$

? It is well known (see "An Inequality for Hyperbolic Polynomials", Lars Garding) that $$(\*\*\*)\qquad \Gamma_{k}^{+} \supset \Gamma_{k+1}^{+}$$ and that $$(****)\qquad \{\sigma_{k+1} > 0\} \cap \Gamma_{k}^{+} = \Gamma_{k+1}^{+}$$ If $(\*)$ is true, then $(\*\*)$ follows. The equation $(\*)$ is simpler to state and appears to be true. Of course $(\*\*)$ may hold with $(\*)$ failing, but I think this is unlikely.

Background and motivation:

Numerical simulations in mathematica suggest that $(\*)$ holds. Certainly $W_{n-k+1} \subset U_{k} \cap U_{k+1}$. I can prove the reverse inclusion in certain cases. For example, denoting $x^{j} = (x_{1}^{j},\dotsb,x_{n}^{j})$ for $x\in\mathbb{R}^{n}$, notice that $W_{n-k+1}$ is exactly the zero set of $\sigma_{k}(x^{2})$. So we have

n arbitrary, k=1: From Newton's identities (see wikipedia), $\sigma_{1}^{2}(x) = \sigma_{1}(x^{2})+2\sigma_{2}(x)$. So if $\sigma_{1}(x) = \sigma_{2}(x) = 0$, then $\sigma_{1}(x^{2}) = 0$ which means $x = (0,\dotsb,0)$ so that $x \in W_{n}$.

n arbitrary, k = 2, $(\*\*)$ only: can be proved similarly to $k=1$ by writing $0 \le \sigma_{2}(x^{2}) = \sigma_{2}^{2}(x) -2\sigma_{1}(x)\sigma_{3}(x)+2\sigma_{4}(x)$. So if $\sigma_{2}(x) = \sigma_{3}(x) = 0$, then $\sigma_{2}(x^{2}) = 2\sigma_{4}(x)$. Now, $\sigma_{4}(x) \le 0$ because of $(\*\*\*)$ and $(\*\*\*\*)$. Hence $\sigma_{2}(x^{2}) = 0$ so that $x \in W_{n-1}$.

n arbitrary, $k=n-2$ and $k =n-1$ can be proved in a similar fashion. However, because of the additional terms in the equation

$$\sigma_{k}(x^{2}) = \sigma_{k}^{2}(x)-2\sigma_{k-1}(x)\sigma_{k+1}(x)+2\sigma_{k-2}(x)\sigma_{k+2}(x)+\dotsb+(-1)^{k}2\sigma_{0}(x)\sigma_{2k}(x)$$

for other values of $k$, the above approach fails in general.

I'm actually trained as a differential geometer, so I may be approaching this problem in the wrong way (perhaps there's a technique in real algebraic geometry?). If you can suggest an alternative approach I would be grateful to hear it. This result seems rather elementary to state so I would be surprised if the result is not known. Thank you.

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To establish that $U_k \cap U_{k+1} \subset W_{n-k+1}$:
Consider the polynomial $P(t)=\prod (t-x_i)$. The elementary symmetric polynomials $\sigma_i$ are its coefficients, up to the sign.
Suppose that $\sigma_k=\sigma_{k+1}=0$. This means that $0$ is a multiple root of the derivative of order $(n-k-1)$ of $P$. Now you can conclude that $0$ is a root of $P$ of multiplicity at least $n-k+1$ (that is: at least $n-k+1$ of the numbers $x_i$ are zero) by means of the following facts:
If a non-constant polynomial $Q$ has all its roots real, then so does its derivative $Q'$. Then the roots of $Q'$ are the multiple roots of $Q$, plus one root between each pair of consecutive distinct roots of $Q$, necessarily simple (otherwise the sum of the multiplicities of the roots of $Q'$ would exceed its degree). In particular, if $x$ is a multiple root of $Q'$ then it is also necessarily a root of $Q$.