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Let $\mathcal{S}\subset\mathbb{R}^2$ such that every point in the real plane is at most at distance $1$ from a point in $\mathcal{S}$. Is it true that if $P\in\mathbb{R}[X,Y]$ is a polynomial that vanishes on $\mathcal{S}$, then $P=0$?

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    Yes. Just find an angular region where the leading monomial dominates the rest. – Emil Jeřábek Jul 03 '14 at 18:22
  • @Emil, I think I see what you mean: the angular (unbounded) region where the leading monomial dominates the rest also contains a unit disk. The question is how to find this angular region. –  Jul 03 '14 at 20:20
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    Write the polynomial as $\sum_{j=0}^m R^j f_j(\theta)$ where $(R,\theta)$ are polar coordinates for $(X,Y)$ and $f_j$ is a trigonometric polynomial. Take an interval for $\theta$ on which $f_m$ is bounded away from $0$, and take $R$ sufficiently large. –  Jul 03 '14 at 21:57

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Following the suggestions in comments, let's expand $P$ into a sum of homogeneous polynomials: $P=P_0+\dots+P_n$, with each $P_k$ homogeneous of degree $k$, and $P_n$ is not identically zero. Pick a closed ball $B\subset \mathbb R^n$ on which $P_n$ is nonzero (this is possible because the zero set of $P_n$ has empty interior). For sufficiently large $\lambda>0$ we have $$\inf_{\lambda B}|P_n|>\sup_{\lambda B}\sum_{k}|P_k|$$ because the left hand side is proportional to $\lambda^n$ while the right hand side is $O(\lambda^{n-1})$. Hence $P\ne 0$ on $\lambda B$, and the radius of this ball can be arbitrarily large.