Every hyperplane is the Chabauty limit of spheres

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Let $X$ be a metric space with distance $d$. Given $A, B \subset X$, we define $$D(A,B) = \sup_{x \in A} \inf_{y \in B} d(x,y).$$ The Hausdorff distance in the set $\mathcal{K}_X$ of compact sets of $X$ is the metric defined by $$d_H(K_1, K_2) = \max\left\{D(K_1, K_2), D(K_2, K_1)\right\}.$$

Now consider $\mathcal{C}_X$ the family of all closed sets of $X$. Given a sequence $C_i$ in $\mathcal{C}_X$, we say that $C_i$ converges to $C \in \mathcal{C}_X$ in the Chabauty sense if for every $K \in \mathcal{K}_X$ we have $$\lim_{i \to \infty} d_H(C_i \cap K, C \cap K) = 0.$$

Given those definitions, set $X = \mathbb{R}^n$. How one goes about proving that every hyperplane is the Chabauty limit of spheres with arbitrarily large radii? I have an intuitive feeling that this is absolutely true, and that the proof should start at the lines of:

Let $H$ be the hyperplane given by the equation $\langle x, a \rangle = 0$, that is, the hyperplane passing $0$ with normal vector $a$. We set $a_i = i \cdot a$ and $R_i = i$. This gives a sequence of spheres $S(a_i, R_i)$ (sphere centered at $a_i$ with radius $R_i$). They all intersect and are tangent to $H$ at $0$.

I feel that this sequence should converge to H, but having to work with arbitrary compact sets is throwing me off, I don't really know how to start. One question that I have, which if answered in a positive way, could help with the proof, is: in euclidean spaces, can I consider the Chabauty limit as defined taking $K$ to be only closed balls with finite radii?

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In the question $D(A,B)$ is undefined when one of the sets $A$ and $B$ is empty. And this makes a problem with the evaluation of $d_H(C_i\cap K,C\cap K)$. Maybe this issue can be fixed somehow, but the other problems remains even when all intersections $C_i\cap K$ and $C\cap K$ are nonempty. For instance, in you proof we can take $K=\{0,b\}$ where $b$ is any nonzero point of $H$. Then for each $i$ we have $S(a_i,R_i)\cap K=\{0\}$ but $H\cap K=\{0,b\}$, so $d_H(S(a_i,R_i)\cap K, H\cap K)=\|b\|>0$.

On the other hand, it seems that if $K$ is a closed ball such that all intersections $S(a_i,R_i)\cap K$ and $H\cap K$ are nonempty then indeed $\lim_{i \to \infty} d_H(S(a_i,R_i) \cap K, H \cap K) = 0.$