The diameter of Voronoi cells in Euclidean spaces Let $A \subset \mathbb{R}^d$ and let $(x_n)_{n \in \mathbb{N}} \subset \mathbb{R}^d$ be a sequence dense in $A$.
For each $n \in \mathbb{N}$, let $V_{1,n},\dots,V_{n,n}$ the sequence of Voronoi cells associated to the points $x_1, \dots, x_n$, where the ties are broken lexicographically, i.e.:
$$V_{1,n} =\{x \in \mathbb{R}^d \mid \forall k\in\{2,\dots,n\}, |x-x_1|\le |x-x_k|\} \\
V_{2,n} =\{x \in \mathbb{R}^d \mid |x-x_2|&lt;|x-x_1| \land \forall k\in\{3,\dots,n\}, |x-x_2|\le |x-x_k|\} \\
\vdots \\
V_{n,n} =\{x \in \mathbb{R}^d \mid \forall k\in\{1,\dots,n-1\}, |x-x_n| &lt; |x-x_k|\}
$$
If $x \in \mathbb{R}^d$ and $n \in \mathbb{N}$, define $V_n(x)$ as the unique element in $V_{1,n},\dots, V_{n,n}$ that contains $x$.

Is it true that
$$\forall x \in A, \operatorname{diam}\big(A \cap V_n(x)\big) \to 0, n \to \infty?$$

I suspect that this result should hold due to the finite dimensionality of $\mathbb{R}^d$, since at least in this case we can obtain bounded sets using a finite number of intersections of half-spaces.
However, it seems quite involved from a geometric point of view to obtain this claim. Has anyone any idea?
EDIT: note that we can WLOG assume that $A$ is the closure of the set whose points are those of the sequence $(x_n)_{n \in \mathbb{N}}$.
Some context: I'm trying to prove the aforementioned result to obtain that if $g \colon A \to \mathbb{R}$ is a continuous function, then
$$\forall x \in A, \sup_{y \in A \cap V_n(x)} |g(x) - g(y)| \to 0, n \to \infty.$$
For additional insight, one key intuition might be that, by definition of density, any part of the set will eventually be dotted by an arbitrarily fine cloud of points.
Note that the definition of $ V_n $ is to choose among $ V_{1,n} \cdots V_{n,n} $ with $ n $ points present from the start. Sufficiently far down the sequence, there should be points arbitrarily close to $ x $ “all around” in order to “force” $ V_n $ into an arbitrarily small corner.
Otherwise, some angle would remain open, in which an open ball could exist, the interior of which must intersect a region of $ A $ where the sequence could then not be dense.
Edit: I see now that this is already plain for the particular case $ A = \mathbb{R}^d $, but does not help understanding what happens when the cells are unbounded. Would it be fair to say you are asking why it is that the unboundedness must resolve when capping by $ A $?
It should be in principle easier to show that the diameter at least does not diverge to infinity. Any point in the sequence is eventually “surrounded” by other points “in the local directions of $ A $” as you so evocatively put it.
Consider any $ a \in A $ and its path connected component of $ A $, then any path emanating from it must encounter points of the sequence within an arbitrarily thin thickness and arbitrarily close to it. Therefore no “local direction” is ever free to go on forever.
This is far from rigorous, I hope it nevertheless contributes positively to the discussion.
For additional insight, one key intuition might be that, by definition of density, any part of the set will eventually be dotted by an arbitrarily fine cloud of points.
Note that the definition of $ V_n $ is to choose among $ V_{1,n} \cdots V_{n,n} $ with $ n $ points present from the start. Sufficiently far down the sequence, there should be points arbitrarily close to $ x $ “all around” in order to “force” $ V_n $ into an arbitrarily small corner.
Otherwise, some angle would remain open, in which an open ball could exist, the interior of which must intersect a region of $ A $ where the sequence could then not be dense.
Edit: I see now that this is already plain for the particular case $ A = \mathbb{R}^d $, but does not help understanding what happens when the cells are unbounded. Would it be fair to say you are asking why it is that the unboundedness must resolve when capping by $ A $?
It should be in principle easier to show that the diameter at least does not diverge to infinity. Any point in the sequence is eventually “surrounded” by other points “in the local directions of $ A $” as you so evocatively put it.
Consider any $ a \in A $ and its path connected component of $ A $, then any path emanating from it must encounter points of the sequence within an arbitrarily thin thickness and arbitrarily close to it. Therefore no “local direction” is ever free to go on forever.
This is far from rigorous, I hope it nevertheless contributes positively to the discussion.