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Let $X$ be a countable set and $d_n,d$ locally finite metrics on $X$. Denote by $R_x^n$ (resp. $R_x$) the radius of the smallest closed ball in the metric $d_n$ (resp. $d$) about $x$ which contains at least two points.

Question: Suppose that $d_n\rightarrow d$ uniformly. Is it true that also $R_x^n\rightarrow R_x$ uniformly?

P.s. In case we can add the hypothesis that the $|C_n(x,R_x^n)|\leq C$, for a universal constant $C$ not depending either on $n$ on $x$ ($C_n(x,R)$ stands for the closed ball in the metric $d_n$ of radius $R$ about $x$).

Sorry, it seems trivial but I am really getting mad for three days.. (I hope it's not terribly trivial)

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Is $R_n$ just the infimum of the $d_n$-distances from $x$ to other points in $X$, or you meant something else? (formally what is written is a bit strange because the smallest ball may fail to exist, etc). – fedja Dec 5 '11 at 23:58

Well, wlog we may assume that $d_n$ is within $1$ of $d$. Then the point at $d_n$-distance $R^n_x$ to $x$ is contained in the punctured ball $B=B_d(x,R_x+2)\setminus x$. By assumption, there are finitely many points in this set. In particular, $R^n_x = \min_{y\in B} d_n(x,y)$ converges.

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Clarification: "by assumption" refers to the assumption of local finiteness, not to the proposed assumption of a uniform bound on the cardinality of balls, which is unnecessary. – Lior Silberman Dec 9 '11 at 17:28

We do not need the assumption. Suppose $|d_n-d|<\epsilon$. I claim that $|R^n_x-R^n|\leq\epsilon$. Proof: Let $p$ be a point of distance $R_x$ (or $R_x+\delta$) from $x$ in $d$. Then its distance in $d_n$ is at most $R_x+\epsilon$ (or $R_x+\epsilon+\delta$), so $R^n_x \leq R_x+\epsilon$.

Similarly, we can fix a point of distance $R^n_x+\delta$ from $x$ in $d_n$, and look at it in $d$.

This bound obviously implies your result.

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