Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls? - MathOverflow

Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls?

2011-12-05T23:35:16Z

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)

Answer by Lior Silberman for Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls?

2011-12-05T23:59:51Z

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.

Answer by Will Sawin for Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls?

2011-12-06T02:22:19Z

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.