Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:53:43Z http://mathoverflow.net/feeds/question/82753 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82753/does-uniform-convergence-of-the-metrics-imply-uniform-convergence-of-the-radii-of Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls? Valerio Capraro 2011-12-05T23:35:16Z 2011-12-06T02:22:19Z <p>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.</p> <p><strong>Question:</strong> Suppose that $d_n\rightarrow d$ uniformly. Is it true that also $R_x^n\rightarrow R_x$ uniformly?</p> <p>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 $x$).</p> <p>Sorry, it seems trivial but I am really getting mad for three days.. (I hope it's not terribly trivial)</p> http://mathoverflow.net/questions/82753/does-uniform-convergence-of-the-metrics-imply-uniform-convergence-of-the-radii-of/82755#82755 Answer by Lior Silberman for Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls? Lior Silberman 2011-12-05T23:59:51Z 2011-12-05T23:59:51Z <p>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.</p> http://mathoverflow.net/questions/82753/does-uniform-convergence-of-the-metrics-imply-uniform-convergence-of-the-radii-of/82761#82761 Answer by Will Sawin for Does uniform convergence of the metrics imply uniform convergence of the radii of the smallest balls? Will Sawin 2011-12-06T02:22:19Z 2011-12-06T02:22:19Z <p>We do not need the assumption. Suppose $|d_n-d|&lt;\epsilon$. 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 $R^n_x \leq R_x+\epsilon$.</p> <p>Similarly, we can fix a point of distance $R^n_x+\delta$ from $x$ in $d_n$, and look at it in $d$.</p> <p>This bound obviously implies your result.</p>