radius-diameter relationship of balls in metric spaces

Question

What necessary and sufficient conditions must a metric $d$ of a metric space $(X,d)$ fulfill so that the open balls of radius $r$ have diameter $2r$?

Answer 1

There is a way of enlarging a metric space $(X,d)$ to a larger metric space $(Y,d)$ so that every ball $B_r(x)$ has diameter $2r$ if and only if there are points $y,z\in Y$ where $d(x,y)=d(x,z)=r,d(y,z)=2r$.

Example: Let $H$ be an infinite dimensional Hilbert space. Let $(e_n)_{n=1}^{\infty}$ be an orthonormal basis for $H$. Let $X=\{(1-1/n)e_n:n>0\}$. Then $X$ is a complete (and even uniformly discrete) metric space. Furthermore, $X=B_1(0)$, but there are no two points $r,s\in X$ with $d(r,s)\geq 2$. This means that the completion of a metric space is not enough of an enlargement to ensure that we get two points $r,s\in X$ with $d(r,s)=2$. In order to obtain our points $r,s\in X$ with $d(r,s)=2$, we need for the enlarged space to have a sort of saturation.

Since we are dealing with open balls, it makes sense to deal solely with bounded metric spaces.

If $(X_i,d_i)_{i\in I}$ is a uniformly bounded collection of metric spaces and $\mathcal{U}$ is an ultrafilter on $I$, then define the pre-ultraproduct of $(X_i,d_i)_{i\in I}$ modulo $\mathcal{U}$ to be the pseudometric space $(\prod_{i\in I}X_i,d)$ where $d((x_i)_{i\in I},(y_i)_{i\in I})$ is the limit of the ultrafilter on $\mathbb{R}$ consisting of all sets $S\subseteq\mathbb{R}$ where $\{i\in I:d(x_i,y_i)\in S\}\in\mathcal{U}$. By compactness, this ultrafilter always converges to a unique real number.

Every pseudometric space $(X,d)$ induces a metric space $(X/\simeq,d)$ where we set $x\simeq y$ precisely when $d(x,y)=0$ and where $d([x],[y])=d(x,y)$. The ultraproduct $\prod_{i\in I}(X_i,d_i)/\mathcal{U}$ is therefore defined to be the metric space induced by the pre-ultraproduct $(X_i,d_i)_{i\in I}$.

If $(X,d)$ is a metric space, then define the ultrapower $(X,d)^{\mathcal{U}}$ to be the ultraproduct $(\prod_{i\in I}(X,d))/\mathcal{U}$. In other words, the ultrapower $(X,d)^{\mathcal{U}}$ is the ultraproduct where all of the factors are the same.

We say that a metric space has maximum diameter balls if $B_r(x)$ always has diameter $2r$. We say that a metric space $(X,d)$ has maximally diametric points on closed balls if for all $x\in X,r>0$, there are $y,z\in X$ with $d(x,y)=d(x,z)=r,d(y,z)=2r$.

Theorem: Suppose that $\mathcal{U}$ is a non-$\sigma$-complete ultrafilter. Let $(X,d)$ be a metric space. Then $(X,d)$ has maximum diameter balls if and only if the ultrapower $(X,d)^\mathcal{U}$ has maximally diametric points on closed balls.

Proof: $\rightarrow$ Suppose that $(X,d)$ has maximum diameter balls. Then let $(R_n)_{n=1}^\infty$ be a partition of $I$ into sets with $R_n\not\in\mathcal{U}$ for all $n$. Then let $(x_i)_{i\in I}\in\prod_{i\in I}X_i$. Then for all $i\in I$, there are $y_i,z_i\in X$ where $d(x_i,y_i)<r,d(x_i,z_i)<r$ and $2r>d(y_i,z_i)>2r-1/n$.

In this case $d((x_i)_{i\in I},(y_i)_{i\in I})\leq r$ and $d((x_i)_{i\in I},(z_i)_{i\in I})\leq r$ and $d((y_i)_{i\in I},(z_i)_{i\in I})=2r$. We therefore conclude that the ultrapower $(X,d)^\mathcal{U}$ has maximally diametric points on closed balls.

$\leftarrow.$ Suppose now that the ultrapower $(X,d)^\mathcal{U}$ has maximally diametric points on closed balls. Let $x\in X$, and let $r>0$. Let $r>s>0$. Then let $(y_i)_{i\in I},(z_i)_{i\in I}$ be tuples where $d((x)_{i\in I},(y_i)_{i\in I})=d((x)_{i\in I},(z_i)_{i\in I})=s$ and $d((y_i)_{i\in I},(z_i)_{i\in I})=2s$. In this case, for all $\epsilon>0$, we have $|d(x,y_i)-s|<\epsilon,|d(x,z_i)-s|<\epsilon$ and $|d(y_i,z_i)-2s|<\epsilon$ for $\mathcal{U}$-almost all $i\in I$. In particular, if $\epsilon<r-s$, then $d(x,y_i)<r,d(x,z_i)<r,d(y_i,z_i)>2s-\epsilon$. Since $s$ can be made arbitrarily close to $r$, we conclude that the diameter of $B_r(x)$ is $2r$. Q.E.D.