1

Suppose that $(X, d)$ is compact metric space. I am thinking about when can we say that every closed ball of $X$ is connected. Intuitively, one can reason in the following way for any closed set:
Let $U\subseteq X$ be closed, and suppose that $U = A\cup B$ for $A, B$ closed. Then, since $U$ closed, $A$ and $B$ are compact (since they are closed subsets of a compact space). Now define $d(A,B) = \inf_{a\in A, b\in B} d(a,b)$. Since $A, B$ are compact, we can find $a^* \in A$, $b^* \in B$ such that $d(a^*, b^*) = d(A, B)$. Hence, if $D(A,B)=0$, then we have a contradiction.

Based on this, I though, one needs to find a closed set $U$, which has the property, that for any $A\subset U$, $d(A,U - A) = 0$.

Now for the question part. I am aware one criterion, that might be helpful, namely that the closure of any open ball of radius $\varepsilon$ is the closed ball of the same radius. Is my reasoning right? Is there a more general, or less lose criteria?

Jakobian
  • 15,280
stomfaig
  • 669
  • You'll need $X$ connected, as well. For example, in Euclidean space, a closed disk minus an open annulus. The set $z\in\mathbb R^2$ with $|z|\in[0,1]\cup[2,3]$ is compact but has disconnected balls. – Thomas Andrews Aug 24 '23 at 19:43
  • 1
    A simpler counter-example is ${0}\cup{1/n\mid n\in\mathbb N^+}.$ No neighborhood of $0$ is connected. – Thomas Andrews Aug 24 '23 at 19:46
  • 2
    You can fairly easily find counterexamples with $X$ connected, too, so this is not true. For example, if $Y={0}\cup{1/n\mid n\in\mathbb N^+}$ then $X=Y\times[0,1]\cup {0}\times[0,1]$ is compact and connected, but no neighborhood radius $<1$ of $(1,0)$ is connected. – Thomas Andrews Aug 24 '23 at 20:03
  • 1
    Whoops, that should be $X=(Y\times [0,1])\cup ([0,1]\times{0}).$ Basically, $Y\times[0,1],$ but with a line segment added at one position to connect all the disconnected copies of $[0,1].$ – Thomas Andrews Aug 24 '23 at 20:44
  • My original source for the question was DJH Garling's A Course in Mathematical Analysis, Problem 16.1.5 Which states: If all open balls are closed, then all the open and closed balls are connected, further the space itself is connected and locally connected. However, you're right, these are definitely counterexamples. Thanks – stomfaig Aug 24 '23 at 20:45
  • Yeah, the condition of all open balls being closed is a very strong statement. It is very hard to imagine such a metric space - it is true in the $p$-adic numbers, I think? – Thomas Andrews Aug 24 '23 at 20:49
  • Does the problem also require compactness? – Thomas Andrews Aug 24 '23 at 20:51
  • Yes. Its the criteria about balls and compactness. But sorry! The criteria is that the closure of the open ball is the closed ball of same radius. – stomfaig Aug 24 '23 at 20:55
  • 1
    @ThomasAndrews Every open ball in a discrete space is closed. On the other hand, every zero-dimensonal perfect compact metric space is homeomorphic to the Cantor set, so I imagine discrete spaces are the only examples. – Ningxin Aug 25 '23 at 04:21
  • It is important that the closure of the open ball be a closed ball of the same radius. A finite discrete space satisfies everything, including the condition that the closures of open balls are closed balls. But if $d(a,b) =1 $ and no other points are closer to $a$, while ${a} = B_a(1) = \overline{B_a(1)} = \overline B_a(1/2)$, so the closure of the open ball is a closed ball. but not of the same radius. Instead, note that $\overline B_a(1) = {a,b}$, which disallows this example. – Paul Sinclair Aug 25 '23 at 19:39

1 Answers1

1

Let me start with a negative result.

Theorem. Let $X$ be a metrizable space with at least two points. Then there is $x\in X$ and a metric $d$ on $X$ such that the closed ball $B_d(x, 1) = \{y\in X : d(x, y) \leq 1 \}$ is disconnected.

Proof: Pick two non-empty open sets $U, V\subseteq X$ such that $\overline{U}\cap \overline{V} = \emptyset$ and some $x\in U\cup V$. There is a metric $d$ on $X$ such that $\{y\in X : d(x, y) < 1\} = U\cup V$, see here. Thus $$U\cup V\subseteq B_d(x, 1) \subseteq\overline{U}\cup\overline{V}$$ so that the closed ball at $x\in X$ is disconnected, as a disjoint union of $B_d(x, 1)\cap\overline{U}$ and $B_d(x, 1)\cap\overline{V}$. $\square$

Now let's see which compact metric spaces admit a metric such that all balls are connected. A compact connected metrizable space is called a continuum. If a continuum is locally connected (or equivalently, locally path-connected), it's called a Peano continuum.

Theorem. Let $X$ be a compact metrizable space. Then $X$ admits a metric with all closed balls connected iff $X$ is a Peano continuum.

Proof: If $X$ is compact and metrizable, let $x\in X$ and $d$ a metric on $X$ with all closed balls connected. Then $\bigcup_{n\in\mathbb{N}} B_d(x, n) = X$ where $B_d(x, n)$ is the closed ball of $X$ of radius $n$ in the metric $d$, so that $X$ is connected as a union of connected sets with common point $x$. Thus $X$ is a continuum. Moreover, for any $y\in X$, the closed balls $B(y, r)$, $r > 0$, form a neighbourhood system at $y$ consisting of connected sets. Thus $X$ is connected im kleinen, or weakly locally connected, at every point of $X$. This is known to be equivalent to $X$ being locally connected. See for example this entry on wikipedia for a proof. Thus $X$ is a Peano continuum.

Conversely, assume that $X$ is a Peano continuum. By this great answer by Moishe Kohan, all locally compact separable connected locally connected metrizable spaces admit a complete geodesic metric $d$, thus making all balls $B_d(x, r)$ path-connected, in fact both open and closed balls. In particular, all Peano continua admit such a metric (note that all compact metric spaces are separable). $\square$

Jakobian
  • 15,280